Question

A wheel having moment of inertia 4kg m2 about its symmetrical axis, rotates at rate of 240 rpm about it. The torque which can stop the rotation of the wheel in one minute is:

& \text{A}\text{. }\dfrac{5\pi }{7}Nm \\\ & \text{B}\text{. }\dfrac{8\pi }{15}Nm \\\ & \text{C}\text{. }\dfrac{2\pi }{9}Nm \\\ & \text{D}\text{. }\dfrac{3\pi }{7}Nm \\\ \end{aligned}$$

Answer 1

Torque is a dimensional quantity; whose direction is perpendicular to the plane of rotation. Mathematically, it is given by the multiplication of moment of inertia with angular acceleration.
$\tau =I\alpha$
To find the torque required, we first need to find the angular acceleration, find it using Newton’s equation of motion in angular motion form. Then we put this value in the formula of torque.

Complete step-by-step answer:
Newton’s equation of motion for rotational motion –
$\omega ={{\omega }_{0}}+\alpha t$
Where,
$\omega$ is the final angular velocity of a rotating body, which is rotating under the influence of angular acceleration $\alpha$, after time t.
Torque is given by
$\tau =I\alpha$
Where, I is the moment of inertia of body with respect to the symmetrical axis
$\alpha$ is the angular acceleration
(1) First, we need to convert angular velocity from rpm to angle per second.
We know that 1 radian makes $2\pi$ degree angle, and there are 60 seconds in 1 minute. So,
${{\omega }_{0}}=\dfrac{240\pi }{60}=4\pi rad/\sec$
(2) Now we need to find the angular acceleration $\alpha$
Initial angular velocity ${{\omega }_{0}}$ is given in the question, which is ${{\omega }_{0}}=4\pi rad/\sec$
Final angular velocity is also given (Wheel stops after applying torque in 60 seconds), which is $\omega =0$
We can put these values in the Newton’s equation as;
$\omega ={{\omega }_{0}}+\alpha t$
The time taken to stop is 1 minute that is 60 seconds.
$\begin{aligned} & 0=4\pi -\alpha (60) \\\ & \Rightarrow \alpha =\dfrac{4\pi }{60}rad/{{\sec }^{2}} \\\ \end{aligned}$
(3) Now we can find required torque as;
$\tau =I\alpha$
I is given in the question, $I=4kg{{m}^{2}}$
$\tau =\dfrac{8\pi }{60}Nm$
The correct answer is B.

Note: Newton’s laws form translational motions are also used for rotational motion. As we have seen in the translational motion, a de-acceleration is required to stop a moving particle. In case of rotational motion, an external torque is also required in the opposite direction of rotation to stop the wheel.
We have used F=ma form to solve the translational motions, same way we can use $\tau =I\alpha$
to solve the rotational motion.
That is torque is analogous to force, angular speed is to velocity, angular acceleration is to acceleration, angle is to position and moment of inertia is to the mass.