Question

A carpet of mass $M$ made of inextensible material is rolled along its length in the form of a cylinder of radius $R$ and kept along a rough floor. The carpet starts unrolling without sliding on the floor when a negligibly small push is given to it. The horizontal velocity of the axis of the cylindrical part of the carpet, when its radius is reduced to $\frac{R}{2}$ is :

• A. $\sqrt{\frac{14}{3}gR}$
• B. $\sqrt{\frac{7}{3}gR}$
• C. $\sqrt{gR}$
• D. $\sqrt{2gR}$

Answer 1

Answer: (A)

$\sqrt{\frac{14}{3}gR}$

The rolling motion of the carpet is due to the lowering of the centre of mass of the carpet in unrolling. The potential energy thus released is converted into kinetic energy of the rolling part, the flat part being at rest.

Initial potential energy: $PE=MgR$

Mass of the carpet of radius $\frac{R}{2}$ is proportional to$M(\frac{1}{2}{)}^2=\frac{M}{4}$

Hence, Potential energy at the point of interest is $P{E}_f=\left(\frac{M}{4}\right)g\left(\frac{R}{2}\right)=\frac{MgR}{8}$

Release of potential energy is $\Delta \left(PE\right)=mgR-\frac{MgR}{8}=\frac{7}{8}MgR$

The kinetic energy at this instant is given by $KE=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2=\frac{1}{2}\frac{M}{4}{v}^2+\frac{1}{2}\ast \frac{M}{4}\ast \frac{(\frac{R}{2}{)}^2}{2}\frac{v^2}{(\frac{R}{2}{)}^2}=\frac{3}{16}M{v}^2$

$\therefore \frac{3}{16}M{v}^2=\frac{7}{8}MgR$

$\Rightarrow v=\sqrt{\frac{14}{3}gR}$