Question

A square plate of mass $120g$ and edge $5.0cm$ rotates about one of the edges. If it has a uniform angular acceleration of $0.2rad/{s^2}$. What torque acts on the plate?

Answer 1

The measurement of force is known as the torque. It makes an object rotate about the axis $5.0cm$. The force accelerates the objects in linear kinetics. Torque is the reason for angular acceleration. Torque is described as the rotational equivalent of linear force

Complete step by step solution:
Mathematically it is described as the product of the applied force and perpendicular distance of the point
Hence the unit of torque is newton meters because the force is measured in newton and the distance is measured in meters then the unit of torque is newton meters (N.m)
The object starts to rotate when we apply the torque over the object.
The angular acceleration is, $a = 0.20$
The mass $M = 120g$
Edge $a = 0.05m$
When the line passing the center and is parallel to the edge then the moment of inertia will be,
$I = \dfrac{{M{a^2}}}{{12}}$
Now substitute the values in the above equation we get as follows,
$I' = \dfrac{{M{a^2}}}{{12}} + \dfrac{{M{a^2}}}{4}$
Now simplify the above equation we get,
$I' = \dfrac{{M{a^2}}}{3}$
Now apply the values in the above equation we get,
$I' = \dfrac{{(0.12 \times 0.05 \times 0.05)}}{3}$
Now solve the above term to find the moment of inertia,
$I' = 0.0001kg.{m^2}$
Now we have to find the torque,
Torque $= I'a$
Now substitute the values in the above equation we get the torque,
Torque $= 0.0001 \times 0.20$
After solving the equation the torque will be,
Torque $= 2 \times {10^{ - 5}}N.m$
So, the torque is $2 \times {10^{ - 5}}N.m$

Note:
The kilogram meter squared is the unit of the moment of inertia.
This moment of inertia fully depends on the mass which is now distributed around the axis of rotation.
The moment of inertia is described as the quantity expressed as the body resisting angular acceleration.