Contents
- 1 What is Kinetic Energy Rotation? Is Angular Momentum Conserved When Torque is Zero?
- 1.1 Rotational Kinetic Energy
- 1.2 Angular Momentum
- 1.3 Unit and dimension of angular momentum:
- 1.4 Relation between Linear Momentum and Angular Momentum
- 1.5 Relation between Angular Momentum and Torque
- 1.6 Law of Conservation of Angular Momentum
- 1.7 Related experiments and practical examples:
- 1.8 Numerical Examples
Physics Topics cover a broad range of concepts that are essential to understanding the natural world.
What is Kinetic Energy Rotation? Is Angular Momentum Conserved When Torque is Zero?
Rotational Kinetic Energy
Let PQR be a rigid body, revolving about the axis AB with angular velocity ω [Fig.], Due to this motion, the body possesses some kinetic energy. This kinetic energy is called its rotational kinetic energy.
The rigid body can be assumed as an aggregate of a number of particles. Let the masses of the particles be m1, m2, m3,……., etc., at distances r1, r2, r3,………, etc., respectively from the axis of rotation AB. Since the body is rigid, the angular velocity of the constituent particles is the same, i.e., ω. But due to the difference in their distances from the axis of rota-tion, linear velocities of different particles are different.
Let the linear velocity of the particle of mass m1 be v1.
So, v1 = ωr1
So, the kinetic energy of the particle of mass m1
= \(\frac{1}{2} m_1 v_1^2\) = \(\frac{1}{2} m_1 \omega^2 r_1^2\)
Similarly, the kinetic energy of the particle of mass m2
= \(\frac{1}{2} m_2 v_2^2\) = \(\frac{1}{2} m_2 \omega^2 r_2^2\), and so on.
In this way, adding the kinetic energies of all particles, the kinetic energy of the entire rigid body is obtained.
So, the rotational kinetic energy of the body
[ri = the perpendicular distance of i -th particle from the axis of rotation, mi = mass of the i-th particle of the rigid body]
= \(\frac{1}{2} I \omega^2\)
Here, I = \(\sum_i m_i r_i^2\) moment of inertia of the body about the axis of rotation.
Comparing torque with force, it is seen that the moment of inertia in rotational motion plays the same role as that played by the mass in translational motion. Now, it is also seen that the same conclusion can be drawn by comparing the translational kinetic energy \(\left(\frac{1}{2} m v^2\right)\) with the rotational kinetic energy \(\left(\frac{1}{2} I \omega^2\right)\).
Relation between rotational kinetic energy with work done; Let a force F on a body of mass m causes a change in its kinetic energy (Δk) only. If the body under this force is displaced linearly from initial position x1 to final position x2, then work done by the force,
W = \(\int_{x_1}^{x_2} F d x\)
According to work-kinetic energy theorem,
Δk = W
i.e., \(\frac{1}{2} m v_2^2\) – \(\frac{1}{2} m v_1^2\) = \(\int_{x_1}^{x_2} F d x\) ….. (1)
[where v1 = initial speed, v2 = final speed]
Now consider a torque r on a body of moment of inertia \(\tau\) (about a certain axis of rotation) causes a change in its rota-tional kinetic energy (Δkr) only. If the body under this torque is displaced from initial angular position θ1 to final angular position θ2 by the torque,
W = \(\int_{\theta_1}^{\theta_2} \tau d \theta\)
Just like above equation (1) now we can relate work and rotational kinetic energy as below:
Δkr = W
i.e., \(\frac{1}{2} I \omega_1^2\) – \(\frac{1}{2} I \omega_2^2\) = \(\int_{\theta_1}^{\theta_2} \tau d \theta\) ……… (2)
[where ω1 = initial angular speed, ω2 = final angular speed] when \(\tau\) is constant, W = \(\tau\)(θ2 – θ1)
Therefore, power, i.e., the rate at which the work is done,
P = \(\frac{d W}{d t}\) = \(\tau \omega\)
Angular Momentum
The rotational analogues of the mass (m) of a body and its linear velocity (v) are moment of inertia (I) and angular velocity (ω), respectively. Hence, the rotational analogue of the linear momentum (mv) of the body is Iω. This physical quantity is called the angular momentum (L) of the body.
