Question

A wheel having moment of inertia 2$kg m^2$ about its vertical axis, rotates at the rate of 60rpm about this axis. The torque which can stop the wheels rotation in one minute would be
Option:
A. $\dfrac{{2\pi }}{{15}}Nm$
B. $\dfrac{\pi }{{12}}Nm$
C. $\dfrac{\pi }{{15}}Nm$
D. $\dfrac{\pi }{{18}}Nm$

Answer 1

The resistance of any physical object to any transition in velocity is known as inertia. Changes in the object's speed or direction of motion are used. The propensity of objects to continue travelling in a straight line at a constant pace while no forces act on them is one example of this property.

Formula used:
$\tau = I\alpha$
$\tau$= Torque
I = Moment of Inertia
$\alpha$= Angular acceleration

Complete step-by-step answer:
The moment of inertia of a solid body, also known as mass moment of inertia, angular mass, second moment of mass, or, more precisely, rotational inertia, is a quantity that calculates the torque required for a desired angular acceleration around a rotational axis, in the same way as mass determines the force required for a desired acceleration. It depends on the mass distribution of the body and the axis chosen, with greater moments necessitating more torque to adjust the rate of rotation.
The force that can cause an object to spin around an axis is measured in torque. In linear kinematics, force is what allows an object to accelerate. Torque is also responsible for angular acceleration. As a result, torque can be defined as the linear force's rotational counterpart.
Given
$I = 2{\text{kg}}{{\text{m}}^2}$
Let angular acceleration be $\alpha$
${\omega _{\text{o}}} = 60{\text{rpm}}$
Converting rpm to rad/s
${\omega _{\text{o}}} = 60{\text{rpm}} = \dfrac{{2\pi 60}}{{60}} = 2\pi {\text{rad}}/{\text{s}}$
Time = 1m = 60 sec
Now using $\omega = {\omega _0} + \alpha t$(Angular form of v = u + at)
$\omega = 0$
${\omega _0} = 2\pi$
So
$\omega = {\omega _0} + \alpha t$
$0 = 2\pi - \alpha 60$
Also
$\alpha = \dfrac{\pi }{{30}}$
Now ${\text{ Torque }} = {\mathbf{I}}\alpha$is applied
$\tau = {\mathbf{I\alpha }} = {\mathbf{2}} \times \dfrac{\pi }{{{\mathbf{30}}}} = \dfrac{\pi }{{{\mathbf{15}}}}{\mathbf{Nm}}$

So, the correct answer is “Option A”.

Note: The torque is a pseudovector in three dimensions, and it is given by the cross product of the position vector (distance vector) and the force vector for point particles. The force applied, the lever arm vector relating the point around which the torque is determined to the point of force application, and the angle between the force and lever arm vectors both influence the degree of torque in a rigid body.