Question

A wheel of moment of inertia is 1\[kg{{m}^{2}}\] is rotating at the speed of 30 rad/s. Due to friction on the axis, it comes to rest in 10 minutes. Calculate the total work done by friction.

Answer 1

Hint: We have been asked to calculate the work done by rotational method. Now, we know that work done in rotational motion is done by the torque applied to some angular displacement. But have not been given the torque or angular displacement in the question. But we also know that, on integrating the work done by torque, we get a relation between work done and moment of inertia and angular velocity. Therefore, we will be using this to calculate the work done.
Formula Used:
\[W=\int_{{{\omega }_{o}}}^{\omega }{I\omega d\omega }=\dfrac{1}{2}I\left( {{\omega }^{2}}-\omega _{o}^{2} \right)\]

Complete answer:
We know that work done in rotational motion is given by the product of torque applied and the angular displacement. In terms on equation
We can write,
\[W=\int \tau \cdot d\theta \]
Upon integrating we get,
\[W=\int_{{{\omega }_{o}}}^{\omega }{I\omega d\omega }=\dfrac{1}{2}I\left( {{\omega }_{i}}^{2}-\omega _{o}^{2} \right)\] ……………. (1)
Now, we have been given that the rotational speed is 30 rad/s which means that at last the wheel comes to rest.
Therefore,
\[{{\omega }_{i}}=30rad/s\] and \[{{\omega }_{o}}=0\]. Moment of inertia is I is given as 1\[kg{{m}^{2}}\]
After substituting the values in equation (1)
 We get,
\[W=\dfrac{1}{2}\times 1\times \left( {{30}^{2}}-0 \right)\]
\[W=450J\]
Therefore, the correct answer is 450 Joules.

Note: We can also solve the above question by using the work – kinetic theorem for rotational motion. This theorem states that the amount of work done by the rotating torque on a body under pure rotation or fixed axis equals the change in rotational kinetic energy. We know that change in rotational kinetic energy is given by,
\[W=\Delta K.E=\dfrac{1}{2}I\left( {{\omega }_{i}}^{2}-\omega _{o}^{2} \right)\]