Question

Two blocks of masses `2m` and `m` are connected by a light string going over a pulley of mass `2m` and radius `R` as shown in the figure. The system is released from rest. Find the angular momentum of system when the mass `2m` has descended through a height `h`. Assume no slipping.

Answer 1

When mass `2m` has descended through a height `h`, the speed of each block is `v` and the angular speed of pulley is `omega=v//r`.
Moment of inertia of pulley `I=(1)/(2)xx2mr^(2)=mr^(2)`

Applying the energy conservation between two situations
`0= -2mgh+mgh+(1)/(2)xx2mv^(2)+(1)/(2)mv^(2)+(1)/(2)Iomega^(2)`
`mgh=(3)/(2)mv^(2)+(1)/(2)xxmr^(2)((V)/(r))^(2)=2mv^(2)`
`v=sqrt((gh)/(2))`
Angular momentum of blocks are `2mvr` and `mvr` and that of pulley is `Iomega=mr^(2)(v//r)=mvr` in the same direction.
The total angular momentum is
`2mvr+mvr+mvr=4mvr`
`=4msqrt((gh)/(2))r`
`=2sqrt2mrsqrtgh`