Question

42. A carpet of mass $M$ made of inextensible material is rolled 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. Calculate the horizontal velocity of the axis of the cylindrical part of the carpet when its radius reduces to $R / 2$.

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

Answer 1

Answer: (B)

$\frac{5}{8}mgr$

Answer 2

The centre of mass of the whole carpet is originally at a height R above the floor. When the carpet unrolls itself and has a radius R/2, the centre of mass is at a height R/2. The mass left over unrolled is $\frac{M\pi\left(R / 2\right)^{2}}{\pi R^{2}}=\frac{M}{4}$ Decrease in potential energy $MgR-\left(\frac{M}{4}\right)g\left(\frac{R}{2}\right)=\frac{7}{8}MgR$Density $\sigma=\frac{M}{\pi r_{2}}=\frac{M_{1}}{\pi\left(\frac{r}{2}\right)^{2}}$ or $M_{1}=\frac{M}{4}$ Potential energy $U_{1}=$M g r and $U_{2}=M_{1} g \frac{r}{2}=\frac{M}{4} g \frac{r}{2}$ The decrease in the potential energy of the system $\Delta U =U_{1}-U_{2}$ $=M g r-\frac{M g r}{8}$ $=\frac{7 M g r}{8}$