Question

A wheel of moment of inertia $1kg.{m^2}$ is rotating at a speed of $30rad/s$. Due to friction on the axis, it comes to rest in $10$ minutes. Calculate the average torque of the friction.

Answer 1

Moment of inertia of a body about an axis of rotation is equal to the torque acting on the body divided by the corresponding angular acceleration, generated about the axis of rotation. Final rotating speed is $0$ as it comes to rest and initial speed is given. Construct the equation as the same as in the case of linear motion. Calculate the angular acceleration from there.
Formula used: Torque, $\tau = I\alpha$ [where, $I$ is the moment of inertia of a body $\alpha$ is the angular acceleration]
Also, ${V_f} = {V_i} - \alpha t$ [Where, ${V_f}$ is final and ${V_i}$ is the initial rotating speed of the wheel and $t$ is time]

Complete step by step solution:
Torque is the tendency of rotational motion, set up in a body by a couple (two equal, opposite, and parallel forces, having different lines of action on the same body). When a force is applied to a body, linear acceleration is produced in that body. Similarly, we can see that when a torque is applied on a body, angular acceleration is produced in it. So, we can say that torque plays the same role in rotational mechanics like that in the case of linear motion.
Now, the moment of inertia of a body about an axis of rotation is equal to the torque acting on the body divided by the corresponding angular acceleration, generated about the axis of rotation.
So, let the moment of inertia of a body is $I$ and the angular acceleration is $\alpha$ .
So, the torque will be $\tau = I\alpha$
Now, it is given that initial rotating speed of the wheel is ${V_i} = 30rad/s$
After $10\min = \left( {10 \times 60} \right) = 600\sec$ it comes to rest.
So, the final rotating speed of the wheel will be ${V_f} = 0$
So, following the equation of motion, we get, ${V_f} = {V_i} - \alpha t$
Putting the values in the equation, we get, $0 = 30 - (\alpha \times 600)$
or, $\alpha = \dfrac{{30}}{{600}} = \dfrac{1}{{20}}$
So, the angular acceleration is $\dfrac{1}{{20}}rad/{\sec ^2}$
Now, it is given that moment of inertia of the wheel is $1kg.{m^2}$
So, the value of the torque is $\tau = I\alpha = \left( {1 \times \dfrac{1}{{20}}} \right) = 0.05N.m$
So, the average torque of the friction is $0.05Nm.$

Note:
The equation $\tau = I\alpha$ is the rotational analogue of the equation $F = ma$ of linear motion. So, we can say that the rotational analogues of the force and linear acceleration are the torque and angular acceleration respectively. So, comparing the two equations we can also say that the rotational analogue of the mass of a body is its moment of inertia.