Chapter 1: Rotational Dynamics
In the middle of the seventeenth century, the Dutch scientist Christiaan Huygens was working on a very practical problem: building a clock accurate enough to be trusted on long sea voyages. That work pushed him to study the swinging pendulum in great detail, and then to try a variation of it that behaved quite differently.
Instead of letting the bob swing to and fro in a vertical plane, Huygens set it revolving, so that the string swept out the curved surface of a cone and the bob traced a steady horizontal circle. He noticed something that seems ordinary until you think about it carefully. The bob was moving with an almost constant speed, yet it was clearly not moving in a straight line, so some force had to be acting on it at every instant. He also found that the string leaned at an angle and could never be brought fully horizontal, no matter how fast the bob was made to revolve.
From such studies Huygens introduced the idea of a “centre-fleeing” or centrifugal tendency, and arrived at a result that is still the backbone of this chapter: the effect grows with the square of the speed and falls off as the radius increases. That single relation, as shown in Fig. 1.0, explains stunt riders, tilted highways, spinning skaters and rolling wheels alike.
Diagram
Fig. 1.0 – A conical pendulum bob tracing a horizontal circle with the string sweeping a cone, shown beside a motorcyclist riding inside a cylindrical well of death.
By the end of this chapter, you will be able to answer each of these questions using the physics of rotational dynamics.
1.1 Introduction
Circular motion is an essential part of our daily life. Every day we come across several revolving or rotating objects, and most of them are rigid — that is, the distance between any two particles of the object does not change while it moves.
Such motion is described using two different words, and the difference between them lies in where the centre or axis of the circular path is located, as shown in Fig. 1.1a.
Diagram
Fig. 1.1a – A planet revolving around the Sun (centre outside the body) shown beside a spinning top rotating about an axis passing through itself.
During revolution, the object moves around some point lying outside the object, or around some other object. Every particle of the revolving object then undergoes circular motion about that outside point.
During rotation, on the other hand, the motion takes place about an axis of rotation that passes through the object itself. The particles of the object again travel in circles, but now the centres of those circles lie on the axis inside the body.
1.2 Characteristics of Circular Motion
Any motion along a circular path shows two features that set it apart from motion along a straight line.
It Is an Accelerated Motion
Velocity is a vector quantity, so it is fixed only when both its magnitude and its direction are fixed. In circular motion the direction of the velocity changes at every instant, because the velocity always points along the tangent to the circle, as seen in Fig. 1.2a. A changing velocity means a non-zero acceleration. Hence circular motion is always an accelerated motion, even when the speed does not change at all.
Diagram
Fig. 1.2a – A particle at two positions on a circle with tangential velocity arrows pointing in different directions.
It Is a Periodic Motion
During the motion, the particle repeats its path along the same trajectory, returning to every point of the circle again and again. The motion is therefore periodic in space — the same set of positions is traced over and over.
1.2.1 Kinematics of Circular Motion
Angular Quantities and Their Linear Counterparts
As seen in Std. XI, describing a circular motion becomes much easier if, instead of tracking how far a particle has travelled along the arc, we track the angle it has turned through. For this we use three angular quantities: angular displacement \(\vec{\theta}\), angular velocity
\[\vec{\omega} = \frac{d\vec{\theta}}{dt}\]
and angular acceleration
\[\vec{\alpha} = \frac{d\vec{\omega}}{dt}\]
Here \(\vec{\theta}\) is measured in radian (rad), \(\vec{\omega}\) in radian per second (rad s\(^{-1}\)) and \(\vec{\alpha}\) in radian per second squared (rad s\(^{-2}\)).
These three are exactly analogous to the quantities used in translational motion — displacement \(\vec{s}\), velocity
\[\vec{v} = \frac{d\vec{s}}{dt}\]
and acceleration
\[\vec{a} = \frac{d\vec{v}}{dt}\]
The word analogous is important here. It means the angular quantities play the same role in rotational motion that the linear quantities play in straight-line motion, and therefore the mathematical relations connecting them have the same form.
Relating the Tangential Velocity to the Angular Velocity
The particle itself, of course, still moves along the circle with an ordinary linear velocity. This velocity is always along the tangent to the circular path, and is therefore called the tangential velocity. It is given by
\[\vec{v} = \vec{\omega} \times \vec{r}\]
where \(\vec{\omega}\) is the angular velocity and \(\vec{r}\) is the position vector.
Here the position vector \(\vec{r}\) is the radius vector drawn from the centre of the circular motion to the particle. Its magnitude is simply the radius of the circle, measured in metre (m).
Since \(\vec{\omega}\) and \(\vec{r}\) are perpendicular to each other in circular motion, the cross product gives the magnitude of \(\vec{v}\) as
\[v = \omega r\]
with \(v\) in metre per second (m s\(^{-1}\)). This single relation is used constantly in numerical problems: it tells us that for the same angular speed, a particle farther from the axis moves faster.
Direction of the Angular Velocity
A beginner often expects \(\vec{\omega}\) to point somewhere along the circle. It does not. The direction of \(\vec{\omega}\) is always along the axis of rotation, and it is given by the right-hand thumb rule. Fig. 1.1 shows the two possible cases.
To know the direction of \(\vec{\omega}\), curl the fingers of the right hand along the sense of rotation, with the thumb outstretched. The outstretched thumb then gives the direction of \(\vec{\omega}\).
Diagram
Fig. 1.1 – Two circles with opposite senses of rotation, each with the angular velocity vector drawn along the axis, upward in one case and downward in the other.
The reason for choosing the axis is practical. Every point of a rotating body has a different direction of linear velocity at every instant, but the axis stays fixed. Only a direction along the axis can represent the rotation of the whole body by a single vector.
Period, Frequency and Angular Speed
If \(T\) is the period of circular motion, that is the time taken for one complete round, and \(n\) is the frequency, that is the number of rounds completed per second, then
\[\omega = 2\pi n = \frac{2\pi}{T}\]
Here \(T\) is in second (s) and \(n\) is in revolution per second (rps) or hertz (Hz). The factor \(2\pi\) appears because one full round corresponds to an angle of \(2\pi\) radian.
Uniform Circular Motion
During circular motion, if the speed of the particle remains constant, it is called Uniform Circular Motion (UCM).
In this case, only the direction of the velocity changes at every instant, and it changes in such a way that the velocity is always tangential to the path.
The acceleration responsible for this continuous change of direction is the centripetal or radial acceleration
\[\vec{a}_r = -\omega^2 \vec{r}\]
The negative sign is not a detail to be ignored. It tells us that the acceleration points opposite to the radius vector \(\vec{r}\), that is, from the particle towards the centre. Fig. 1.2 shows the directions of \(\vec{\omega}\), \(\vec{r}\), \(\vec{v}\) and \(\vec{a}_r\) together.
Diagram
Fig. 1.2 – A circle with centre C, showing at point P the outward radius vector, the tangential velocity, the inward radial acceleration, and the angular velocity at C.
For UCM, the magnitude of this acceleration is constant and is given by
\[a = \omega^2 r = \frac{v^2}{r} = v\omega\]
with \(a\) in metre per second squared (m s\(^{-2}\)). All three forms are equal; which one you use in a problem depends only on whether the data gives you \(\omega\), \(v\), or both.
It is always directed towards the centre of the circular motion (along \(-\hat{r}\)), and hence it is called centripetal, meaning centre-seeking.
An everyday illustration of UCM is the circular motion of any particle of a fan rotating uniformly. Each particle of each blade moves in a circle of its own radius, and once the fan has settled to a steady speed, that speed no longer changes.
Non-Uniform Circular Motion
When a fan is switched ON or OFF, the speeds of the particles of the fan go on increasing or decreasing for some time, however their directions are always tangential to their circular trajectories.
During this time, it is a non-uniform circular motion. As the velocity is still tangential, the centripetal or radial acceleration \(\vec{a}_r\) is still there. However, for non-uniform circular motion, the magnitude of \(\vec{a}_r\) is not constant, because \(a_r\) depends on the speed, and the speed itself is now changing.
Tangential Acceleration
Something must be changing the magnitude of the velocity as well. The acceleration responsible for changing the magnitude of velocity is directed along or opposite to the velocity, hence always tangential, and is called the tangential acceleration \(\vec{a}_T\).
So in non-uniform circular motion two accelerations act together: the radial one turning the velocity, and the tangential one speeding it up or slowing it down. Being mutually perpendicular, they never interfere with each other’s job.
As the magnitude of tangential velocity \(\vec{v}\) is changing during a non-uniform circular motion, the corresponding angular velocity \(\vec{\omega}\) is also changing at every instant. This is due to the angular acceleration
\[\vec{\alpha} = \frac{d\vec{\omega}}{dt}\]
Direction of the Angular Acceleration
Though the motion is non-uniform, the particles are still in the same plane. Hence, the direction of \(\vec{\alpha}\) is still along the axis of rotation, as shown in Fig. 1.3.
Diagram
Fig. 1.3 – Two rotating circles showing the angular acceleration vector along the axis, parallel to the angular velocity in one case and antiparallel in the other.
For increasing speed, it is along the direction of \(\vec{\omega}\), while during decreasing speed, it is opposite to that of \(\vec{\omega}\). This is the rotational version of a familiar idea: acceleration along the velocity speeds a body up, acceleration opposite to it slows the body down.
Kinematical Equations for Rotational Motion
If the angular acceleration \(\vec{\alpha}\) is constant and along the axis of rotation, all of \(\vec{\theta}\), \(\vec{\omega}\) and \(\vec{\alpha}\) will be directed along the axis. Since all three vectors then lie along one and the same line, their directions can be handled by simple plus and minus signs.
This makes it possible to use scalar notation and write the kinematical equations of motion analogous to those for translational motion, as given in Table 1.1 at the end of the chapter. Note the condition carefully — if \(\vec{\alpha}\) is not constant, or not axial, these equations cannot be used.
1.2.2 Dynamics of Circular Motion (Centripetal Force and Centrifugal Force)
i) Centripetal Force (CPF)
Kinematics tells us how a particle moves in a circle. Dynamics asks what force makes it do so. Since an acceleration can never appear on its own, there must be a force behind the centripetal acceleration.
As seen above, the acceleration responsible for circular motion is the centripetal or radial acceleration \(\vec{a}_r = -\omega^2 \vec{r}\). The force providing this acceleration is the centripetal or radial force,
\[\text{CPF} = -m\omega^2 \vec{r}\]
where \(m\) is the mass of the particle in kilogram (kg), \(\omega\) is the angular speed in rad s\(^{-1}\) and \(\vec{r}\) is the radius vector in metre (m). The force is measured in newton (N). The minus sign carries the same meaning as before: the force points from the particle towards the centre, as shown in Fig. 1.4a.
Diagram
Fig. 1.4a – A particle on a circular path with the centripetal force arrow directed towards the centre and the velocity arrow along the tangent.
ii) Centrifugal Force (CFF)
Visualize yourself on a merry-go-round rotating uniformly. If you close your eyes, you will not know that you are performing a circular motion but you will feel that you are at rest.
This feeling of being at rest is genuine for you, because you are sitting in a frame of reference that is itself rotating, and such a frame is not inertial. Newton’s laws in their usual form do not hold in it unless we make a correction.
In order to explain that you are at rest, you need to consider a force equal in magnitude to the resultant real force, but directed opposite, i.e., away from the centre. This force,
\[\left(+m\omega^2 \vec{r}\right)\]
is the centrifugal (away from the centre) force. It is a pseudo force arising due to the centripetal acceleration of the frame of reference.
Is the Centrifugal Force Real?
It must be understood that centrifugal force is a non-real force, but NOT an imaginary force. Remember, before the merry-go-round reaches its uniform speed, you were really experiencing an outward pull (because, centrifugal force is greater than the resultant force towards the centre). A force measuring instrument can record it as well.
On reaching the uniform speed, in the frame of reference of merry-go-round, this centrifugal force exactly balances the resultant of all the real forces. The resultant force in that frame of reference is thus zero.
Thus, only in such a frame of reference we can say that the centrifugal force balances the centripetal force. It must be remembered that in this case, centrifugal force means the ‘net pseudo force’ and centripetal force means the ‘resultant of all the real forces’.
Two Ways of Writing the Force Equation
There are two ways of writing force equation for a circular motion. Which one you choose depends entirely on the frame you decide to work in, and mixing the two in a single equation is a common source of mistakes.
