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

Hint: 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.