Q. A mass of $10\ kg$ connected at the end of a rod of negligible mass is rotating in a circle of radius $30\ cm$ with an angular velocity of $10\ rad/sec$. If this mass is brought to rest in $10\ sec$ by a brake, what is the magnitude of the torque applied
A) $0.9\ N$-$m$
B) $1.2\ N$-$m$
C) $2.3\ N$-$m$
D) $0.5\ N$-$m$
Explanation
(a) $\omega_{1} = 10\ Rad/s$, $\omega_{2} = 0,t = 10s$
$\therefore\alpha = \frac{\omega_{2} - \omega_{1}}{\tau} = \frac{0 - 10}{10} = - 1rad/s^{2}$
Negative sign means retardation
Now $I = mr^{2} = 10 \times (0.3)^{2} = 0.9\ kg$-$m^{2}$
$\therefore$ Torque $\tau = I\alpha = 0.9 \times (1) = 0.9\ N$-$m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.