Question

Moment of inertia of a ring is $3\,{\text{kg}} \cdot {{\text{m}}^2}$. It is rotated for $20\,{\text{s}}$ from its rest position by a torque of $6\,{\text{N}} \cdot {\text{m}}$. Calculate the work done.

Answer 1

Use the formula for torque in terms of angular acceleration and moment of inertia and determine the angular acceleration of ring. Using third kinematic equation for angular motion of an object, determine the angular displacement of the ring. Use the formula for work done in terms of torque and angular displacement and determine the required work done.

Formulae used:
The torque $\tau$ acting on an object is
$\tau = I\alpha$ …… (1)
Here, $I$ is the moment of inertia of the object and $\alpha$ is angular acceleration of the object.
The kinematic equation for angular displacement $\theta = {\omega _0}t + \dfrac{1}{2}\alpha {t^2}$ of the object is
$\theta = {\omega _0}t + \dfrac{1}{2}\alpha {t^2}$ …… (2)
Here, ${\omega _0}$ is the initial angular speed of the object, $\alpha$ is the angular acceleration of the object and $t$ is time.
The work done $W$ is given by
$W = \tau \theta$ …… (3)
Here, $\tau$ is the torque acting on the object and $\theta$ is the angular displacement of the object.

Complete step by step answer:
We have given that the moment of inertia of the ring is $3\,{\text{kg}} \cdot {{\text{m}}^2}$ and torque acting on it is $6\,{\text{N}} \cdot {\text{m}}$.
$I = 3\,{\text{kg}} \cdot {{\text{m}}^2}$
$\tau = 6\,{\text{N}} \cdot {\text{m}}$

The ring starts rotating from rest and rotates for $20\,{\text{s}}$. Hence, the initial angular speed of the ring is zero.
${\omega _0} = 0\,{\text{rad/s}}$
$t = 20\,{\text{s}}$

Let us first determine the angular acceleration of the ring.
Rearrange equation (1) for angular acceleration of the ring.
$\alpha = \dfrac{\tau }{I}$

Substitute $6\,{\text{N}} \cdot {\text{m}}$ for $\tau$ and $3\,{\text{kg}} \cdot {{\text{m}}^2}$ for $I$ in the above equation.
$\alpha = \dfrac{{6\,{\text{N}} \cdot {\text{m}}}}{{3\,{\text{kg}} \cdot {{\text{m}}^2}}}$
$\Rightarrow \alpha = 2\,{\text{rad/}}{{\text{s}}^2}$
Hence, the angular acceleration of the ring is $2\,{\text{rad/}}{{\text{s}}^2}$.

Now let us determine the angular displacement of the ring.
Substitute $0\,{\text{rad/s}}$ for ${\omega _0}$, $2\,{\text{rad/}}{{\text{s}}^2}$ for $\alpha$ and $20\,{\text{s}}$ for $t$ in equation (2).
$\theta = \left( {0\,{\text{rad/s}}} \right)\left( {20\,{\text{s}}} \right) + \dfrac{1}{2}\left( {2\,{\text{rad/}}{{\text{s}}^2}} \right){\left( {20\,{\text{s}}} \right)^2}$
$\Rightarrow \theta = 400\,{\text{rad}}$
Hence, the angular displacement of the ring in 20 seconds is $400\,{\text{rad}}$.

We can determine the work done using equation (3).
Substitute $6\,{\text{N}} \cdot {\text{m}}$ for $\tau$and $400\,{\text{rad}}$ for $\theta$ in equation (3).
$W = \left( {6\,{\text{N}} \cdot {\text{m}}} \right)\left( {400\,{\text{rad}}} \right)$
$\therefore W = 2400\,{\text{J}}$

Therefore, the work done is $2400\,{\text{J}}$.

Note: The students should keep in mind that the ring is starting to move from rest. Hence, the initial angular speed of the ring must be taken as zero. If this value of initial angular speed is not taken correctly, the final answer for the work done will be incorrect. The students should also remember that the kinematic equations are not only applicable for linear motion but also applicable for the rotational motion with constant angular acceleration.