Question

In the figure, two blocks, of masses 2.00 kg and 3.00 kg, are connected by a light string that passes over a frictionless pulley of moment of inertia 0.006kg - m^2 and radius 7.00 cm. Find: a) The acceleration of the blocks b) The angular acceleration of the pulley

Answer 1

in such kind of scenario

acceleration of hanging block m1 is determined by following formula ;

$a=\frac{m_{1}g}{m_{1}+m_{2}+\frac{I}{R^2}}$

where I is moment of inertia of pulley of radius R

here

m1= 2kg, m2=3kg and I= 0.006kg-m^2, R=0.07 m

thus

$a=\frac{2g}{3+2+\frac{0.006}{0.07^2}}=0.32g=3.15m/s^2$

Part B;

Angular acceleration of pulley

$\alpha_{pulley}=\frac{a}{R}=45rad/s^2$