Answer to Question #179446 in Physics for john vincent burlas

Question

What is the average power dissipated by a 2kg grinding wheel of radius 0.1 m brought to rest in 10 rev from an initial velocity of 3000 rpm? Assume constant acceleration. 

Answer 1

$n=10$

$\omega=3000\ rpm=314\ rad/s$

$I=mr^2/2=2\cdot0.1^2/2=0.01\ kg\cdot m^2$

$W=\Delta KE=\frac{I\omega^2}{2}=\frac{0.01\cdot314^2}{2}\approx493\ (J)$

$P=W/t$

$\phi=\omega t-\epsilon t^2/2$

$\omega_f=\omega-\epsilon t\to \epsilon=(\omega-\omega_f)/t=(\omega-0)/t=\omega/t$

$\phi=\omega t-\epsilon t^2/2=\omega t-\frac{\omega t^2}{2t}=\frac{\omega t}{2}$

$2\pi n=\frac{\omega t}{2}\to t=\frac{4\pi n}{\omega}=\frac{4\cdot 3.14\cdot10}{314}=0.4\ (s)$

$P=W/t=493/0.4=1232.5\ (W)$ . Answer