Question

If $I=50kg.{{m}^{2}}$ , then how much torque will be applied to stop it in $10\sec$ . Its initial angular speed is $20rad/\sec$
(A) $100N.m$
(B) $150N.m$
(C) $200N.m$
(D) $250N.m$

Answer 1

In the given question, we have been given the moment of inertia of a body and we have to find the torque needed to stop the angular rotation of the body; the angular velocity also has been provided to us. We can calculate the angular acceleration with the given angular velocity and the time and then we can use the obtained angular acceleration and the moment of inertia to find the torque. Let’s see the detailed solution.
Formula Used: $\alpha =\underset{\Delta \to 0}{\mathop{\lim }}\,\dfrac{\Delta \omega }{\Delta t}$ , $\tau =I.\alpha$

Complete step by step solution:
We have been provided with the following values
The angular velocity of the body $(\omega )=20rad/\sec$
The time for which the body rotates, or the time after which the body is supposed to stop $(t)=10\sec$
The moment of inertia of the body $(I)=50kg.{{m}^{2}}$
As discussed in the hint section, our first move would be to find the angular acceleration or retardation of the body.
The angular acceleration of a body is given as $\alpha =\underset{\Delta \to 0}{\mathop{\lim }}\,\dfrac{\Delta \omega }{\Delta t}$ where $\Delta \omega$ is the change in the angular velocity and $\Delta t$ is the change in the time
Since we are given only one value of angular velocity and time, we can resolve the limit in the above equation and say that $\alpha =\dfrac{\omega }{t}$
Substituting the values in the above equation, we get
$\alpha =\dfrac{20rad/\sec }{10\sec }=2rad/{{\sec }^{2}}$
As discussed above, the torque applied on a body is the product of its moment of inertia and the angular acceleration, that is $\tau =I.\alpha$
Substituting the values in the above equation, we can say
$\tau =\left( 50kg.{{m}^{2}} \right)\left( 2rad/{{\sec }^{2}} \right)=100N.m$

Hence we can say that option (A) is the correct answer to the given question.

Note:
In the above question, the units of the moment of inertia and the angular acceleration, upon multiplying will furnish the unit $kg.{m^2}.rad/{\sec ^2}$ but we have written the unit of torque as $N.m$. This is correct because all the values that we substituted in the equation to find the value of torque were in SI units, and we know that the SI unit of torque is newton metres. Thus, if we always use standard units for our calculations, we won’t have to worry about unit conversion and the units of individual quantities.