Question

A ring of radius 1m has moment of inertia of $10kg-m^2$ about an axis through its centre and perpendicular to its plane. If a constant force of 50N is applied to it tangentially as shown, what will be its angular acceleration?

• A. $5rad/s^2$
• B. $10rad/s^2$
• C. $2.5rad/s^2$
• D. $\pi rad/s^2$

Answer 1

The correct option is A $5rad/s^2$

This Force will produce torque on this ring as discussed by Aanand in the previous video clip.
Given,$I=10kg-m^2$
r = 1m
F = 50N
$\left|\overrightarrow{\tau }\right|=rFsin\theta =rFsin{90}^o$
= r F = 50 N-m
Recall, $\left|\overrightarrow{\tau }\right|=I\left|\overrightarrow{\alpha }\right|$
$\left|\overrightarrow{\alpha }\right|=\frac{\left|\overrightarrow{\tau }\right|}{I}=\frac{50}{10}=5\frac{rad}{s^2}$