Question

A wheel of moment of inertia $5\times {{10}^{-3}}kg/{{m}^{2}}$ is rotating at the speed of 20 cycles per second. Calculate the torque required to stop it in 10 second.

Answer 1

In this question we have been asked to calculate the torque required to stop the wheel. We know that torque is the product of moment of inertia and the angular acceleration of the wheel. We have been given the moment of inertia of the wheel. Therefore, we will first calculate the angular acceleration of the wheel and thus the torque of the wheel.
Formula Used:
$T=I\alpha$
Where,
T is the torque in Nm
I is the moment of inertia in $kg/{{m}^{2}}$
$\alpha$is the angular acceleration in rad/sec

Complete answer:
As shown in the figure we are given the moment of inertia of the wheel for the given axis. Assume the direction of rotation of the wheel is as shown in the figure.

We are given that speed of wheel is 20 cycles per second
We know that,
$1rev=2\pi rad/s$
Therefore, the given angular speed of wheel is $40\pi rad/s$
Now we know that
$T=I\alpha$
Therefore, solving for $\alpha$
We know,
$\alpha =\dfrac{\omega }{\Delta t}$
We have been given $\Delta t$as 10 sec
After substituting the given values in above equation
We get,
$\alpha =\dfrac{40}{10}rad/{{s}^{2}}$
Therefore,
$\alpha =4\pi rad/{{s}^{2}}$
Now, substituting the value of in $\alpha$ in equation (1)
We get,
$T=I\alpha$
$T=5\times {{10}^{-3}}\times 4\pi$
Therefore,
$T=2\pi \times {{10}^{-2}}Nm$

Therefore, the correct answer is $T=2\pi \times {{10}^{-2}}Nm$

Note:
The tendency of a body to resist angular acceleration as called the moment of inertia. It is the sum of the products of mass of each particle with the square of its distance from the axis. Torque also determines the torque necessary for angular acceleration about a rotational axis. Assuming the angular momentum of a system is constant, then as the moment of inertia gets smaller, the angular velocity will increase.