Question

A carpet of mass $M$ is rolled along its length in the form of a cylinder of radius $R$ and kept on a rough floor. The decrease in potential energy, if the carpet is unrolled without sliding to a radius $\frac{R}{2}$ is equal to

• A. $\frac{1}{2}MgR$
• B. $\frac{3}{4}MgR$
• C. $\frac{5}{8}MgR$
• D. $\frac{7}{8}MgR$

Answer 1

The correct option is D

$\frac{7}{8}MgR$

The entire mass $M$ of the carpet can be assumed to be concentrated at its centre of mass which is originally at a height $R$ above the floor. So its original potential energy (P.E.) $=MgR$.

When the carpet is unrolled to a radius $\frac{R}{2}$, its centre of mass will be at a height $\frac{R}{2}$ above the floor, but the mass left over unrolled is

$m=\frac{M(R/2{)}^2}{R^2}=\frac{M}{4}$

and its P.E. $=mg\frac{R}{2}=\frac{M}{4}\times g\times \frac{R}{2}=\frac{MgR}{8}.$

$\therefore$ Drop in P.E. $=MgR-\frac{MgR}{8}=\frac{7}{8}MgR$