Definition: The dynamical property generated in a body under rotational motion, due to moment of inertia about an axis and angular velocity, is called the angular momentum of the body about that axis.
Angular momentum is measured by the product of moment of inertia and angular velocity, i.e., L = Iω.
Since I is a scalar and ω is an axial vector, angular momen-tum L is also an axial vector whose direction is along the axis of rotation, and in the direction of ω.
Unit and dimension of angular momentum:
Dimension of L = dimension of I × dimension of ω
= ML2 × T-1 = ML2T-1
Relation between Linear Momentum and Angular Momentum
Suppose a body is revolving with an angular velocity ω about an axis. If m1, m2, m3, …. are the constituent parti-cles of that body and they are at distances r1, r2, r3,…… respectively from the axis of rotation, then the moment of inertia of the body,
I = \(m_1 r_1^2\) + \(m_2 r_2^2\) + \(m_3 r_3^2\) + ….. = \(\sum_i m_i r_i^2\)
In case of pure rotation, the angular velocity of each particle becomes equal to the angular velocity of the body.
So, angular momentum of the body,
[pi = mivi = linear momentum of i -th particle]
For the particles, the quantities r1 × m1v1, r2 × m2v2,………., etc., can be called the moments of linear momentum, or in brief, moments of momentum (in analogy with the moment of force).
So, the angular momentum of a body about an axis is the algebraic sum of the moments of linear momentum about the same axis, of all particles constituting the body.
Thus, for a particle rotating about a circle of radius r and having a linear momentum p, the angular momentum will be L = rp.
Vector representation: The vector representation for the relation between linear and angular momentum is \(\vec{L}\) = \(\vec{r} \times \vec{p}\). This is often referred to as the defining equation of i.
We know the vector representation for the relation between linear velocity and angular velocity is \(\vec{v}\) = \(\vec{\omega} \times \vec{r}\).
In Fig.(b) [Chapter Circular Motion] if \(\vec{v}\) and \(\vec{\omega}\) are replaced by \(\vec{p}\) and \(\vec{L}\), respectively, the geometric form for the relation of \(\vec{L}\), \(\vec{p}\) and \(\vec{r}\) is obtained.
Relation between Angular Momentum and Torque
In case of rotational motion, when a torque is applied on a body, an angular acceleration is produced in it. If the initial angular velocity of the body is ω1 and its angular velocity after time t is ω2, then angular acceleration of the body,
α = \(\frac{\omega_2-\omega_1}{t}\)
Again, torque = moment of inertia × angular acceleration
or, \(\tau\) = Iα = I × \(\frac{\omega_2-\omega_1}{t}\) = \(\frac{I \omega_2-I \omega_1}{t}\)
or, \(\tau t\) = Iω2 – Iω1
Hence, torque × time = change in angular momentum of the body during that interval
This is the relation between torque and angular acceleration. From this relation, it is evident that a change in angular momentum takes place about the axis along which the torque acts on the body.
We know that in case of translational motion, Ft = mv – mu and the rotational analogue of this equation is \(\tau t\) = Iω2 – Iω1. The quantity Ft is known as the impulse of force. Similarly, the quantity \(\tau t\) is known as the angular impulse or the impulse of torque.
Law of Conservation of Angular Momentum
Suppose the moment of inertia of a body changes from I1 to I2 in time t. In this case, the equation \(\tau t\) = Iω2 – Iω1 changes to \(\tau t\) = I2ω2 – I1ω1. Now, if no external torque acts on the body, i.e., if \(\tau\) = 0, then from the equation, \(\tau\)t = I2ω2 – I1ω1 we get,
I2ω2 – I1ω1 = 0 or, I2ω2 = I1ω1
It means that the final angular momentum of the body is equal to its initial angular momentum, i.e., the angular momentum is conserved.
Law: If the net external torque on a body is zero, the angular momentum of the body rotating about an axis always remains conserved.
So, this law is nothing but the rotational analogue of the law of conservation of linear momentum.
Again we know, \(\frac{d L}{d t}\) = \(\tau_{\text {ext }}\)
From this it is clear that, if total external torque acts on a body is zero; its angular velocity decreases with the increase of its moment of inertia and vice versa i.e., angular momen-tum remains constant.