In an inertial frame, we simply say that the resultant of all real forces provides the centripetal force:
\[\text{Resultant force} = -m\omega^2 \vec{r}\]
In the rotating frame of the body, the body is at rest, so the real forces and the pseudo force together add up to zero:
\[m\omega^2 \vec{r} + \sum \left(\text{real forces}\right) = 0\]
1.3 Applications of Uniform Circular Motion
The ideas developed so far — centripetal force, centrifugal force, and the freedom to work either in an inertial frame or in the rotating frame of the body — are not merely theoretical. They decide how fast a car may safely turn, how a stunt rider stays pinned to a vertical wall, and why highway engineers tilt curved roads.
In the sections that follow, four standard situations of uniform circular motion are analysed: a vehicle moving along a horizontal circular track, a vehicle inside the well of death, a vehicle on a banked road, and the conical pendulum.
The method of solving all four is the same, and it is worth fixing in mind before starting. First identify every real force acting on the body. Then remember that the motion is along a horizontal circle, so the resultant of these forces must be horizontal and directed towards the centre, while all the vertical forces must cancel one another. Resolving the forces along these two directions and applying \(F = mv^2/r\) along the radius gives the required result in every case.
1.3.1 Vehicle Along a Horizontal Circular Track
Forces Acting on the Vehicle
Figure 1.4 shows vertical section of a car on a horizontal circular track of radius \(r\). The plane of the figure is a vertical plane, perpendicular to the track but including only centre C of the track. This is worth picturing carefully: the car itself is moving into or out of the page, while the centre of its circular path lies to one side, in the plane of the figure.
Diagram
Fig. 1.4 – Vertical section of a car on a horizontal circular road of radius r, showing weight mg downwards, normal reaction N upwards, force of static friction towards centre C.
Forces acting on the car (considered to be a particle) are (i) weight \(mg\), vertically downwards, (ii) normal reaction \(N\), vertically upwards that balances the weight \(mg\) and (iii) force of static friction \(f_s\) between road and the tyres.
This is static friction because it prevents the vehicle from outward slipping or skidding. The tyres are rolling, not sliding, so the point of the tyre in contact with the road is momentarily at rest with respect to the road — which is precisely the situation in which static friction acts.
The first two forces are vertical and cancel each other. The friction alone is horizontal and points towards the centre of the track. This is the resultant force which is centripetal.
Working in the Frame of the Vehicle
While working in the frame of reference attached to the vehicle, the vehicle is at rest and the friction balances the centrifugal force. Either way of looking at it gives the same equations.
\[\therefore mg = N \quad \text{and} \quad f_s = mr\omega^2 = \frac{mv^2}{r}\]
Dividing the second equation by the first removes the mass altogether and leaves a neat ratio:
\[\therefore \frac{f_s}{N} = \frac{r\omega^2}{g} = \frac{v^2}{rg}\]
Here \(v\) is the speed of the vehicle in m s\(^{-1}\), \(r\) the radius of the track in m, and \(g\) the acceleration due to gravity in m s\(^{-2}\).
Why There Is an Upper Limit to the Speed
For a given track, radius \(r\) is constant. For a given vehicle, \(mg = N\) is constant. Thus, as the speed \(v\) increases, the force of static friction \(f_s\) also increases — the road simply grips harder as more grip is demanded of it.
However, \(f_s\) has an upper limit
\[\left(f_s\right)_{max} = \mu_s \cdot N\]
where \(\mu_s\) is the coefficient of static friction between road and tyres of the vehicle. This is a pure number with no unit. Beyond this value the road cannot supply any more sideways force, and the vehicle begins to skid outwards. This imposes an upper limit to the speed \(v\).
At the maximum possible speed \(v_s\), we can write
\[\frac{\left(f_s\right)_{max}}{N} = \mu_s = \frac{v_{max}^2}{rg} \quad \therefore v_{max} = \sqrt{\mu_s r g}\]
1.3.2 Well (or Wall) of Death
This is a vertical cylindrical wall of radius \(r\) inside which a vehicle is driven in horizontal circles. This can be seen while performing stunts.
Forces Acting on the Vehicle
As shown in the Fig. 1.5, the forces acting on the vehicle (assumed to be a point) are (i) Normal reaction \(N\) acting horizontally and towards the centre, (ii) Weight \(mg\) acting vertically downwards, and (iii) Force of static friction \(f_s\) acting vertically upwards between vertical wall and the tyres.
Diagram
Fig. 1.5 – A vehicle inside a vertical cylindrical well, with normal reaction N pointing horizontally towards centre C, weight mg downwards and static friction upwards along the wall.
It is static friction because it has to prevent the downward slipping. Its magnitude is equal to \(mg\), as this is the only upward force.
Compare this carefully with the previous case. On a horizontal road the normal reaction was vertical and friction was horizontal; here the wall is vertical, so the two forces have exchanged their directions and therefore their roles.
Normal reaction \(N\) is thus the resultant centripetal force (or the only force that can balance the centrifugal force). Thus, in magnitude,
\[N = mr\omega^2 = \frac{mv^2}{r} \quad \text{and} \quad mg = f_s\]
Condition for the Minimum Speed
Force of static friction \(f_s\) is always less than or equal to \(\mu_s N\). This single inequality decides whether the stunt is possible at all.
\[\therefore f_s \le \mu_s N \quad \therefore mg \le \mu_s \left(\frac{mv^2}{r}\right)\]
\[\therefore g \le \frac{\mu_s v^2}{r} \quad \therefore v^2 \ge \frac{rg}{\mu_s}\]
\[\therefore v_{min} = \sqrt{\frac{rg}{\mu_s}}\]
Read the result physically. The friction that holds the rider up is proportional to \(N\), and \(N\) itself is produced only by the circular motion. Go slower, and the wall presses less, so the available friction falls below the weight and the rider slides down. Speed is what generates the very grip that supports him.
Notice also that \(\mu_s\) sits in the denominator here, whereas on a horizontal road it appeared in the numerator. A rougher wall therefore lowers the minimum speed needed, and a smooth, oily wall makes the stunt far more demanding.
1.3.3 Vehicle on a Banked Road
Why Curved Roads Are Banked
As seen earlier, while taking a turn on a horizontal road, the force of static friction between the tyres of the vehicle and the road provides the necessary centripetal force (or balances the centrifugal force).
However, the frictional force is having an upper limit. Also, its value is usually not constant as the road surface is not uniform. A patch of oil, loose gravel, rain water or a worn-out tyre can reduce it without warning. Thus, in real life, we should not depend upon it, as far as possible.
For this purpose, the surfaces of curved roads are tilted with the horizontal with some angle \(\theta\). This is called banking of a road, or the road is said to be banked.
The idea behind banking is simple. If the road is tilted, the normal reaction is no longer vertical, and a part of it points horizontally towards the centre of the turn. That component can then supply the centripetal force without any help from friction.
Forces on a Vehicle on a Banked Road
Figure 1.6 shows the vertical section of a vehicle on a curved road of radius \(r\) banked at an angle \(\theta\) with the horizontal.
Diagram
Fig. 1.6 – Vertical section of a vehicle on a road banked at angle θ, showing weight mg downwards and normal reaction N resolved into N cos θ and N sin θ.
Considering the vehicle to be a point and ignoring friction (not eliminating) and other non-conservative forces like air resistance, there are two forces acting on the vehicle, (i) weight \(mg\), vertically downwards and (ii) normal reaction \(N\), perpendicular to the surface of the road.
The words ignoring, not eliminating are deliberate. Friction has not been wished away; we are only deriving the result for the ideal case in which its help is not needed at all. Its effect is brought back later, while finding the speed limits.
As the motion of the vehicle is along a horizontal circle, the resultant force must be horizontal and directed towards the centre of the track. It means, the vertical force \(mg\) must be balanced. Thus, we have to resolve the normal reaction \(N\) along the vertical and along the horizontal.
Its vertical component \(N\cos\theta\) balances the weight \(mg\). The horizontal component \(N\sin\theta\), being the resultant force, must be the necessary centripetal force (or balance the centrifugal force). Thus, in magnitude,
\[N\cos\theta = mg \quad \text{and}\]
\[N\sin\theta = mr\omega^2 = \frac{mv^2}{r} \therefore \tan\theta = \frac{v^2}{rg} \quad \text{--- (1.1)}\]
Dividing one equation by the other has quietly removed both \(N\) and \(m\) from the result. The banking condition therefore involves only the speed, the radius and \(g\) — it is the same for a scooter and for a loaded truck.
Most Safe Speed
For a particular road, \(r\) and \(\theta\) are fixed. Thus, this expression gives us the expression for the most safe speed (not a minimum or a maximum speed) on this road as
\[v_s = \sqrt{rg\tan\theta}\]
At exactly this speed, no friction is called upon at all. A vehicle would negotiate the curve safely even on a sheet of ice.
Banking Angle
While designing a road, this expression helps us in knowing the angle of banking as
\[\theta = \tan^{-1}\left(\frac{v^2}{rg}\right) \quad \text{--- (1.2)}\]
Here \(v\) is in m s\(^{-1}\), \(r\) in m and \(g\) in m s\(^{-2}\), so the quantity inside the bracket has no unit, as it must be for a tangent.
Speed Limits on a Rough Banked Road
Figures 1.7 and 1.8 show the vertical section of a vehicle on a rough curved road of radius \(r\), banked at an angle \(\theta\). If the vehicle is running exactly at the speed \(v_s = \sqrt{rg\tan\theta}\), the forces acting on the vehicle are (i) weight \(mg\) acting vertically downwards and (ii) normal reaction \(N\) acting perpendicular to the road.
As seen above, only at this speed, the resultant of these two forces (which is \(N\sin\theta\)) is the necessary centripetal force (or balances the centrifugal force). In practice, vehicles never travel exactly with this speed. For speeds other than this, the component of the force of static friction between road and the tyres helps us, up to a certain limit.
Lower Speed Limit
For speeds \(v_1 < \sqrt{rg\tan\theta}\), we have \(\dfrac{mv_1^2}{r} < N\sin\theta\) (or \(N\sin\theta\) is greater than the centrifugal force \(\dfrac{mv_1^2}{r}\)).
The vehicle now needs less centripetal force than the road is offering, so the excess would push it down the slope, towards the inner edge. Friction must therefore act to stop it sliding down.
In this case, the direction of force of static friction \(f_s\) between road and the tyres is directed along the inclination of the road, upwards (Fig. 1.7). Its horizontal component is parallel and opposite to \(N\sin\theta\). These two forces take care of the necessary centripetal force (or balance the centrifugal force).
Diagram
Fig. 1.7 – Banked road at lower speed limit, with static friction acting up the incline and its components f_s sin θ and f_s cos θ marked.
\[\therefore mg = f_s\sin\theta + N\cos\theta \quad \text{and}\]
\[\frac{mv_1^2}{r} = N\sin\theta - f_s\cos\theta\]
For minimum possible speed, \(f_s\) is maximum and equal to \(\mu_s N\). Using this in the equations above and solving for minimum possible speed, we get
\[\left(v_1\right)_{min} = v_{min} = \sqrt{rg\left(\frac{\tan\theta - \mu_s}{1 + \mu_s\tan\theta}\right)} \quad \text{--- (1.3)}\]
For \(\mu_s \geq \tan\theta\), \(v_{min} = 0\). This is true for most of the rough roads, banked at smaller angles. Physically it means that friction alone is strong enough to hold the vehicle on the slope, so the vehicle may even stand still on the road without sliding down.
Upper Speed Limit
For speeds \(v_2 > \sqrt{rg\tan\theta}\), we have \(\dfrac{mv_2^2}{r} > N\sin\theta\) (or \(N\sin\theta\) is less than the centrifugal force \(\dfrac{mv_2^2}{r}\)).
Now the road is not offering enough centripetal force, and the vehicle tends to slide up the slope, towards the outer edge. Friction therefore reverses its direction and acts down the incline.
In this case, the direction of force of static friction \(f_s\) between road and the tyres is directed along the inclination of the road, downwards (Fig. 1.8). Its horizontal component is parallel to \(N\sin\theta\). These two forces take care of the necessary centripetal force (or balance the centrifugal force).
