Answer to Question #280175 in Physics for Nishat

Question

A uniform 2.5 kg disk of radius 30 cm is rotating around its central axis at an

angular speed of 72 rad/s. At time t=0, a man begins to slow it at a uniform rate until it stops

at t = 8 s.

(a) By time t = 5 s, how much work had the man done?

(b) For full 8 s, at what average rate did the man do work?

Answer 1

$I=\frac{1}{2}mr^2=0.5\cdot2.5\cdot0.3^2=0.1125\ (kg\cdot m^2)$

(a) $KE_0=\frac{I\omega^2}{2}=\frac{0.1125\cdot72^2}{2}=291.6\ (J)$

$\omega=\omega_0-\epsilon t$

$\epsilon=\omega_0/t=72/8=9\ (rad/s^2)$

$\omega_5=72-9\cdot5=27\ (rad/s)$

$KE_5=\frac{I\omega^2}{2}=\frac{0.1125\cdot27^2}{2}=41\ (J)$

$W=\Delta KE=291.6-41=250.6\ (J)$

(b) $P=291.6/8=36.45\ (W)$