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 tension in the string?

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

Answer 1

The correct option is A $\frac{Mmg}{(M+2m)}$

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 tension of the string is

$T=\frac{M}{2}\times \frac{2mg}{M+2m}$

$T=\frac{Mmg}{M+2m}$