Diagram
Fig. 1.8 – Banked road at upper speed limit, with static friction acting down the incline and its components f_s cos θ and f_s sin θ marked.
\[\therefore mg = N\cos\theta - f_s\sin\theta \quad \text{and}\]
\[\frac{mv_2^2}{r} = N\sin\theta + f_s\cos\theta\]
For maximum possible speed, \(f_s\) is maximum and equal to \(\mu_s N\). Using this in the equations above, and solving for maximum possible speed, we get
\[\left(v_2\right)_{max} = v_{max} = \sqrt{rg\left(\frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}\right)} \quad \text{--- (1.4)}\]
If \(\mu_s = \cot\theta\), \(v_{max} = \infty\). But \(\left(\mu_s\right)_{max} = 1\). Thus, for \(\theta \geq 45°\), \(v_{max} = \infty\).
The infinity here only means that the mathematics no longer sets any ceiling; in practice, toppling, tyre strength and engine power do. However, for a heavily banked road, the minimum limit may be important. Try to relate the concepts used while explaining the well of death — a wall of death is nothing but a road banked at 90°, where only a lower limit survives.
The Frictionless Case
For \(\mu_s = 0\), both the equations 1.3 and 1.4 give us \(v = \sqrt{rg\tan\theta}\), which is the safest speed on a banked road as we don’t take the help of friction. The upper and the lower limits close in on each other and merge into this single speed.
1.3.4 Conical Pendulum
From a Simple Pendulum to a Conical Pendulum
A tiny mass (assumed to be a point object and called a bob) connected to a long, flexible, massless, inextensible string, and suspended to a rigid support is called a pendulum.
If the string is made to oscillate in a single vertical plane, we call it a simple pendulum (to be studied in the Chapter 5).
We can also revolve the string in such a way that the string moves along the surface of a right circular cone of vertical axis and the point object performs a (practically) uniform horizontal circular motion. In such a case the system is called a conical pendulum.
The name comes from the shape swept by the string. The bob stays in one horizontal plane, while the string always leans away from the vertical by a fixed angle.
Forces Acting on the Bob
Figure 1.9 (a) shows the vertical section of a conical pendulum having bob (point mass) of mass \(m\) and string of length \(L\).
Diagram
Fig. 1.9 (a) – Vertical section of a conical pendulum in an inertial frame, showing string of length L from support A, bob at B, centre C, tension T₀ resolved into T₀ cos θ and T₀ sin θ, and weight mg.
In a given position B, the forces acting on the bob are (i) its weight \(mg\) directed vertically downwards and (ii) the force \(T_0\) due to the tension in the string, directed along the string, towards the support A.
As the motion of the bob is a horizontal circular motion, the resultant force must be horizontal and directed towards the centre C of the circular motion. For this, all the vertical forces must cancel. Hence, we shall resolve the force \(T_0\) due to the tension.
If \(\theta\) is the angle made by the string with the vertical, at any position (semi-vertical angle of the cone), the vertical component \(T_0\cos\theta\) balances the weight \(mg\). The horizontal component \(T_0\sin\theta\) then becomes the resultant force which is centripetal.
\[\therefore T_0\sin\theta = \text{centripetal force} = mr\omega^2 \quad \text{--- (1.5)}\]
Also,
\[T_0\cos\theta = mg \quad \text{--- (1.6)}\]
Here \(T_0\) is in newton (N), \(r\) is the radius of the horizontal circle in metre (m) and \(\omega\) is the angular speed in rad s\(^{-1}\).
Period and Frequency of a Conical Pendulum
Dividing eq (1.5) by Eq. (1.6), the unknown tension \(T_0\) disappears, and we get
\[\omega^2 = \frac{g\sin\theta}{r\cos\theta}\]
Radius \(r\) of the circular motion is \(r = L\sin\theta\). This is clear from the geometry of Fig. 1.9 (a): the string of length \(L\) is the slant side of the cone, and the radius is its horizontal projection.
Substituting this value of \(r\) cancels \(\sin\theta\) from the expression and leaves
\[\omega^2 = \frac{g}{L\cos\theta}\]
If \(T\) is the period of revolution of the bob,
\[\omega = \frac{2\pi}{T} = \sqrt{\frac{g}{L\cos\theta}}\]
\[\therefore \text{Period } T = 2\pi\sqrt{\frac{L\cos\theta}{g}} \quad \text{--- (1.7)}\]
Frequency of revolution,
\[n = \frac{1}{T} = \frac{1}{2\pi}\sqrt{\frac{g}{L\cos\theta}} \quad \text{--- (1.8)}\]
Here \(T\) is in second (s), \(L\) in metre (m) and \(n\) in revolution per second.
Working in the Non-Inertial Frame
The same problem can be handled from the bob’s own rotating frame, shown in Fig. 1.9 (b).
Diagram
Fig. 1.9 (b) – The same conical pendulum in a non-inertial frame, with the outward centrifugal force mrω² shown balancing the horizontal component of the tension.
In the frame of reference attached to the bob, the centrifugal force should balance the resultant of all the real forces (which we call CPF) for the bob to be at rest.
\[\therefore T_0\sin\theta = mr\omega^2 \quad \text{--- (in magnitude)}\]
This is the same as the Eq. (1.5). The two frames, as always, lead to identical equations.
1.4 Vertical Circular Motion
All the situations studied so far involved horizontal circles, where gravity acted perpendicular to the motion and could be balanced separately. Once the circle is turned into a vertical plane, gravity begins to act along the direction of motion for most of the path, and the speed can no longer stay constant on its own.
Two Types of Vertical Circular Motion
Two types of vertical circular motions are commonly observed in practice:
A controlled vertical circular motion such as a giant wheel or similar games. In this case the speed is either kept constant or NOT totally controlled by gravity. An external agency — a motor, in the case of a giant wheel — keeps supplying or removing energy as needed.
Vertical circular motion controlled only by gravity. In this case, we initially supply the necessary energy (mostly) at the lowest point. Then onwards, the entire kinetics is governed by the gravitational force. During the motion, there is interconversion of kinetic energy and gravitational potential energy.
The second type is the more interesting of the two, and it is the one analysed in detail in the sub-sections that follow.
1.4.1 Point Mass Undergoing Vertical Circular Motion Under Gravity
Case I: Mass Tied to a String
The figure 1.10 shows a bob (treated as a point mass) tied to a (practically) massless, inextensible and flexible string. It is whirled along a vertical circle so that the bob performs a vertical circular motion and the string rotates in a vertical plane.
Diagram
Fig. 1.10 – A bob on a string in vertical circular motion, marking the uppermost point A, lowermost point B, and the two horizontal-string positions C and D, with tension and weight shown at each.
At any position of the bob, there are only two forces acting on the bob: (a) its weight \(mg\), vertically downwards, which is constant and (b) the force due to the tension along the string, directed along the string and towards the centre. Its magnitude changes periodically with time and location.
The word flexible matters here. A string can only pull, never push, so the tension can fall to zero but can never become negative. This single restriction is what creates the minimum speed condition derived below.
As the motion is non-uniform, the resultant of these two forces is not directed towards the centre except at the uppermost and the lowermost positions of the bob. At all the other positions, part of the resultant is tangential and is used to change the speed.
Uppermost Position (A)
Both, weight \(mg\) and force due to tension \(T_A\) are downwards, i.e., towards the centre. In this case, their resultant is used only as the centripetal force. Thus, if \(v_A\) is the speed at the uppermost point, we get,
\[mg + T_A = \frac{mv_A^2}{r} \quad \text{--- (1.9)}\]
Radius \(r\) of the circular motion is the length of the string. For minimum possible speed at this point (or if the motion is to be realized with minimum possible energy), \(T_A = 0\)
\[\therefore \left(v_A\right)_{min} = \sqrt{rg} \quad \text{--- (1.10)}\]
Read this carefully. Putting \(T_A = 0\) means the string has gone completely slack, and the weight alone is doing the whole job of bending the path into a circle. Below this speed, gravity would be more than enough to curve the path, and the bob would leave the circular track and fall inward.
Lowermost Position (B)
Force due to the tension, \(T_B\), is vertically upwards, i.e., towards the centre, and opposite to \(mg\). In this case also their resultant is the centripetal force. If \(v_B\) is the speed at the lowermost point, we get,
\[T_B - mg = \frac{mv_B^2}{r} \quad \text{--- (1.11)}\]
To connect the speeds at the two ends we cannot use the kinematical equations, since the acceleration is not constant. Energy conservation is the tool that works.
While coming down from the uppermost to the lowermost point, the vertical displacement is \(2r\) and the motion is governed only by gravity. Hence the corresponding decrease in the gravitational potential energy is converted into the kinetic energy.
\[\therefore mg(2r) = \frac{1}{2}mv_B^2 - \frac{1}{2}mv_A^2\]
\[\therefore v_B^2 = v_A^2 + 4rg \quad \text{--- (1.12)}\]
Using this in the eq (1.11), and using \(\left(v_A\right)_{min}\) from Eq. (1.10) we get,
\[\left(v_B\right)_{min} = \sqrt{5rg} \quad \text{--- (1.13)}\]
This is the number most often needed in problems: to complete a full vertical circle on a string, the bob must be given a speed of at least \(\sqrt{5rg}\) at the bottom.
Relation Between the Extreme Tensions
Subtracting eq (1.9) from eq (1.11), we can write,
\[T_B - T_A = 2mg + \frac{m}{r}\left(v_B^2 - v_A^2\right) \quad \text{--- (1.14)}\]
Using eq (1.12) and rearranging, we get,
\[T_B - T_A = 6mg \quad \text{--- (1.15)}\]
Positions When the String Is Horizontal (C and D)
Force due to the tension is the only force towards the centre as weight \(mg\) is perpendicular to the tension. Thus, force due to the tension is the centripetal force used to change the direction of the velocity and weight \(mg\) is used only to change the speed.
Using similar mathematics, it can be shown that
\[T_C - T_A = T_D - T_A = 3mg \quad \text{and}\]
\[\left(v_C\right)_{min} = \left(v_D\right)_{min} = \sqrt{3rg}\]
Arbitrary Positions
Force due to the tension and weight are neither along the same line, nor perpendicular. Tangential component of weight is used to change the speed. It decreases the speed while going up and increases it while coming down.
The radial component of the weight, meanwhile, adds to or subtracts from the tension in providing the centripetal force. This is why the tension is largest at the bottom and smallest at the top.
Case II: Mass Tied to a Rod
Consider a bob (point mass) tied to a (practically massless and rigid) rod and whirled along a vertical circle.
The basic difference between the rod and the string is that the string needs some tension at all the points, including the uppermost point. Thus, a certain minimum speed, Eq. (1.10), is necessary at the uppermost point in the case of a string.
In the case of a rod, as the rod is rigid, such a condition is not necessary. A rod can push as well as pull, so at the top it can hold the bob up instead of merely pulling it down. Thus (practically) zero speed is possible at the uppermost point.
Using similar mathematics, it is left to the readers to show that
\[\left(v_{lowermost}\right)_{min} = \sqrt{4rg} = 2\sqrt{rg}\]
\[v_{min} \text{ at the } rod\ horizontal \text{ position} = \sqrt{2rg}\]
\[T_{lowermost} - T_{uppermost} = 6mg\]
Notice that the last relation is unchanged from the string case. That result came purely from energy conservation over a height \(2r\), and a rigid rod does not alter that argument in any way.
1.4.2 Sphere of Death
This is a popular show in a circus. During this, a two-wheeler rider (or riders) undergo rounds inside a hollow sphere. Starting with small horizontal circles, they eventually perform revolutions along vertical circles, as shown in Fig. 1.10a.
Diagram
Fig. 1.10a – A hollow sphere with a two-wheeler rider on its inner surface, showing an early small horizontal circle near the bottom and a full vertical circle, with the normal reaction N directed towards the centre.
The dynamics of this vertical circular motion is the same as that of the point mass tied to the string, except that the force due to tension \(T\) is replaced by the normal reaction force \(N\).
The replacement is exact because both forces act along the same line — from the rider towards the centre of the sphere — and both can only push or pull inwards. Every result of the previous sub-section therefore carries over unchanged. In particular, the rider must have a speed of at least \(\sqrt{rg}\) at the topmost point and at least \(\sqrt{5rg}\) at the lowermost point, \(r\) being the radius of the sphere in metre.
