Question

ind the acceleration of the block of mass 2M attached to a light string which passes over a disc shaped pulley of mass M and is connected to another block of mass Assume there is no slipping between the string and pulley.

Answer 1

Answer:

When mass 2m has descended through a height h, the speed of each block is v and the angular speed of pulley is ω=v/r.

Moment of inertia of pulley I=12×2mr^2=mr^2

Explanation:

Applying the energy conservation between two situations

0=−2mgh+mgh+12×2mv2+12mv2+12Iω2

mgh=32mv2+12×mr2(Vr)2=2mv2

v=gh2−−−√

Angular momentum of blocks are 2mvr and mvr and that of pulley is Iω=mr2(v/r)=mvr in the same direction.

The total angular momentum is

2mvr+mvr+mvr=4mvr

=4mgh2−−−√r

=22–√mrg√h