Related experiments and practical examples:
i) A man is sitting on a turntable holding a pair of dumb-bells of equal mass, one in each hand with his arms out-stretched while the turntable rotates with a definite angular velocity [Fig.].
If the man suddenly draws the dumbbells towards his chest, the speed of rotation of the turntable is found to increase. This is due to the fact that when the man draws the dumb-bells towards his chest, the moment of inertia of the man about the axis of rotation decreases and his angular velocity increases due to conservation of angular momentum. If the man again stretches his arms, his angular velocity decreases due to increase in moment of inertia, and the turntable consequently rotates slowly.
ii) In a diving event, when a competitor dives from a high platform or spring board into water, he keeps his legs and arms outstretched and starts descending with less angular velocity [Fig.].
After that when he curls his body by rolling the legs and arms inwards, his moment of inertia decreases. As angular momentum is conserved, his angular velocity goes on increasing rapidly. As a result, his body begins to spin rapidly and before reaching the surface of water, he can perform a good number of somersaults.
In case of skating on the surface of ice or during the performance of acrobatics, the principle of conservation of angular momentum can be applied in a similar way.
Numerical Examples
Example 1.
If the radius of the earth decreases by \(\frac{1}{2}\)%, then what will be the change in the length of a day? Assume that the earth is a uniform sphere and its moment of inertia, I = \(\frac{2}{5}\)MR2, where M and R are the mass and the radius of the earth.
Solution:
If the mass of a solid sphere remains unaltered, then its moment of inertia ∝ (radius)2. Here, the changed radius = \(\frac{100-\frac{1}{2}}{100} R\) = \(\frac{199}{200} R\). So, if the moment of inertia of the earth for its present radius R is I and the moment of inertia for its changed radius is I’, then
\(\frac{I}{I^{\prime}}\) = \(\frac{R^2}{\left(\frac{199 R}{200}\right)^2}\) = \(\left(\frac{200}{199}\right)^2\)
If the present angular velocity of the earth is ω and its changed angular velocity is ω’, then according to the prin-ciple of conservation of angular momentum,
∴ The length of the day will decrease by (24 – 23.76) = 0.24 h = 14 min 24 s .
Example 2.
A solid sphere of mass 1 kg and of radius 10 cm is rotating about one of its diameter with an angular velocity of π rad ᐧ s-1. Calculate the kinetic energy of the sphere by using the relevant formula. [HS 08]
Solution:
Let the moment of inertia of the sphere about its diameter I = \(\frac{2}{5}\)MR2, M = mass of the sphere and R = radius of the sphere.
Kinetic energy of the body
= rotational kinetic energy of the body
= \(\frac{1}{2} I \omega^2\) = \(\frac{1}{2}\) × \(\frac{2}{5} M R^2 \cdot \omega^2\)
= \(\frac{1}{5}\) × 1000 × (10)2 × π2
= 197392.09 erg.
Example 3.
A thin rod of length l and mass m per unit length is rotating about an axis passing through the mid-point of its length and perpendicular to it. Prove that its kinetic energy = \(\frac{1}{24} m \omega^2 l^3\); ω = angular velocity of the rod.
Solution:
Kinetic energy of the rod = \(\frac{1}{2} I \omega^2\)
According to the problem,
I = \(\frac{1}{12} M l^2\) [M = mass of the rod = ml]
= \(\frac{1}{12}\) × ml × l2 = \(\frac{n l^3}{12}\)
∴ Kinetic energy of the rod
= \(\frac{1}{2}\) × \(\frac{m l^3}{12}\) × ω2 = \(\frac{1}{24} m \omega^2 l^3\)
Example 4.
Calculate the moment of inertia of a solid cylinder of length 10 cm and of radius 20 cm about its own axis. Density of the material of the cylinder = 9 g ᐧ cm-3.
Solution:
L = length of the cylinder,
R = radius of the cylinder and
M = mass of the cylinder
= volume of the cylinder × density
= πR2L × density
= π × (20)2 × 10 × 9 g
Moment of inertia of a solid cylinder about its own axis,
I = \(\frac{1}{2}\)MR2
∴ I = \(\frac{1}{2}\) × π × (20)2 × 10 × 9 × (20)2
= 22.6 × 106 g ᐧ cm2.
