Question

A mass m hangs from a string. The string is attached to a frictionless pulley of mass M and is wrapped around it many times around it. The hanging mass is released from rest from a height h above the floor. The pulley is a uniform disk. use the rotational and linear second laws to find the acceleration of the mass as it falls.

I got a = 2mg/(2m+M). Is this correct? If, so please explain

Answer 1

ㄒ 0 mma

from the force diagram for the mass "m", force equation is given as

mg - T = ma

T = mg - ma                                               eq-1

For the pulley :

I = moment of inertia of disk = (0.5) Mr²

Torque equation is given as

$\tau$ = T r = I $\alpha$

Using eq-1

(mg - ma) r = (0.5) Mr² (a/r)

mg - ma = (0.5) M a

2mg - 2ma = Ma

a = 2mg/(2m + M )