Question

A disc of radius 2m and mass 200kg is acted upon by a torque 100N-m. Its angular acceleration would be
A) $1rad/{\sec ^2}$
B) $0.5rad/{\sec ^2}$
C) $0.25rad/{\sec ^2}$
D) $2rad/{\sec ^2}$

Answer 1

Here, first find moment of inertia I. There is a relation of moment of inertia I, angular acceleration $\alpha$ and torque $\tau$. Write the relation and find out the angular acceleration. The formula for moment of inertia for a circular disk is $\dfrac{{M{r^2}}}{2}$ where M = mass; r = radius.

Formula Used:
Here we are writing a formula for torque and angular acceleration
$\tau = I\alpha$
Where:
I = Moment of Inertia.
$\tau$= Torque.
$\alpha$= Angular Acceleration.

Complete step by step answer:
Calculate Angular Acceleration:
$I = \dfrac{{M{r^2}}}{2}$
Put the value
$I = \dfrac{{200 \times 4}}{2}$
Solve,
$I = 400$
Put the value of I in the equation
$\tau = I\alpha$
Take I to LHS
$\dfrac{\tau }{I} = \alpha$
$\dfrac{{100}}{{400}} = \alpha$
Do calculation,
$\alpha = \dfrac{1}{4}$
$\alpha = 0.25rad/{s^2}$
Final Answer: A disc of radius 2m and mass 200kg is acted upon by a torque 100N-m. Its angular acceleration would be Option C, $0.25rad/{s^2}$

Note: First carefully find out the formula for moment of inertia. The moment of inertia is different for different objects. Make sure to write the formula of moment of inertia for a circular disk. Apply the formula that relates angular acceleration and torque.