Example 5.
A solid sphere of diameter 2 cm and of mass 20 g is rolling with a velocity of 3 cm ᐧ s-1. What is the total kinetic energy of the sphere?
Solution:
Let M = mass of the sphere, R = radius of the sphere, V = linear velocity of the sphere, I = \(\frac{2}{5}\)MR2 (moment of inertia of the sphere about its diameter), ω = \(\frac{V}{R}\)
Total kinetic energy of the sphere
= translational kinetic energy + rotational kinetic energy
Example 6.
A stone of mass m tied with a thread is rotating along a horizontal circular path (force of gravity is neglected). The length of the thread is decreasing gradually in such a manner that the angular momentum of the stone remains constant with respect to the centre of the circle. If the tension in the thread is T = Arn, where A = constant, r = instantaneous radius of the circle, then find the value of n.
Solution:
If the instantaneous angular velocity of the stone is ω, then angular momentum,
L = Iω = mr2ω = constant (according to the problem)
or, ω = \(\frac{L}{m r^2}\)
Here, the tension in the thread provides the necessary centripetal force for rotation.
So, T = Arn = mω2r = m ᐧ \(\frac{L^2}{m^2 r^4}\)r = \(\frac{L^2}{m} r^{-3}\)
= Ar-3 (A = \(\frac{L^2}{m}\) = constant)
∴ n = -3
Example 7.
Two ends of a uniform rod weighing W, are placed on supports so that the rod remains horizontal. If a support at one end is suddenly removed, what will be the force exerted on the horizontal rod by the support at the other end? [WBIEE 2000]
Solution:
Let the length of the rod = l cm, its weight = W = Mg [Fig.], where M is the mass of the rod. When the support at one end is removed suddenly, centre of gravity of the rod falls downwards with an acceleration a.
Let R = reaction force at the end with the support. Hence, if the C.G. now falls with an acceleration a, the rod will turn about the point P.
The torque on the rod = \(\frac{M g-R}{M}\)
Also, Mg – R = Ma or, a = \(\frac{M g-R}{M}\)
Here moment of inertia, I = \(\frac{1}{3}\)Ml2 = moment of inertia of the rod about the perpendicular axis passing through the end of the rod and the angular acceleration, α = \(\frac{a}{l / 2}\) = \(\frac{2 a}{l}\)
Therefore, when one support is removed, the support at the other end will exert a reaction force of \(\frac{W}{4}\).
Example 8.
A rod of length L and mass M is attached with a hinge on a wall at a point O. After releasing the rod from its
vertical position OA, when it comes to the position OA’, then what is the reaction on the point O of the rod by the hinge?
Solution:
Let, the angular velocity of the rod at the horizon-tal position OA’ is ω.
∴ At that instant its kinetic energy
= \(\frac{1}{2} I \omega^2\) = \(\frac{1}{2} \cdot \frac{M L^2}{3} \cdot \omega^2\) = \(\frac{M L^2 \omega^2}{6}\)
The centre of mass of the rod shifts down by \(\frac{L}{2}\) from OA to OA’.
So, decrease in potential energy of the rod = Mg\(\frac{L}{2}\)
According to the kinetic energy conservation law,
\(M g \frac{L}{2}\) = \(\frac{M L^2 \omega^2}{6}\) or, ω = \(\sqrt{\frac{3 g}{L}}\) …….. (1)
Two forces act on the rod at position OA’ –
(i) gravitational force (Mg) vertically downward direction and
(ii) reaction force (n) of the hinge.
Let, the horizontal and the vertical components of n are nx and ny respectively; the horizontal and the vertical components of the acceleration of the centre of mass of the rod are ax and ay respectively.
∴ According to Fig.,
Mg – ny = May
and nx = Max = \(M \omega^2 \cdot \frac{L}{2}\) [∵ ax = centripetal acceleration]
= \(M \cdot \frac{3 g}{L} \cdot \frac{L}{2}\) = \(\frac{3}{2}\)Mg
[putting the value of ω from equation(1)]
The rod starts to rotate due to the action of torque created by ny and Mg.
If the angular acceleration of the rod is α,