Two masses M and m are connect by a light string gong over a pulley of radis r. The pulley is free to rotate about its axis which is kept horizontal. The moment of inertia of the pulley about the axis is I. The system is releaed from rest. Find the angular momentum fo teh system when teh mass Mhas descended through a height h. The string does not slip over the pulley.

Explanation

The situation is shown in ure. Let the speed of the masses be v at time t. This will also be the speed of a point on the rim of the wheel and hence the angular velocity of the wheel at time t will be v/r. If the height descended by the mass M is h the loss in the potential energy of the mass plus the pulley system is `mgh-mgh.` The gain in kinetic energy is `1/2Mv^2+1/2mv^2+1/2I(v/r)^2`. An no energy is lost,
`1/2(M=m+1/r^2)v^2=(M-m)gh`
`or, v^2=(2(M-m)gh)/(M+m+1/r^2)`
The angular momentum of the mass M is Mvr and that of the mass is mvr in the same direction. The angular momentum of the pulley is `Iomega=Iv/r.` The total angular momentum is
`[(M+m)r+I/r]v=[(M+m+I/r)r]sqrt((2(M-m)gh)/(M+m+I/r^2)) `
`=sqrt(2(M-m)(M+M+1/r^2)r^2gh)`