If you have seen this show, try to visualize that initially there are nearly horizontal circles. The linear speed is more for larger circles but angular speed (frequency) is more for smaller circles (while starting or stopping). This is as per the theory of conical pendulum.
1.4.3 Vehicle at the Top of a Convex Over-Bridge
Forces at the Topmost Point
Figure 1.11 shows a vehicle at the top of a convex over-bridge, during its motion (part of vertical circular motion). The road here is curved upwards, so its topmost point behaves exactly like the uppermost point A of a vertical circle, with the centre of curvature C lying below the road.
Diagram
Fig. 1.11 – A vehicle at the top of a convex over-bridge, with weight mg downwards, normal reaction N upwards, and the centre of curvature C marked below.
Forces acting on the vehicle are (a) Weight \(mg\) and (b) Normal reaction force \(N\), both along the vertical line (topmost position). The resultant of these two must provide the necessary centripetal force (vertically downwards) if the vehicle is at the uppermost position.
Thus, if \(v\) is the speed at the uppermost point,
\[mg - N = \frac{mv^2}{r}\]
where \(r\) is the radius of curvature of the bridge in metre (m).
The Upper Speed Limit
Rearranged, this equation reads \(N = mg - \dfrac{mv^2}{r}\). As the speed is increased, \(N\) goes on decreasing — the faster the vehicle moves, the less it presses on the road.
Normal reaction is an indication of contact. Thus, for just maintaining contact, \(N = 0\). This imposes an upper limit on the speed as
\[v_{max} = \sqrt{rg}\]
1.5 Moment of Inertia as an Analogous Quantity for Mass
Why Mass Needs a Rotational Replacement
In XI\(^{th}\) Std. we saw that angular displacement, angular velocity and angular acceleration respectively replace displacement, velocity and acceleration for various kinematical equations. Also, torque is an analogous quantity for force.
Expressions of linear momentum, force (for a fixed mass) and kinetic energy include mass as a common term. In order to have their rotational analogues, we need a replacement for mass.
Mass, in translational motion, measures inertia — the reluctance of a body to change its state of motion. The question here is what plays that role when the motion is rotational.
What the Door Tells Us
If we open a door (with hinges), we give a certain angular displacement to it. The efforts needed for this depend not only upon the mass of the door, but also upon the (perpendicular) distance from the axis of rotation, where we apply the force.
Everyone knows this from experience: pushing near the handle, far from the hinges, opens the door easily, while pushing close to the hinges is hard work even though the door’s mass has not changed.
Thus, the quantity analogous to mass includes not only the mass, but also takes care of the distance-wise distribution of the mass around the axis of rotation. To know the exact relation, let us derive an expression for the rotational kinetic energy which is the sum of the translational kinetic energies of all the individual particles.
Deriving the Rotational Kinetic Energy
Figure 1.12 shows a rigid object rotating with a constant angular speed \(\omega\) about an axis perpendicular to the plane of paper.
Diagram
Fig. 1.12 – A rigid body of N particles rotating about an axis perpendicular to the plane of the paper, with particles of masses m₁, m₂, … m_N at perpendicular distances r₁, r₂, … r_N from the axis.
For theoretical simplification let us consider the object to be consisting of \(N\) particles of masses \(m_1, m_2, \ldots m_N\) at respective perpendicular distances \(r_1, r_2, \ldots r_N\) from the axis of rotation.
As the object rotates, all these particles perform UCM with the same angular speed \(\omega\), but with different linear speeds
\[v_1 = r_1\omega,\ v_2 = r_2\omega,\ \ldots,\ v_N = r_N\omega\]
This is the crucial point of the whole derivation. The angular speed is common to every particle of a rigid body, but the linear speed is not — a particle farther from the axis moves faster, and therefore carries more kinetic energy.
Translational K.E. of the first particle is
\[\text{K.E.}_1 = \frac{1}{2}m_1v_1^2 = \frac{1}{2}m_1r_1^2\omega^2\]
Similar will be the case of all the other particles. Rotational K.E. of the object is the sum of individual translational kinetic energies. Thus, rotational K.E.
\[= \frac{1}{2}m_1r_1^2\omega^2 + \frac{1}{2}m_2r_2^2\omega^2 + \ldots + \frac{1}{2}m_Nr_N^2\omega^2\]
Since \(\omega\) is the same for all the particles, it can be taken out as a common factor:
\[\therefore \text{Rotational K.E.} = \frac{1}{2}\left(m_1r_1^2 + m_2r_2^2 + \ldots + m_Nr_N^2\right)\omega^2 = \frac{1}{2}I\omega^2\]
Where
\[I = m_1r_1^2 + m_2r_2^2 + \ldots + m_Nr_N^2 = \sum_{i=1}^{N}m_ir_i^2\]
Moment of Inertia Defined
If \(I = \sum m_ir_i^2\) replaces mass \(m\) and angular speed \(\omega\) replaces linear speed \(v\), rotational K.E. \(= \dfrac{1}{2}I\omega^2\) is analogous to translational K.E. \(\dfrac{1}{2}mv^2\).
Thus, \(I\) is defined to be the rotational inertia or moment of inertia (M.I.) of the object about the given axis of rotation.
Its SI unit is kilogram metre squared (kg m\(^2\)), and it is a scalar quantity. Note that the distance enters as a square, so shifting mass twice as far from the axis increases its contribution four times over.
What the Moment of Inertia Depends Upon
It is clear that the moment of inertia of an object depends upon (i) individual masses and (ii) the distribution of these masses about the given axis of rotation.
For a different axis, it will again depend upon the mass distribution around that axis and will be different if there is no symmetry. Unlike mass, therefore, moment of inertia is never a property of the object alone — the axis must always be specified along with it.
Moment of Inertia by Integration
During this discussion, for simplicity, we assumed the object to be consisting of a finite number of particles. In practice, usually, it is not so — a real body has its mass spread continuously.
For a homogeneous rigid object of mathematically integrable mass distribution, the moment of inertia is to be obtained by integration as
\[I = \int r^2\,dm\]
Here \(dm\) is the mass of an infinitesimally small element of the object and \(r\) is the perpendicular distance of that element from the axis. The summation of the earlier expression has simply become an integration.
If integrable mass distribution is not known, it is not possible to obtain the moment of inertia theoretically, but it can be determined experimentally.
1.5.1 Moment of Inertia of a Uniform Ring
An object is called a uniform ring if its mass is (practically) situated uniformly on the circumference of a circle, as shown in Fig. 1.13. Obviously, it is a two dimensional object of negligible thickness.
Diagram
Fig. 1.13 – A uniform ring of mass M and radius R rotating about its central axis, perpendicular to its plane and passing through the centre.
If it is rotating about its own axis (line perpendicular to its plane and passing through its centre), its entire mass \(M\) is practically at a distance equal to its radius \(R\) from the axis.
This makes the ring the simplest case of all. Every mass element \(dm\) has the same perpendicular distance \(R\), so \(R^2\) can be taken out of the integral:
\[I = \int r^2\,dm = R^2\int dm = MR^2\]
Hence, the expression for the moment of inertia of a uniform ring of mass \(M\) and radius \(R\) is \(I = MR^2\).
Since no part of a ring’s mass lies closer to the axis than \(R\), no other object of the same mass and same radius can have a moment of inertia larger than this about its central axis.
1.5.2 Moment of Inertia of a Uniform Disc
Surface Density
A disc is a two dimensional circular object of negligible thickness. It is said to be uniform if its mass per unit area and its composition is the same throughout.
The ratio
\[\sigma = \frac{m}{A} = \frac{\text{mass}}{\text{area}}\]
is called the surface density. Its SI unit is kilogram per square metre (kg m\(^{-2}\)).
Consider a uniform disc of mass \(M\) and radius \(R\) rotating about its own axis, which is the line perpendicular to its plane and passing through its centre.
\[\therefore \sigma = \frac{M}{\pi R^2}\]
since the total area of the disc is \(\pi R^2\).
Splitting the Disc into Rings
Unlike a ring, the mass of a disc is not all at one distance from the axis, so we cannot simply write \(MR^2\). The way out is to break the disc into pieces whose moment of inertia we already know.
As it is a uniform circular object, it can be considered to be consisting of a number of concentric rings of radii increasing from (practically) zero to \(R\). One of such rings of mass \(dm\) is shown by the shaded portion in the Fig. 1.14.
Diagram
Fig. 1.14 – A uniform disc of radius R with a thin concentric ring of average radius r and width dr shaded, rotating about its central axis.
Width of this ring is \(dr\), which is so small that the entire ring can be considered to be of average radius \(r\). (In a practical sense, \(dr\) is less than the least count of the instrument that measures \(r\), so that \(r\) is constant for that ring.)
If such a thin ring is imagined cut open and straightened, it becomes a strip of length \(2\pi r\) and width \(dr\). Area of this ring is \(A = 2\pi r\,dr\).
\[\therefore \sigma = \frac{dm}{2\pi r\,dr} \quad \therefore dm = 2\pi\sigma r\,dr\]
As it is a ring, this entire mass is at a distance \(r\) from the axis of rotation. Thus, the moment of inertia of this ring is
\[I_r = dm\left(r^2\right)\]
Integrating Over the Whole Disc
Moment of inertia (\(I\)) of the disc can now be obtained by integrating \(I_r\) from \(r = 0\) to \(r = R\), which adds up the contributions of all such rings from the centre right up to the rim.
\[\therefore I = \int_0^R dm\ r^2 = \int_0^R 2\pi\sigma r\,dr\ r^2 = 2\pi\sigma\int_0^R r^3\,dr\]
\[\therefore I = 2\pi\sigma\left(\frac{R^4}{4}\right) = 2\pi\left(\frac{M}{\pi R^2}\right)\left(\frac{R^4}{4}\right) = \frac{1}{2}MR^2\]
Using a similar method, expressions for moment of inertias of objects of several integrable geometrical shapes can be derived. Some of those are given in the Table 3 at the end of the chapter.
1.6 Radius of Gyration
Why Such a Quantity Is Needed
As stated earlier, theoretical calculation of moment of inertia is possible only for mathematically integrable geometrical shapes. However, experimentally we can determine the moment of inertia of any object.
It depends upon the mass of that object and how that mass is distributed from or around the given axis of rotation. These two factors are mixed together in a single number \(I\), and often we wish to separate them — to set the mass aside and look only at the distribution.
If we are interested in knowing only the mass distribution around the axis of rotation, we can express the moment of inertia of any object as
\[I = MK^2\]
where \(M\) is the mass of that object in kilogram (kg) and \(K\) is measured in metre (m).
It means that the mass of that object is effectively at a distance \(K\) from the given axis of rotation. In this case, \(K\) is defined as the radius of gyration of the object about the given axis of rotation.
In other words, if \(K\) is the radius of gyration for an object, \(I = MK^2\) is the moment of inertia of that object. Larger the value of \(K\), farther is the mass from the axis.
Comparing a Ring and a Disc
The idea becomes clearest when two objects of the same mass and same radius, but different shapes, are compared, as in Fig. 1.15a.
Diagram
Fig. 1.15a – A uniform ring and a uniform disc of the same mass and radius side by side, with the radius of gyration K marked as a dashed circle on each.
1.7 Theorem of Parallel Axes and Theorem of Perpendicular Axes
Expressions of moment of inertias of regular geometrical shapes given in the table 3 are about their axes of symmetry. These are derived by integration.
However, every time the axis need not be the axis of symmetry. Real objects are rotated about all sorts of axes — a door about its hinges, a rod about one end — and carrying out a fresh integration for each new axis would be tedious, and often impossible.
Fortunately, two shortcuts exist. In simple transformations the new axis may be parallel or perpendicular to the symmetrical axis.
For example, if a rod is rotated about one of its ends, the axis is parallel to its axis of symmetry. If a disc or a ring is rotated about its diameter, the axis is perpendicular to the central axis.
In such cases, simple transformations are possible in the expressions of moment of inertias. These are called theorem of parallel axes and theorem of perpendicular axes.
Together, these two theorems let us obtain the moment of inertia about a very large number of axes from just one known result for each shape, which is why they are so heavily used in numerical problems.
1.7.1 Theorem of Parallel Axes
Setting Up the Two Axes
In order to apply this theorem to any object, we need two axes parallel to each other with one of them passing through the centre of mass of the object.
