Consider two masses with `m_(1) gt m_(2)` connected by a light inextensible string that passes over a pulley of radius R and moment of inertia I about its axis of rotation. The string does not slip on the pulley and the pulley turns without friction. The two masses are released from rest separated by a vertical distance 2h. When the two masses pass each other, the speed of the masses is proportional to

Options

A. `sqrt((m_(1)-m_(2))/(m_(1)+m_(2)+(I)/(R^(2))))`
B. `sqrt(((m_(1)+m_(2))(m_(1)-m_(2)))/(m_(1)+m_(2)+(I)/(R^(2))))`
C. `sqrt((m_(1)+m_(2)+(I)/(R^(2)))/(m_(1)-m_(2)))`
D. `sqrt(((I)/(R^(2)))/(m_(1)+m_(2)))`

Answer

A

Explanation

`m_(1)g-T_(1)=m_(1)a`
`T_(2)-m_(2)g=m_(2)a`
`(T_(1)-T_(2))R=(I)/(R)a`
`a=((m_(1)-m_(2))g)/(((I)/(R^(2))+m_(1)+m_(2)))impliesv_("rel")^(2)=(8hg(m_(1)-m_(2)))/(((I)/(R^(2))+m_(1)+m_(2)))`
As `v_("rel")=v-(-v)=2vimpliesv=(1)/(2)sqrt((8gh(m_(1)-m_(2)))/(((I)/(R^(2))+m_(1)+m_(2))))`