Question

A constant torque acting on a uniform circular wheel changes its angular momentum from ${{A}_{0}}$ to 4${{A}_{0}}$ in 4 sec. The magnitude of this torque is:
$\text{A}\text{. 4}{{A}_{0}}$
$\text{B}\text{. }{{A}_{0}}$
$\text{C}\text{. }\dfrac{3}{4}{{A}_{0}}$
$\text{D}\text{. 12}{{A}_{0}}$

Answer 1

Hint: Constant torque is given as $\tau =\dfrac{\Delta L}{\Delta t}$, where, $\Delta L$ is the change in angular velocity in time interval of $\Delta t$. Calculate the change in angular momentum of the wheel for the given time interval of 4 seconds. Then substitute it in $\tau =\dfrac{\Delta L}{\Delta t}$.

Formula used:
$\tau =\dfrac{\Delta L}{\Delta t}$

Complete step by step answer:
When a body is rotating about a fixed axis, we define a term called angular velocity. Angular velocity is the change of angle of rotation of the body about the axis of rotation in per unit time.
A rotating body has a momentum called angular momentum. It is defined as the product of the moment of inertia and the angular velocity about the axis of rotation.
i.e. $L=I\omega $, where L is the angular momentum, I is the moment of inertia and $\omega $ is the angular velocity of the body about the axis of rotation.
When a torque is applied on a body, its angular momentum changes with time and torque is defined as the rate of change of angular momentum with time.
i.e. $\tau =\dfrac{dL}{dt}$.
If the applied torque is constant, then the angular momentum will change uniformly with time. Constant torque is given as $\tau =\dfrac{\Delta L}{\Delta t}$ …. (i).
Here, $\Delta L$ is the change in angular velocity in the time interval of $\Delta t$.
Let us calculate the torque on the wheel.
It is given that the angular momentum of the wheel changes from ${{A}_{0}}$ to 4${{A}_{0}}$. Therefore, $\Delta L=4{{A}_{0}}-{{A}_{0}}=3{{A}_{0}}$.
And the time interval for which this change takes place is given as 4 seconds. Hence, $\Delta t$=4sec.
Substitute the values of $\Delta L$ and $\Delta t$ in equation (i).
$\Rightarrow \tau =\dfrac{3{{A}_{0}}}{4}$.
Therefore, the torque on the wheel is equal to $\dfrac{3{{A}_{0}}}{4}$.
Hence, the correct answer is option C.

Note:
Note that rotational motion is analogous to translational motion.
Torque is analogous to force (F).
Angular momentum is analogous to momentum (P).
We know that $F=\dfrac{dP}{dt}$.
Therefore, $\tau =\dfrac{dL}{dt}$.
The conclusion is that if we know the formulas of translation motion then we can write the formulas of rotational motion.