Question

Two masses M and m are connected by a light string going over a pulley of radius 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 released from rest. Find the angular momentum of the system when the mass M has descended through a height h. The string does not slip over the pulley.

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

The situation is shown in figure. Let the speed for 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 $\frac{v}{r}$. If the height descended by the mass M is h, the loss in the potential energy of the "masses plus the pulley” system is Mgh - mgh. The gain in kinetic energy is $\frac{1}{2}M{v}^2+\frac{1}{2}M{v}^2+\frac{1}{2}l{\left(\frac{v}{r}\right)}^2$ . As no energy is lost,

$\frac{1}{2}(M+m+\frac{l}{r^2})v^2=(M-m)ghor,v^2=\frac{2(M-m)gh}{M+m+\frac{l}{r^2}}$.

The angular momentum of the mass M is Mvr and that of the mass m is mvr in the same direction. The angular momentum of the pulley is l$\omega$ $=\frac{lv}{r}$. The total angular momentum is $\sqrt{2(M-m)(M+m+\frac{l}{r^2})r^{2}gh}$