Question

A carpet of mass $M$ made of inextensible material is along its length in the form of a cylinder of radius $R$ and is kept on a rough floor. The carpet starts unrolling without sliding on the floor when a negligibly small push is given to it. If the horizontal velocity of the axis of the cylindrical part of the carpet when its radius reduces to $R/2$ is $\sqrt{\frac{x}{3}gR}$, find the value of $x$.

Answer 1

The figure shows the forces acting on the carpet in two cases:
i. when the radius of the roll is $R$
ii. when the radius is $R/2$
Mass of the roll of radius $R/2=M/4$
(Mass is proportional to the area of cross section)
The mass of the part of the carpet spread on the floor$=\left(3/4\right)M$. There has to be frictional force that is too static in nature on the carpet once its starts unrolling. Otherwise, why should the centre of mass of the system (carpet of mass $m$) move towards the right.
Here, $MgR=\left(\frac{M}{4}\right)g\left(\frac{R}{2}\right)+\frac{1}{2}\left(\frac{M}{4}\right)v_{CM}^2+\frac{1}{2}\left(\frac{1}{2}\frac{M}{4}{\left(\frac{R}{2}\right)}^2\right){\omega }^2$
$\Rightarrow \frac{7}{8}MgR=\frac{M}{8}{v}_{CM}^2+\frac{1}{16}M{\left(\frac{R}{2}\right)}^2{\left(\frac{v_{CM}}{R/2}\right)}^2(v_{CM}=\frac{R}{2}\omega )$
$\Rightarrow \frac{7}{8}MgR=\frac{3}{8}m{v}_{CM}^2\Rightarrow v_{CM}=\sqrt{\frac{14}{3}gR}$