This requirement is not a formality. The theorem simply fails if neither axis passes through the centre of mass, and forgetting this is one of the commonest mistakes in numericals.
Figure 1.15 shows an object of mass \(M\). Axis MOP is any axis passing through point O. Axis ACB is passing through the centre of mass C of the object, parallel to the axis MOP, and at a distance \(h\) from it ($h = $ CO).
Diagram
Fig. 1.15 – An irregular object of mass M with axis ACB through its centre of mass C and a parallel axis MOP through point O at distance h, showing mass element dm at D and perpendicular DN on OC produced.
Consider a mass element \(dm\) located at point D. Perpendicular on OC (produced) from point D is DN.
Moment of inertia of the object about the axis ACB is
\[I_C = \int (\text{DC})^2\,dm\]
and about the axis MOP it is
\[I_O = \int (\text{DO})^2\,dm\]
Derivation
Since DN is perpendicular to the line ON, the triangle DNO is right-angled at N, and Pythagoras’ theorem may be applied to it. The same triangle argument applied to DNC lets us bring DC into the expression.
\[\therefore I_O = \int (\text{DO})^2\,dm = \int \left([\text{DN}]^2 + [\text{NO}]^2\right)dm\]
\[= \int \left([\text{DN}]^2 + [\text{NC}]^2 + 2 \cdot \text{NC} \cdot \text{CO} + [\text{CO}]^2\right)dm\]
\[= \int \left([\text{DC}]^2 + 2\,\text{NC}.h + h^2\right)dm\]
\[= \int (\text{DC})^2\,dm + 2h\int \text{NC}.dm + h^2\int dm\]
Now, \(\int (\text{DC})^2\,dm = I_C\) and \(\int dm = M\).
The middle term is the interesting one. NC is the distance of a point from the centre of mass. Any mass distribution is symmetric about the centre of mass. Thus, from the definition of the centre of mass,
\[\int \text{NC}.dm = 0\]
For every mass element on one side of the centre of mass there is a matching contribution on the other side, with the opposite sign, so the whole integral cancels out. This is exactly why one of the axes must pass through the centre of mass.
\[\therefore I_O = I_C + M.h^2\]
This is the mathematical form of the theorem of parallel axes.
Two consequences follow immediately. Since \(Mh^2\) can never be negative, the moment of inertia about the axis through the centre of mass is always the smallest among all parallel axes. And as the axis is shifted farther away, \(I\) grows as the square of the shift.
1.7.2 Theorem of Perpendicular Axes
The Set-Up
This theorem relates the moment of inertias of a laminar object about three mutually perpendicular and concurrent axes, two of them in the plane of the object and the third perpendicular to the object.
A laminar object is like a leaf, or any two dimensional object, e.g., a ring, a disc, any plane sheet, etc. The word concurrent means that all three axes must meet at one common point.
Figure 1.16 shows a rigid laminar object able to rotate about three mutually perpendicular axes \(x\), \(y\) and \(z\). Axes \(x\) and \(y\) are in the plane of the object while the \(z\) axis is perpendicular to it, and all are concurrent at O.
Diagram
Fig. 1.16 – A laminar object in the x–y plane with the z axis perpendicular to it, all three concurrent at O, showing a mass element dm at point P with perpendiculars PM = y and PN = x.
Consider a mass element \(dm\) located at any point P. PM = \(y\) and PN = \(x\) are the perpendiculars drawn from P respectively on the \(x\) and \(y\) axes.
The respective perpendicular distances of point P from \(x\), \(y\) and \(z\) axes will then be \(y\), \(x\) and \(\sqrt{y^2 + x^2}\).
The third distance comes straight from Pythagoras’ theorem: the foot of the \(z\) axis is at O, and OP is the hypotenuse of a right-angled triangle whose other two sides are \(x\) and \(y\).
Derivation
If \(I_x\), \(I_y\) and \(I_z\) are the respective moments of inertia of the body about \(x\), \(y\) and \(z\) axes, we can write,
\[I_x = \int y^2\,dm, \quad I_y = \int x^2\,dm \quad \text{and}\]
\[I_z = \int \left(y^2 + x^2\right)dm\]
Splitting this last integral into two parts gives the result at once:
\[\therefore I_z = \int y^2\,dm + \int x^2\,dm = I_x + I_y\]
This is the mathematical form of the theorem of perpendicular axes.
1.8 Angular Momentum or Moment of Linear Momentum
The quantity in rotational mechanics, analogous to linear momentum, is angular momentum or moment of linear momentum. It is similar to the torque being the moment of a force.
Just as a force produces a turning effect only when it acts at some distance from the axis, linear momentum has a turning effect about an axis only when the particle is moving at some distance from it. That turning effect is what angular momentum measures.
If \(\vec{p}\) is the instantaneous linear momentum of a particle undertaking a circular motion, its angular momentum at that instant is given by
\[\vec{L} = \vec{r} \times \vec{p}\]
where \(\vec{r}\) is the position vector from the axis of rotation, as shown in Fig. 1.16a.
Diagram
Fig. 1.16a – A particle moving along a curved path, with position vector r from the axis, linear momentum p, and angular momentum L perpendicular to both by the right-hand rule.
In magnitude, it is the product of linear momentum and its perpendicular distance from the axis of rotation.
\[\therefore L = p \times r\sin\theta\]
where \(\theta\) is the smaller angle between the directions of \(\vec{p}\) and \(\vec{r}\).
Being a cross product, \(\vec{L}\) is a vector perpendicular to the plane containing \(\vec{r}\) and \(\vec{p}\), and its direction is given by the right-hand rule. Its SI unit is kg m\(^2\) s\(^{-1}\).
1.8.1 Expression for Angular Momentum in Terms of Moment of Inertia
Setting Up the Particle Model
Figure 1.12 in the section 1.5 shows a rigid object rotating with a constant angular speed \(\omega\) about an axis perpendicular to the plane of paper. The same model is used again here, because the reasoning that gave us the rotational kinetic energy will now give us the angular momentum.
For theoretical simplification let us consider the object to be consisting of \(N\) number of particles of masses \(m_1, m_2, \ldots m_N\) at respective perpendicular distances \(r_1, r_2, \ldots r_N\) from the axis of rotation.
As the object rotates, all these particles perform UCM with the same angular speed \(\omega\), but with different linear speeds
\[v_1 = r_1\omega,\ v_2 = r_2\omega,\ \ldots\ v_N = r_N\omega\]
Angular Momentum of Each Particle
Directions of individual velocities \(\vec{v}_1\), \(\vec{v}_2\), etc., are along the tangents to their respective tracks. Linear momentum of the first particle is of magnitude
\[p_1 = m_1v_1 = m_1r_1\omega\]
Its direction is along that of \(\vec{v}_1\).
For a particle in circular motion, the radius vector and the tangential velocity are perpendicular to each other, so \(\sin\theta = 1\) and the magnitude of the cross product reduces to a simple product.
Its angular momentum is thus of magnitude
\[L_1 = p_1r_1 = m_1r_1^2\omega\]
Similarly,
\[L_2 = m_2r_2^2\omega,\quad L_3 = m_3r_3^2\omega,\quad \ldots\quad L_N = m_Nr_N^2\omega\]
Adding Them Up
Angular momentum is a vector, so in general such quantities cannot simply be added as numbers. Here, however, they can.
For a rigid body with a fixed axis of rotation, all these angular momenta are directed along the axis of rotation, and this direction can be obtained by using the right hand thumb rule. As all of them have the same direction, their magnitudes can be algebraically added.
Thus, magnitude of angular momentum of the body is given by
\[L = m_1r_1^2\omega + m_2r_2^2\omega + \ldots + m_Nr_N^2\omega\]
\[= \left(m_1r_1^2 + m_2r_2^2 + \ldots + m_Nr_N^2\right)\omega = I\omega\]
Where, \(I = m_1r_1^2 + m_2r_2^2 + \ldots + m_Nr_N^2\) is the moment of inertia of the body about the given axis of rotation.
The expression for angular momentum \(L = I\omega\) is analogous to the expression \(p = mv\) of linear momentum, if the moment of inertia \(I\) replaces mass, which is its physical significance.
1.9 Expression for Torque in Terms of Moment of Inertia
Setting Up the Particle Model
Figure 1.17 shows a rigid object rotating with a constant angular acceleration \(\alpha\) about an axis perpendicular to the plane of paper.
Diagram
Fig. 1.17 – A rigid rotating object made up of N particles at perpendicular distances r₁, r₂, … r_N from the axis, with tangential forces acting on them.
For theoretical simplification let us consider the object to be consisting of \(N\) number of particles of masses \(m_1, m_2, \ldots m_N\) at respective perpendicular distances \(r_1, r_2, \ldots r_N\) from the axis of rotation.
Note the change from the previous two derivations. There the body rotated with a constant angular speed; here it is being angularly accelerated, so forces must be acting on its particles.
As the object rotates, all these particles perform circular motion with the same angular acceleration \(\alpha\), but with different linear (tangential) accelerations
\[a_1 = r_1\alpha,\quad a_2 = r_2\alpha,\quad \ldots\quad a_N = r_N\alpha \quad \text{etc.}\]
Torque on Each Particle
Force experienced by the first particle is
\[f_1 = m_1a_1 = m_1r_1\alpha\]
by Newton’s second law applied to that particle alone.
As these forces are tangential, their respective perpendicular distances from the axis are \(r_1, r_2, \ldots r_N\). A tangential force is always perpendicular to the radius, so the moment arm is simply the radius itself.
Thus, the torque experienced by the first particle is of magnitude
\[\tau_1 = f_1r_1 = m_1r_1^2\alpha\]
Similarly,
\[\tau_2 = m_2r_2^2\alpha,\quad \tau_3 = m_3r_3^2\alpha,\quad \ldots\quad \tau_N = m_Nr_N^2\alpha\]
The Resultant Torque
If the rotation is restricted to a single plane, directions of all these torques are the same, and along the axis. Being collinear, they may again be added as plain numbers.
Magnitude of the resultant torque is then given by
\[\tau = \tau_1 + \tau_2 + \ldots + \tau_N\]
\[= \left(m_1r_1^2 + m_2r_2^2 + \ldots + m_Nr_N^2\right)\alpha = I\alpha\]
where, \(I = m_1r_1^2 + m_2r_2^2 + \ldots + m_Nr_N^2\) is the moment of inertia of the object about the given axis of rotation.
Here \(\tau\) is measured in newton metre (N m) and \(\alpha\) in rad s\(^{-2}\).
The relation \(\tau = I\alpha\) is analogous to \(f = ma\) for the translational motion if the moment of inertia \(I\) replaces mass, which is its physical significance.
1.10 Conservation of Angular Momentum
In the article 4.7 of XI\(^{th}\) Std. we have seen the conservation of linear momentum, which says that the linear momentum of an isolated system is conserved in the absence of an external unbalanced force.
As seen earlier, torque and angular momentum are the respective analogous quantities to force and linear momentum in rotational dynamics. With suitable changes this can be transformed into the conservation of angular momentum.
Deriving the Principle
As seen in the section 1.8, angular momentum or the moment of linear momentum of a system is given by
\[\vec{L} = \vec{r} \times \vec{p}\]
where \(\vec{r}\) is the position vector from the axis of rotation and \(\vec{p}\) is the linear momentum.
Differentiating with respect to time, and using the product rule for a cross product, we get,
\[\frac{d\vec{L}}{dt} = \frac{d}{dt}\left(\vec{r} \times \vec{p}\right) = \frac{d\vec{r}}{dt} \times \vec{p} + \vec{r} \times \frac{d\vec{p}}{dt}\]
Now, \(\dfrac{d\vec{r}}{dt} = \vec{v}\) and \(\dfrac{d\vec{p}}{dt} = \vec{F}\), the second being Newton’s second law in its momentum form.
\[\therefore \frac{d\vec{L}}{dt} = \vec{v} \times m\vec{v} + \vec{r} \times \vec{F}\]
Now \(\vec{v} \times \vec{v} = 0\), since the cross product of any vector with itself is zero.
\[\therefore \frac{d\vec{L}}{dt} = \vec{r} \times \vec{F}\]
But \(\vec{r} \times \vec{F}\) is the moment of force or torque \(\vec{\tau}\).
\[\therefore \frac{d\vec{L}}{dt} = \vec{\tau}\]
Thus, if \(\vec{\tau} = 0\), \(\dfrac{d\vec{L}}{dt} = 0\) or $ = $ constant.
