Question

The instantaneous angular position of a point on a rotating wheel is given by the equation $\theta \left( t \right) = 2{t^3} - 6{t^2}$. The torque on the wheel becomes zero at:
$\begin{gathered} {\text{A}}{\text{. }}t = 1s \\\ {\text{B}}{\text{. }}t = 0.5s \\\ {\text{C}}{\text{. }}t = 0.25s \\\ {\text{D}}{\text{. }}t = 2s \\\ \end{gathered}$

Answer 1

Hint : The torque acting on a point is equal to the first order time derivative of the angular momentum of the point. The angular momentum is equal to the product of the moment of inertia and the angular velocity and angular velocity is equal to the first order time derivative of the angular position. By equating the torque to zero, we can obtain the required answer.
Formula used:
Torque acting on a body can be calculated by the following formula:
$\tau = \dfrac{{dL}}{{dt}}$
Here L represents the angular momentum of the body which can be calculated by the following formula:
$L = I\omega$
Here I represent the moment of inertia of the body while $\omega$ is the angular momentum of the body which can be given in terms of the angular displacement $\theta$ in the following way.
$\omega = \dfrac{{d\theta }}{{dt}}$
Using these formulas in the expression for torque, we get
$\tau = I\dfrac{{{d^2}\theta }}{{d{t^2}}}{\text{ }}...{\text{(i)}}$

Detailed step by step solution:
We are given the angular displacement of a point on a rotating wheel. The expression for the angular position is given as
$\theta \left( t \right) = 2{t^3} - 6{t^2}$
As we know that the first order time derivative of angular displacement is equal to the angular velocity of the point, angular momentum is equal to the product of moment of inertia of the point and torque acting on a point is equal to first order time derivative of the angular momentum of the point. Therefore, we can directly calculate the torque on the point from the angular position using equation (i).
Let us first find the second order time derivative of the angular position of the particle. This can be done in the following way.
$\begin{gathered} \dfrac{{{d^2}\theta }}{{d{t^2}}}{\text{ }} = \dfrac{{{d^2}}}{{d{t^2}}}\left( {2{t^3} - 6{t^2}} \right) \\\ = \dfrac{d}{{dt}}\left( {6{t^2} - 12t} \right) \\\ = 12t - 12 \\\ \end{gathered}$
Therefore torque is equal to
$\tau = I\dfrac{{{d^2}\theta }}{{d{t^2}}} = I\left( {12t - 12} \right)$
Equating the torque to zero, we get
$\begin{gathered} I\left( {12t - 12} \right) = 0 \\\ 12t = 12 \\\ t = 1s \\\ \end{gathered}$
This implies that torque is equal to zero at t = 1s.

Hence, the correct answer is option A.

Note: When torque acting on a rotating body is zero, it means that the angular momentum of the body is constant. This implies that the body is in the state of uniform circular motion when external torque acting on it is equal to zero.