Answer to Question #254540 in Physics for Dcgfxv

Question

 a 1.5 kg block is hung by a light string which is wound around a 

 smooth pulley of radius 20 cm. The moment of inertia of the pulley is 2 kg m2

. 

(i) Sketch the free body diagram of the 1.5 kg block.

(ii) When the mass is released from rest, calculate the angular velocity and number of 

revolutions of the pulley at t = 4.2 s.

Answer 1

$\alpha=\frac{mg}{r(m+0.5M)}\\\omega=\frac{1.5(9.8)}{0.2(1.5+0.5(2))}4.2=123.5\frac{rad}{s}\\ n=\frac{1.5(9.8)}{0.2(1.5+0.5(2))}\frac{4.2^2}{2(2\pi)}=41.3$