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.

Answer

A, B, D

Explanation

According to the question
`04g-T_1=00.4a`……..1
`T_2-0.2g=0.2a`……2
`(T_1-T_2)r=(la)/r`……..3
From eq. 1, 2, and 3
`rarr a=((0.4-0.2)g)/((0.4+0.2+1.6/0.4))=g/5` ltb. Therefoee b. `V=sqrt(2gh)=sqrt(2xgxx1/20`
`rarr sqrt((g/5))=sqrt((9.8/5))=1.4m/s`
a. total kinetic energy of the system
`=1/2m_1V^2+1/2m_2V^2+1/2 15^`
` =(1/2xx0.4xx1.4^2)+(1/2xx0.2xx1.4^2)+ {1/2xx(1.6/4xx1.4^2)}`
`=(0.2+0.1+0.2)(1.4)^2`
`=0.54x1.96=0.98Joule`