Hence, angular momentum \(\vec{L}\) is conserved in the absence of an external unbalanced torque \(\vec{\tau}\).
This is the principle of conservation of angular momentum, analogous to the conservation of linear momentum. Note the condition carefully — internal forces and torques within the system may do as they please; only an external unbalanced torque can change \(\vec{L}\).
Examples of Conservation of Angular Momentum
During some shows of ballet dance, acrobats in a circus, sports like ice skating, diving in a swimming pool, etc., the principle of conservation of angular momentum is realized.
In all these applications the product
\[L = I\omega = I\left(2\pi n\right)\]
is constant (once the players acquire a certain speed). Thus, if the moment of inertia \(I\) is increased, the angular speed and hence the frequency of revolution \(n\) decreases. Also, if the moment of inertia is decreased, the frequency increases.
Diagram
Fig. 1.17a – A performer spinning with arms and legs outstretched shown beside the same performer with limbs folded close to the body, marking the larger I with smaller n and the smaller I with larger n.
Ballet Dancers
During ice ballet, the dancers have to undertake rounds of smaller and larger radii.
The dancers come together while taking the rounds of smaller radius (near the centre). In this case, the moment of inertia of their system becomes minimum and the frequency increases, to make it thrilling.
While taking the outer rounds, the dancers outstretch their legs and arms. This increases their moment of inertia, which reduces the angular speed and hence the linear speed. This is essential to prevent slipping.
Diving in a Swimming Pool
While on the diving board, the divers stretch their body so as to increase the moment of inertia.
Immediately after leaving the board, they fold their body. This reduces the moment of inertia considerably. As a result, the frequency increases and they can complete more rounds in air to make the show attractive.
Again, while entering into the water they stretch their body into a streamlined shape. This allows them a smooth entry into the water.
1.11 Rolling Motion
Rolling as Two Motions Combined
The objects like a cylinder, sphere, wheels, etc. are quite often seen to perform rolling motion.
In the case of pure rolling, two motions are undertaken simultaneously: circular motion and linear motion. Individual motion of the particles (except the one at the centre of mass) is too difficult to describe — a point on the rim traces a looping curve rather than a circle or a straight line.
However, for theory considerations we can consider the actual motion to be the result of
rotational motion of the body as a whole, about its own symmetric axis and
linear motion of the body assuming it to be concentrated at its centre of mass. In other words, the centre of mass performs purely translational motion.
Diagram
Fig. 1.17b – A rolling wheel shown as the sum of two separate motions: pure translation of the centre of mass and pure rotation about the central axis.
Accordingly, the object possesses two types of kinetic energies, rotational and translational. The sum of these two is its total kinetic energy.
Total Kinetic Energy of a Rolling Body
Consider an object of moment of inertia \(I\), rolling uniformly. The following quantities can be related.
\(v\) = Linear speed of the centre of mass, in m s\(^{-1}\)
\(R\) = Radius of the body, in m
\(\omega\) = Angular speed of rotation of the body \(\therefore \omega = \dfrac{v}{R}\) for any particle
\(M\) = Mass of the body, in kg
\(K\) = Radius of gyration of the body \(\therefore I = MK^2\)
Total kinetic energy of rolling = Translational K.E. + Rotational K.E.
Substituting \(I = MK^2\) and \(\omega = v/R\) turns the rotational term into a second expression in \(v\), allowing both terms to be combined.
\[\therefore E = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}Mv^2 + \frac{1}{2}MK^2\frac{v^2}{R^2} = \frac{1}{2}Mv^2\left(1 + \frac{K^2}{R^2}\right) \quad \text{--- (1.18)}\]
The bracket is worth noticing. A rolling body always carries more kinetic energy than a body of the same mass sliding at the same speed, and the factor \(\left(1 + \dfrac{K^2}{R^2}\right)\) measures exactly how much extra. It depends only on the shape of the body, not on its mass, radius or speed.
The Role of Static Friction
It must be remembered that static friction is essential for a purely rolling motion. In this case, it prevents the sliding motion.
You might have noticed that many a time while rolling down, the motion is initially a purely rolling motion that later on turns out to be a sliding motion. Similarly, if you push a sphere-like object along a horizontal surface, initially it slips for some distance and then starts rolling.
1.11.1 Linear Acceleration and Speed While Pure Rolling Down an Inclined Plane
Figure 1.18 shows a rigid object of mass \(M\) and radius \(R\), rolling down an inclined plane, without slipping. Inclination of the plane with the horizontal is \(\theta\).
Diagram
Fig. 1.18 – A circular object rolling down an inclined plane of inclination θ, showing the vertical drop h and the distance s travelled along the incline.
Using Energy Conservation
As the object starts rolling down, its gravitational P.E. is converted into K.E. of rolling. Since the friction here is static and the contact point does not slide, this friction does no work, and mechanical energy is conserved.
Starting from rest, let \(v\) be the speed of the centre of mass as the object comes down through a vertical distance \(h\).
From Eq. (1.18),
\[E = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}Mv^2\left(1 + \frac{K^2}{R^2}\right)\]
\[\therefore E = Mgh = \frac{1}{2}Mv^2\left(1 + \frac{K^2}{R^2}\right)\]
\[\therefore v = \sqrt{\frac{2gh}{\left(1 + \dfrac{K^2}{R^2}\right)}} \quad \text{--- (1.19)}\]
The mass \(M\) has cancelled from both sides. The speed at the bottom therefore does not depend on how heavy the object is, nor on its radius — only on the height fallen and on the shape factor \(K^2/R^2\).
Finding the Acceleration
Linear distance travelled along the plane is
\[s = \frac{h}{\sin\theta}\]
as can be seen from the geometry of the incline in Fig. 1.18.
During this distance, the linear velocity has increased from zero to \(v\). Since the acceleration along the incline is constant, the ordinary equation of motion may be used here. If \(a\) is the linear acceleration along the plane,
\[2as = v^2 - u^2 \quad \therefore 2a\left(\frac{h}{\sin\theta}\right) = \frac{2gh}{\left(1 + \dfrac{K^2}{R^2}\right)} - 0\]
\[\therefore a = \frac{g\sin\theta}{\left(1 + \dfrac{K^2}{R^2}\right)} \quad \text{--- (1.20)}\]
Comparison with Pure Sliding
For pure sliding, without friction, the acceleration is \(g\sin\theta\) and the final velocity is \(\sqrt{2gh}\).
Thus, during pure rolling, the factor \(\left(1 + \dfrac{K^2}{R^2}\right)\) is effective for both the expressions. A rolling body is always slower than a sliding one, because part of the energy released by the fall has gone into spinning the body rather than into moving it forward.
Remarks
- For a rolling object, if the expression for moment of inertia is of the form \(n\left(MR^2\right)\), the numerical factor \(n\) gives the value of \(\dfrac{K^2}{R^2}\) for that object.
For example, for a uniform solid sphere,
\[I = \frac{2}{5}MR^2 = MK^2 \quad \therefore \frac{K^2}{R^2} = \frac{2}{5}\]
Similarly,
\[\frac{K^2}{R^2} = 1, \text{ for a ring or a hollow cylinder}\]
\[\frac{K^2}{R^2} = \frac{1}{2} \text{ for a uniform disc or a solid cylinder}\]
\[\frac{K^2}{R^2} = \frac{2}{3} \text{ for a thin walled hollow sphere}\]
When a rod rolls, it is actually a cylinder that is rolling.
While rolling, the ratio ‘Translational K.E. : Rotational K.E. : Total K.E.’ is
\[1 : \frac{K^2}{R^2} : \left(1 + \frac{K^2}{R^2}\right)\]
For example, for a hollow sphere, \(\dfrac{K^2}{R^2} = \dfrac{2}{3}\)
Thus, for a rolling hollow sphere,
Translational K.E. : Rotational K.E. : Total K.E. \(= 1 : \dfrac{2}{3} : \left(1 + \dfrac{2}{3}\right) = 3 : 2 : 5\)
Percentage wise, 60% of its kinetic energy is translational and 40% is rotational.
1.12 Summary
Circular Motion — Basics
During revolution, a body moves in a circle about a point or axis lying outside it. During rotation, the motion is about an axis passing through the body itself.
Circular motion is always an accelerated motion, because the direction of velocity changes at every instant. It is also periodic in space, as the same path is retraced.
Angular quantities: angular displacement \(\vec{\theta}\) (rad), angular velocity \(\vec{\omega} = \dfrac{d\vec{\theta}}{dt}\) (rad s\(^{-1}\)) and angular acceleration \(\vec{\alpha} = \dfrac{d\vec{\omega}}{dt}\) (rad s\(^{-2}\)).
Tangential velocity: \(\vec{v} = \vec{\omega} \times \vec{r}\), with magnitude \(v = \omega r\).
Direction of \(\vec{\omega}\) is always along the axis of rotation, given by the right-hand thumb rule.
Period \(T\) and frequency \(n\): \(\omega = 2\pi n = \dfrac{2\pi}{T}\).
Uniform circular motion (UCM): the speed remains constant; only the direction of velocity changes. The centripetal (radial) acceleration is \(\vec{a}_r = -\omega^2\vec{r}\), with magnitude \(a = \omega^2r = \dfrac{v^2}{r} = v\omega\), always directed towards the centre.
Non-uniform circular motion: \(\vec{a}_r\) is still present but its magnitude is not constant. An additional tangential acceleration \(\vec{a}_T\), along or opposite to \(\vec{v}\), changes the speed.
Direction of \(\vec{\alpha}\) is along \(\vec{\omega}\) for increasing speed and opposite to \(\vec{\omega}\) for decreasing speed.
Dynamics of Circular Motion
Centripetal force (CPF) \(= -m\omega^2\vec{r}\); in magnitude \(mr\omega^2 = \dfrac{mv^2}{r} = mv\omega\). It is the resultant of all real forces, always directed towards the centre. “Centripetal” describes the direction of a force, not its type.
Centrifugal force (CFF) \(= \left(+m\omega^2\vec{r}\right)\) is a pseudo force, experienced only in the rotating (non-inertial) frame of the body. It is non-real, but not imaginary.
Two equivalent ways of writing the force equation:
\[\text{Resultant force} = -m\omega^2\vec{r} \quad \text{or} \quad m\omega^2\vec{r} + \sum\left(\text{real forces}\right) = 0\]
Applications of Uniform Circular Motion
Vehicle on a horizontal circular track: \(mg = N\), \(f_s = \dfrac{mv^2}{r}\), \(\dfrac{f_s}{N} = \dfrac{r\omega^2}{g} = \dfrac{v^2}{rg}\). Since \(\left(f_s\right)_{max} = \mu_sN\), there is an upper speed limit \(v_{max} = \sqrt{\mu_s rg}\).
Well (or wall) of death: \(N = \dfrac{mv^2}{r}\) and \(mg = f_s\). Since \(f_s \leq \mu_sN\), there is a lower speed limit \(v_{min} = \sqrt{\dfrac{rg}{\mu_s}}\), and \(N_{min} = \dfrac{mg}{\mu_s}\).
Banked road (friction ignored): \(N\cos\theta = mg\) and \(N\sin\theta = \dfrac{mv^2}{r}\), giving
\[\tan\theta = \frac{v^2}{rg} \quad \text{--- (1.1)}\]
- Most safe speed \(v_s = \sqrt{rg\tan\theta}\); banking angle
\[\theta = \tan^{-1}\left(\frac{v^2}{rg}\right) \quad \text{--- (1.2)}\]
- Banked rough road, speed limits:
\[v_{min} = \sqrt{rg\left(\frac{\tan\theta - \mu_s}{1 + \mu_s\tan\theta}\right)} \quad \text{--- (1.3)}\]
\[v_{max} = \sqrt{rg\left(\frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}\right)} \quad \text{--- (1.4)}\]
For \(\mu_s \geq \tan\theta\), \(v_{min} = 0\). For \(\theta \geq 45°\), \(v_{max} = \infty\). For \(\mu_s = 0\), both reduce to \(v = \sqrt{rg\tan\theta}\).
