Question

1. Two wheels of M.I $2\;kg-m^2$ and $4\;kg-m^2$ rotate with angular velocities of $120\;rpm$ and $60\;rpm$ respectively in the same direction. If the two wheels are coupled so as to rotate with the same angular velocity, find the frequency of rotation of the coupled system.
   2)A circular disc of MI $10\;kg-m^2$ rotates about its own axis at a constant speed of $60\;rpm$ under the action of an electric motor of power $31.4\;W$. If the motor is switched off, how many rotations will it cover before coming to rest?

Answer 1

1. First find the initial angular momentum of the system when the motion of the two wheels is independent of each other. Then find final angular momentum of the system when the two wheels are coupled together and rotate with the same angular speed, by taking the moment of inertia MI of the system as the collective sum of their individual MI. Use the law of conservation of angular momentum to equate the two expressions to arrive at the collective angular velocity which is nothing but $2\pi$ times the frequency of rotation.

2. First find the torque developed on the disc due to the driving motor using power and angular speed magnitudes. Then find the angular acceleration of the disc, which you can substitute in an equation of motion to arrive at the angular displacement of the disc after the motor is turned off. Use the resultant value to determine the number of rotations.

Formula Used:
Angular momentum: $L=I\omega$
Conservation of angular momentum: $L_i=L_f$
Angular speed $\omega = 2\pi f$
Power $P=\tau \omega$
Torque $\tau = \alpha I$
Angular equation of motion: $\omega_f^2 = \omega_i^2+2\alpha \theta$

Complete Step-by-Step Solution:
1)We are given that for,
Wheel $W_1$: $I_1=2\;kgm^2$, $\omega_1=120\;rpm = 120 \times \dfrac{2\pi}{60} = 4\pi\;rads^{-1}$
Wheel $W_2$: $I_2=4\;kgm^2$, $\omega_2=60\;rpm = 60 \times \dfrac{2\pi}{60} = 2\pi\;rads^{-1}$

Now, the angular momentum of a rotating wheel is given as:
$L=I\omega$

The total initial angular momentum of system will be the sum of the individual angular momenta of the wheels, i.e.,
$L_i = I_1\omega_1 + I_2\omega_2$
$\Rightarrow L_i =(2\times 4\pi)+ (4\times 2\pi) = 8\pi + 8\pi = 16\pi \;kgm^2s^{-1}$

Now, the two wheels are coupled so as to rotate with the same angular velocity, say $\omega$.. This means that the total moment of inertia of the system becomes the sum of their individual moments of inertia.

The total final angular momentum of the system of two wheels coupled with each other will be,
$L_f = (I_1+I_2)\omega = (2+4)\omega = 6\omega\; kgm^2s^{-1}$

We now need to determine the final angular velocity in order to subsequently determine the frequency of rotation of the coupled system. For this, we proceed as follows.

According to the law of conservation of angular momentum, for any isolated system, the total initial and final angular momentum will remain constant, i.e.,
$L_i=L_f$
$\Rightarrow 16\pi = 6\omega$
$\Rightarrow \omega=\dfrac{16\pi}{6} = \dfrac{8\pi}{3}\;rads^{-1}$

Now, the frequency f of the coupled system is related to its angular speed as:
$\omega = 2\pi f \Rightarrow f = \dfrac{\omega}{2\pi}$
$\Rightarrow f = \dfrac{\left(\dfrac{8\pi}{3}\right)}{2\pi} = \dfrac{8\pi}{3\times 2\pi} = 1.33\;s^{-1} = 1.33\;Hz$

Therefore, the frequency of the coupled system is found to be $1.33\;Hz$.

2. We are given that the moment of inertia of the circular disc is $I = 10\;kg-m^2$.
   The angular speed of the disc is $60\;rpm = 60\times \dfrac{2\pi}{60} = 2\pi\;rads^{-1}$
   The power of the motor is $P = 31.4\;W$.

We need to find the number of rotations executed by the disc before coming to a stop on cutting off the motor power.

For this we need to find the angular displacement $\theta$ of the disc after the motor is turned off, which we can obtain from the angular acceleration $\alpha$ of the disc under the action of the motor.

Now, we know that the power and torque are related to the angular speed as:
$P=\tau \omega \Rightarrow \tau = \dfrac{P}{\omega} = \dfrac{31.4}{2\pi} = \dfrac{10\pi}{2\pi} = 5\;Nm$

The torque developed produces an inertial angular acceleration which is given by:
$\alpha = \dfrac{\tau}{I} = \dfrac{5}{10} = 0.5\;rads^{-2}$
Now, when the motor is switched off, the angular speed of the disc finally becomes $\omega_f = 0$.

Using the following equation of angular motion and solving for angular displacement $\theta$:
$\omega_f^2 = \omega_i^2+2\alpha \theta$
$\Rightarrow 0 = (2\pi)^2 + 2 \times 0.5 \times \theta$
$\Rightarrow \theta = 4\pi^2$

Now, we know that 1 rotation entails an angular displacement of $2\pi$. Therefore, the number of rotations N for $\theta = 4\pi^2$ will be:
$N = \dfrac{\theta}{2\pi}=\dfrac{4\pi^2}{2\pi} = 2\pi = 2\times 3.14 = 6.28$

Therefore, the circular disc rotates nearly 6 times after the motor is turned off before it comes to rest.

Note:
Regarding the first problem, it is important to understand the collective behaviour of components and the factors that get influenced under their non-individualistic behaviour. We add up the moments of inertia of the two wheels in the second case to signify the repositioning of the centre of mass of the entire system, since it now behaves as a collective unit with the two wheels working together. This also influences their rotatory motion which is why we get a kind of a superposed final angular speed and frequency.

Remember that the basic kinematic equations of linear motion can be extrapolated to be used in case of angular and rotatory motion, as we did in the second question. The only difference is that we consider the angular parameters in contrast to linear quantities.