Question

A solid cylinder of mass $M$ and radius $R$ is mounted on frictionless horizontal axle so that it can freely rotate about its axis. A string of negligible mass is wrapped round the cylinder and a body of mass $m$ is hung from the string, what will be the rate of change of angular momentum immediately after the mass is allowed to fall?

• A. $\frac{2MmgR}{(M+2m)}$
• B. $\frac{MmgR}{(M+2m)}$
• C. $\frac{2Mmg}{(M+m)}$
• D. none of the above

Answer 1

The correct option is B $\frac{MmgR}{(M+2m)}$

Let the mass 'm' fall with acceleration 'a'. So its equation of motion is

$mg-T=ma$ (1)

Let the cylinder rotates with angular acceleration $\alpha$,

$\alpha =\frac{a}{R}$ (2)

I= Moment of inertia of the cylinder about center of mass of cylinder = $\frac{M{R}^2}{2}$

$TR=I\alpha$

$T=\frac{aM{R}^2}{2R^2}$

$T=\frac{Ma}{2}$ (3)

From (1) and (3), we can write

$mg-\frac{Ma}{2}=ma$

$a=\frac{2mg}{M+2m}$

So the rate of change of angular momentum = torque applied by the body

$I\alpha$ = $\frac{MRa}{2}$

$=\frac{2mgMR}{2(M+2m)}$

$=\frac{mgMR}{M+2m}$

So the answer is option B.