Conical pendulum: \(T_0\sin\theta = mr\omega^2\) (1.5) and \(T_0\cos\theta = mg\) (1.6), with \(r = L\sin\theta\).
\[\text{Period } T = 2\pi\sqrt{\frac{L\cos\theta}{g}} \quad \text{--- (1.7)}\]
\[n = \frac{1}{T} = \frac{1}{2\pi}\sqrt{\frac{g}{L\cos\theta}} \quad \text{--- (1.8)}\]
The string can never become horizontal, since nothing would then balance \(mg\).
Vertical Circular Motion
Uppermost point A: \(mg + T_A = \dfrac{mv_A^2}{r}\) (1.9); for minimum speed \(T_A = 0\), so \(\left(v_A\right)_{min} = \sqrt{rg}\) (1.10).
Lowermost point B: \(T_B - mg = \dfrac{mv_B^2}{r}\) (1.11); energy conservation over a drop \(2r\) gives \(v_B^2 = v_A^2 + 4rg\) (1.12), hence \(\left(v_B\right)_{min} = \sqrt{5rg}\) (1.13).
\(T_B - T_A = 2mg + \dfrac{m}{r}\left(v_B^2 - v_A^2\right)\) (1.14), which reduces to \(T_B - T_A = 6mg\) (1.15) — independent of \(v\) and \(r\). Since \(T_A\) can never be exactly zero for a string, \(T_B > 6mg\).
String horizontal (C and D): \(T_C - T_A = T_D - T_A = 3mg\) and \(\left(v_C\right)_{min} = \left(v_D\right)_{min} = \sqrt{3rg}\).
Mass tied to a rod: a rigid rod can push as well as pull, so practically zero speed is possible at the top. Then \(\left(v_{lowermost}\right)_{min} = \sqrt{4rg} = 2\sqrt{rg}\), \(v_{min}\) at the rod-horizontal position \(= \sqrt{2rg}\), and \(T_{lowermost} - T_{uppermost} = 6mg\).
Sphere of death: same as the string case, with the tension \(T\) replaced by the normal reaction \(N\).
Convex over-bridge: \(mg - N = \dfrac{mv^2}{r}\). Contact is just maintained when \(N = 0\), giving the upper limit \(v_{max} = \sqrt{rg}\).
Moment of Inertia and Radius of Gyration
Rotational K.E. \(= \dfrac{1}{2}I\omega^2\), analogous to \(\dfrac{1}{2}mv^2\).
Moment of inertia: \(I = \sum m_ir_i^2\), or \(I = \int r^2\,dm\) for a continuous body. SI unit: kg m\(^2\). It depends on the individual masses and on their distribution about the chosen axis.
Uniform ring about its own axis: \(I = MR^2\). Uniform disc about its own axis: \(I = \dfrac{1}{2}MR^2\), obtained with surface density \(\sigma = \dfrac{M}{\pi R^2}\) and \(dm = 2\pi\sigma r\,dr\).
Radius of gyration: \(I = MK^2\); \(K\) is the effective distance of the whole mass from the axis. For a ring \(K_r = R\); for a disc \(K_d = \dfrac{R}{\sqrt{2}}\), so \(K_d < K_r\). Among objects of the same mass and radius, a ring has the largest \(K\) and hence the maximum M.I.
Theorems of Parallel and Perpendicular Axes
Theorem of parallel axes: \(I_O = I_C + Mh^2\), where \(I_C\) is about a parallel axis through the centre of mass and \(h\) is the distance between the two axes. One of the axes must pass through the centre of mass.
Theorem of perpendicular axes: \(I_z = I_x + I_y\), valid only for a laminar object, with the three mutually perpendicular axes concurrent and \(x\), \(y\) lying in the plane of the object.
Angular Momentum, Torque and Conservation
Angular momentum: \(\vec{L} = \vec{r} \times \vec{p}\); in magnitude \(L = p\,r\sin\theta\). SI unit: kg m\(^2\) s\(^{-1}\).
For a rigid body rotating about a fixed axis, \(L = I\omega\) — analogous to \(p = mv\).
Torque: \(\tau = I\alpha\) — analogous to \(f = ma\). SI unit: N m.
\(\dfrac{d\vec{L}}{dt} = \vec{\tau}\). Hence if \(\vec{\tau} = 0\), then $ = $ constant.
Principle of conservation of angular momentum: in the absence of an external unbalanced torque, \(L = I\omega = I\left(2\pi n\right)\) stays constant. If \(I\) increases, \(n\) decreases, and vice versa. Seen in ice ballet, diving, circus acrobatics.
Rolling Motion
Pure rolling = translation of the centre of mass + rotation about the central axis, with \(\omega = \dfrac{v}{R}\). Static friction is essential for pure rolling.
Total kinetic energy of rolling:
\[E = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}Mv^2\left(1 + \frac{K^2}{R^2}\right) \quad \text{--- (1.18)}\]
- Rolling down an incline through a vertical height \(h\):
\[v = \sqrt{\frac{2gh}{\left(1 + \dfrac{K^2}{R^2}\right)}} \quad \text{--- (1.19)}\]
\[a = \frac{g\sin\theta}{\left(1 + \dfrac{K^2}{R^2}\right)} \quad \text{--- (1.20)}\]
For pure sliding without friction these become \(\sqrt{2gh}\) and \(g\sin\theta\).
If \(I = n\left(MR^2\right)\), then \(\dfrac{K^2}{R^2} = n\). Thus \(\dfrac{K^2}{R^2} = 1\) for a ring or hollow cylinder, \(\dfrac{1}{2}\) for a disc or solid cylinder, \(\dfrac{2}{5}\) for a solid sphere and \(\dfrac{2}{3}\) for a thin-walled hollow sphere.
Translational K.E. : Rotational K.E. : Total K.E. \(= 1 : \dfrac{K^2}{R^2} : \left(1 + \dfrac{K^2}{R^2}\right)\). For a hollow sphere this is \(3 : 2 : 5\), i.e. 60% translational and 40% rotational.
Table 1.1: Analogous Kinematical Equations
(\(\omega_0\) is the initial angular velocity)
| Equation for translational motion | Analogous equation for rotational motion |
|---|---|
| \(v_{av} = \dfrac{u + v}{2}\) | \(\omega_{av} = \dfrac{\omega_0 + \omega}{2}\) |
| \(a = \dfrac{dv}{dt} = \dfrac{v - u}{t}\ \therefore v = u + at\) | \(\alpha = \dfrac{d\omega}{dt} = \dfrac{\omega - \omega_0}{t}\ \therefore \omega = \omega_0 + \alpha t\) |
| \(s = v_{av}\cdot t = \left(\dfrac{u + v}{2}\right)t = ut + \dfrac{1}{2}at^2\) | \(\theta = \omega_{av}\cdot t = \left(\dfrac{\omega_0 + \omega}{2}\right)t = \omega_0t + \dfrac{1}{2}\alpha t^2\) |
| \(v^2 = u^2 + 2as\) | \(\omega^2 = \omega_0^2 + 2\alpha\theta\) |
These apply only when \(\vec{\alpha}\) is constant and along the axis of rotation.
Table 2: Analogous Quantities
| Translational motion | Rotational motion | Inter-relation |
|---|---|---|
| Linear displacement \(\vec{s}\) | Angular displacement \(\vec{\theta}\) | \(\vec{s} = \vec{\theta} \times \vec{r}\) |
| Linear velocity \(\vec{v} = \dfrac{d\vec{s}}{dt}\) | Angular velocity \(\vec{\omega} = \dfrac{d\vec{\theta}}{dt}\) | \(\vec{v} = \vec{\omega} \times \vec{r}\) |
| Linear acceleration \(\vec{a} = \dfrac{d\vec{v}}{dt}\) | Angular acceleration \(\vec{\alpha} = \dfrac{d\vec{\omega}}{dt}\) | \(\vec{a} = \vec{\alpha} \times \vec{r}\) |
| Inertia or mass \(m\) | Moment of inertia \(I = \int r^2\,dm = \sum m_ir_i^2\) | — |
| Linear momentum \(\vec{p} = m\vec{v}\) | Angular momentum \(\vec{L} = I\vec{\omega}\) | \(\vec{L} = \vec{r} \times \vec{p}\) |
| Force \(\vec{f} = \dfrac{d\vec{p}}{dt}\) | Torque \(\vec{\tau} = \dfrac{d\vec{L}}{dt}\) | \(\vec{\tau} = \vec{r} \times \vec{f}\) |
| Work \(W = \vec{f}\cdot\vec{s}\) | Work \(W = \vec{\tau}\cdot\vec{\theta}\) | — |
| Power \(P = \dfrac{dW}{dt} = \vec{f}\cdot\vec{v}\) | Power \(P = \dfrac{dW}{dt} = \vec{\tau}\cdot\vec{\omega}\) | — |
Table 3: Moments of Inertia of Some Symmetric Objects
| Object | Axis | Moment of inertia |
|---|---|---|
| Thin ring or hollow cylinder | Central | \(I = MR^2\) |
| Thin ring | Diameter | \(I = \dfrac{1}{2}MR^2\) |
| Annular ring or thick walled hollow cylinder | Central | \(I = \dfrac{1}{2}M\left(r_1^2 + r_2^2\right)\) |
| Uniform disc or solid cylinder | Central | \(I = \dfrac{1}{2}MR^2\) |
| Uniform disc | Diameter | \(I = \dfrac{1}{4}MR^2\) |
| Thin walled hollow sphere | Central | \(I = \dfrac{2}{3}MR^2\) |
| Solid sphere | Central | \(I = \dfrac{2}{5}MR^2\) |
| Uniform symmetric spherical shell | Central | \(I = \dfrac{2}{5}M\left(\dfrac{r_2^5 - r_1^5}{r_2^3 - r_1^3}\right)\) |
| Thin uniform rod or rectangular plate | Perpendicular to length and passing through centre | \(I = \dfrac{1}{12}ML^2\) |
| Thin uniform rod or rectangular plate | Perpendicular to length and about one end | \(I = \dfrac{1}{3}ML^2\) |
| Uniform plate or rectangular parallelepiped | Central | \(I = \dfrac{1}{12}M\left(L^2 + b^2\right)\) |
| Uniform solid right circular cone | Central | \(I = \dfrac{3}{10}MR^2\) |
| Uniform hollow right circular cone | Central | \(I = \dfrac{1}{2}MR^2\) |
1.13 Objective Questions
- Choose the correct option.
- When seen from below, the blades of a ceiling fan are seen to be revolving anticlockwise and their speed is decreasing. Select the correct statement about the directions of its angular velocity and angular acceleration.
- Angular velocity upwards, angular acceleration downwards.
- Angular velocity downwards, angular acceleration upwards.
- Both, angular velocity and angular acceleration, upwards.
- Both, angular velocity and angular acceleration, downwards.
- A particle of mass 1 kg, tied to a 1.2 m long string is whirled to perform vertical circular motion, under gravity. Minimum speed of the particle is 5 m/s. Consider the following statements.
- Maximum speed must be \(5\sqrt{5}\) m/s.
- Difference between maximum and minimum tensions along the string is 60 N. Select the correct option.
- Only the statement P is correct.
- Only the statement Q is correct.
- Both the statements are correct.
- Both the statements are incorrect.
- Select the correct statement about the formula (expression) of moment of inertia (M.I.) in terms of mass \(M\) of the object and some of its distance parameter/s, such as \(R\), \(L\), etc.
- Different objects must have different expressions for their M.I.
- When rotating about their central axis, a hollow right circular cone and a disc have the same expression for the M.I.
- Expression for the M.I. for a parallelepiped rotating about the transverse axis passing through its centre includes its depth.
- Expression for M.I. of a rod and that of a plane sheet is the same about a transverse axis.
- In a certain unit, the radius of gyration of a uniform disc about its central and transverse axis is \(\sqrt{2.5}\). Its radius of gyration about a tangent in its plane (in the same unit) must be
- \(\sqrt{5}\) (B) 2.5 (C) \(2\sqrt{2.5}\) (D) \(\sqrt{12.5}\)
- Consider the following cases:
- A planet revolving in an elliptical orbit.
- A planet revolving in a circular orbit. Principle of conservation of angular momentum comes in force in which of these?
- Only for (P)
- Only for (Q)
- For both, (P) and (Q)
- Neither for (P), nor for (Q)
- A thin walled hollow cylinder is rolling down an incline, without slipping. At any instant, the ratio “Rotational K.E. : Translational K.E. : Total K.E.” is
- 1:1:2 (B) 1:2:3 (C) 1:1:1 (D) 2:1:3
1.14 Check Your Concepts
A body moves along a circle with a perfectly constant speed. Explain why it is still said to be accelerating, and identify the direction of that acceleration.
Why is “centripetal” described as an adjective rather than the name of a type of force? Name the real force that acts as the centripetal force in each of the following: a stone whirled on a string, a car taking a turn on a level road, and a rider inside a well of death.
A passenger standing in a turning bus insists that an outward force pushed her sideways, while an observer standing on the road insists no such force existed. Explain how both descriptions can be correct, referring to the frames of reference used.
The maximum speed for turning on an unbanked road does not depend on the mass of the vehicle, yet a heavily loaded truck is more dangerous on a curve than a small car. Explain this apparent contradiction.
Why does a flat circular road impose an upper speed limit, while the vertical wall of a well of death imposes a lower speed limit? What does this suggest about a road banked at some intermediate angle?
While deriving the expression for the banking angle, friction is deliberately ignored although real roads are rough. Explain why this is done and what advantage the resulting design offers.
What happens to the lower speed limit on a banked road when \(\mu_s \geq \tan\theta\)? Interpret this result physically.
Give two independent arguments — one based on forces and one based on the expressions for period and tension — to show that the string of a conical pendulum can never become horizontal.
Two conical pendulums have bobs of very different masses but identical string lengths and identical semi-vertical angles. Compare their periods and explain your reasoning.
Why is a minimum speed necessary at the uppermost point when a bob is whirled in a vertical circle using a string, but not when a rigid rod is used instead? What property of the string is responsible?
The difference between the maximum and minimum tensions during vertical circular motion is \(6mg\), whatever the speed or the radius. Explain how this speed-independent result arises.
A car crosses the crest of a convex bridge at increasing speeds. Describe what happens to the normal reaction, to the passengers’ sensation of weight, and to the steering control of the vehicle.
Mass is a fixed property of a body, but its moment of inertia is not. Explain why, and state what additional information must always accompany a value of moment of inertia.
A ring and a disc have equal masses and equal radii. Equal torques are applied to both about their central axes. Compare their angular accelerations and explain the difference in terms of mass distribution.
Why must one of the two axes pass through the centre of mass while applying the theorem of parallel axes? Point out the step in the derivation that fails if this condition is not met.
Why can the theorem of perpendicular axes be applied to a disc but not to a solid sphere? Which step of the derivation restricts it to laminar objects?
Among all objects of the same mass and same radius, a ring has the largest radius of gyration. Justify this statement using the meaning of radius of gyration.
A diver folds her body immediately after leaving the diving board and stretches it again before entering the water. Discuss the changes in her moment of inertia, angular speed and angular momentum during this sequence, and state which of the three remains constant.
Why is static friction, and not kinetic friction, essential for pure rolling? What is the instantaneous velocity of the point of contact during pure rolling, and why?
Two solid spheres of very different masses and radii are released together from the top of the same incline and roll down without slipping. Predict which one reaches the bottom first, and justify your prediction using the expression for acceleration.
An object rolling down an incline always reaches the bottom later than an identical object sliding down the same incline without friction. Explain where the “missing” speed has gone.
A hollow sphere and a solid sphere of equal mass and radius roll with the same speed of the centre of mass. Compare the fractions of their total kinetic energy that are rotational, and explain the difference.
Answer in brief.
- Why are curved roads banked?
- Do we need a banked road for a two-wheeler? Explain.
- On what factors does the frequency of a conical pendulum depend? Is it independent of some factors?
- Why is it useful to define radius of gyration?
- A uniform disc and a hollow right circular cone have the same formula for their M.I., when rotating about their central axes. Why is it so?
While driving along an unbanked circular road, a two-wheeler rider has to lean with the vertical. Why is it so? With what angle does the rider have to lean? Derive the relevant expression. Why is such leaning not necessary for a four wheeler?
Using energy conservation, derive the expressions for the minimum speeds at different locations along a vertical circular motion controlled by gravity. Is zero speed possible at the uppermost point? Under what condition/s? Also prove that the difference between the extreme tensions (or normal forces) depends only upon the weight of the object.
Discuss the necessity of radius of gyration. Define it. On what factors does it depend and on what does it not depend? Can you locate some similarity between the centre of mass and radius of gyration? What can you infer if a uniform ring and a uniform disc have the same radius of gyration?
State the conditions under which the theorems of parallel axes and perpendicular axes are applicable. State the respective mathematical expressions.
Derive an expression that relates angular momentum with the angular velocity of a rigid body.
Obtain an expression relating the torque with angular acceleration for a rigid body.
State and explain the principle of conservation of angular momentum. Use a suitable illustration. Do we use it in our daily life? When?
Discuss the interlink between translational, rotational and total kinetic energies of a rigid object that rolls without slipping.
A rigid object is rolling down an inclined plane. Derive expressions for the acceleration along the track and the speed after falling through a certain vertical distance.
1.15 Practice with Numericals
Use \(g = 10\) m/s\(^2\), unless otherwise stated.
A particle performs uniform circular motion along a circle of radius 0.5 m at 120 rpm. Calculate its angular speed, its linear speed, its period and the magnitude of its centripetal acceleration.
A wheel rotating at 300 rpm is brought to rest in 60 revolutions by a constant retarding torque. Calculate its angular retardation and the time taken to come to rest.
A car of mass 1200 kg negotiates a level circular turn of radius 25 m at a steady speed of 10 m s\(^{-1}\). Calculate the centripetal force required and the minimum coefficient of static friction needed between the tyres and the road.
A circular track of radius 90 m is unbanked and the coefficient of static friction between the tyres and the road is 0.4. Calculate the maximum speed at which a vehicle can turn without skidding. Express your answer in km h\(^{-1}\) as well.
A stunt rider performs horizontal circles inside a well of death of inner radius 5 m. The coefficient of static friction between the tyres and the wall is 0.5. Calculate the minimum speed and the minimum frequency of revolution needed to perform the stunt.
A curved road of radius 60 m is to be banked so that a vehicle moving at 20 m s\(^{-1}\) can negotiate the turn without any help from friction. Calculate the angle of banking. If the road is 12 m wide, find the height of its outer edge above the inner edge.
A road of radius 100 m is banked at an angle \(\tan^{-1}(0.4)\). The coefficient of static friction between the tyres and the road is 0.3. Calculate the minimum and maximum speeds with which a vehicle can safely negotiate this turn.
A conical pendulum has a string of length 0.5 m carrying a bob of mass 0.2 kg. The string makes an angle of 30° with the vertical during revolution. Calculate the radius of the circular path, the period of revolution, the frequency and the tension in the string.
A stone of mass 0.4 kg is tied to a string of length 1 m and whirled in a vertical circle with the minimum possible speed. Calculate its speed at the uppermost point, its speed at the lowermost point, and the tension in the string at the lowermost point.
A motorcyclist performs vertical circles inside a sphere of death of radius 5 m. Calculate the minimum possible speed at the uppermost point and the corresponding minimum speed at the lowermost point.
A car of mass 800 kg crosses the top of a convex over-bridge whose radius of curvature at the crest is 50 m. Calculate the normal reaction on the car when its speed is 10 m s\(^{-1}\), and the maximum speed at which it can cross the crest without losing contact with the road.
Four point masses of 1 kg, 2 kg, 3 kg and 4 kg are placed at the four corners of a square of side 1 m. Calculate the moment of inertia of this system about an axis passing through the centre of the square and perpendicular to its plane.
A solid sphere of mass 10 kg and radius 0.3 m rotates about its central axis. Calculate its moment of inertia about this axis, and about a tangent to the sphere, using the theorem of parallel axes.
A thin uniform ring of mass 2 kg and radius 0.25 m lies in a horizontal plane. Calculate its moment of inertia about (i) its own central axis, (ii) one of its diameters, and (iii) a tangent lying in its plane.
A wheel of moment of inertia 4 kg m\(^2\) starts from rest and attains an angular speed of 20 rad s\(^{-1}\) in 5 seconds under a constant torque. Calculate the angular acceleration, the torque applied, the angular momentum at the end of 5 seconds and the power delivered at that instant.
A turntable of moment of inertia 1.2 kg m\(^2\) is rotating freely at 3 revolutions per second. A lump of clay of mass 0.5 kg is gently dropped on it and sticks at a distance of 0.4 m from the axis. Calculate the new frequency of rotation.
A uniform disc rolls down an incline of inclination 30° without slipping, starting from rest. Calculate its linear acceleration along the incline and the speed of its centre of mass after it has descended through a vertical height of 2 m.
A thin walled hollow sphere of mass 3 kg rolls without slipping with the speed of its centre of mass equal to 4 m s\(^{-1}\). Calculate its translational kinetic energy, rotational kinetic energy and total kinetic energy, and verify the ratio 3 : 2 : 5.
A ceiling fan of moment of inertia 1.5 kg m\(^2\) reaches a frequency of 2 revolutions per second in 4 seconds from rest, with uniform angular acceleration. Calculate the angular acceleration, the torque supplied by the motor and the number of revolutions completed in these 4 seconds.
A solid cylinder and a hollow cylinder of the same mass and same radius are released together from the top of an incline of height 1.8 m and allowed to roll down without slipping. Calculate the speed of each at the bottom of the incline, and find the ratio of these speeds.
Somehow, an ant is stuck to the rim of a bicycle wheel of diameter 1 m. While the bicycle is on a central stand, the wheel is set into rotation and it attains the frequency of 2 rev/s in 10 seconds, with uniform angular acceleration. Calculate (i) Number of revolutions completed by the ant in these 10 seconds. (ii) Time taken by it for the first complete revolution and the last complete revolution.
Coefficient of static friction between a coin and a gramophone disc is 0.5. Radius of the disc is 8 cm. Initially the centre of the coin is 2 cm away from the centre of the disc. At what minimum frequency will it start slipping from there? By what factor will the answer change if the coin is almost at the rim? (Use \(g = \pi^2\) m/s\(^2\))
Part of a racing track is to be designed for a radius of curvature of 72 m. We are not recommending the vehicles to drive faster than 216 kmph. With what angle should the road be tilted? At what height will its outer edge be, with respect to the inner edge, if the track is 10 m wide?
The road in question 23 above is constructed as per the requirements. The coefficient of static friction between the tyres of a vehicle and this road is 0.8. Will there be any lower speed limit? By how much can the upper speed limit exceed in this case?
During a stunt, a cyclist (considered to be a particle) is undertaking horizontal circles inside a cylindrical well of radius 6.05 m. If the necessary friction coefficient is 0.5, how much minimum speed should the stunt artist maintain? Mass of the artist is 50 kg. If she/he increases the speed by 20%, how much will the force of friction be?
A pendulum consisting of a massless string of length 20 cm and a tiny bob of mass 100 g is set up as a conical pendulum. Its bob now performs 75 rpm. Calculate the kinetic energy and the increase in the gravitational potential energy of the bob. (Use \(\pi^2 = 10\))
A motorcyclist (as a particle) is undergoing vertical circles inside a sphere of death. The speed of the motorcycle varies between 6 m/s and 10 m/s. Calculate the diameter of the sphere of death. What are the minimum values possible for these two speeds?
A metallic ring of mass 1 kg has moment of inertia 1 kg m\(^2\) when rotating about one of its diameters. It is molten and remoulded into a thin uniform disc of the same radius. How much will its moment of inertia be, when rotated about its own axis?
A big dumb-bell is prepared by using a uniform rod of mass 60 g and length 20 cm. Two identical solid thermocol spheres of mass 25 g and radius 10 cm each are at the two ends of the rod. Calculate the moment of inertia of the dumb-bell when rotated about an axis passing through its centre and perpendicular to the length.
A flywheel used to prepare earthenware pots is set into rotation at 100 rpm. It is in the form of a disc of mass 10 kg and radius 0.4 m. A lump of clay (to be taken equivalent to a particle) of mass 1.6 kg falls on it and adheres to it at a certain distance \(x\) from the centre. Calculate \(x\) if the wheel now rotates at 80 rpm.
Starting from rest, an object rolls down along an incline that rises by 3 units in every 5 units (along it). The object gains a speed of 10 m/s as it travels a distance of \(\dfrac{5}{3}\) m along the incline. What can be the possible shape/s of the object?