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. If R = 1.05 m then calculate the horizontal velocity of the axis (in m/s) of the cylindrical part of the carpet when its radius reduces to R/2. (Take g = 10 m/s²) 

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 = (3/4) M.

The mechanical energy of the system will remains constant.

The part of the carpet spread on the ground will have no kinetic energy. 

According to conservation of energy, $\mathrm{\Delta K}+\mathrm{\Delta U}=0$

$\Rightarrow \left({\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}\right)+\mathrm{\Delta U}=0$              …(i)

As the roll of the carpet is not sliding, we can consider that the roll of carpet as a rolling cylinder. We can calculate the kinetic energy of the moving cylindrical roll by the relation

$\mathrm{K}=(1+\mathrm{\beta })\frac{{\mathrm{mv}}^2}{2}=\left(1+\frac{1}{2}\right)\frac{{\mathrm{mv}}^2}{2}=\frac{3{\mathrm{mv}}^2}{4}$

M = mass of the roll at any time

Hence the kinetic energy of the roll when its radius is R/2

${\mathrm{K}}_{\mathrm{f}}=\frac{3(\mathrm{M}/4){\mathrm{v}}^2}{4}=\frac{3{\mathrm{Mv}}^2}{16}$

Change in kinetic energy, $\mathrm{\Delta K}={\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}=\frac{3{\mathrm{Mv}}^2}{16}$

Change in gravitational potential energy.

$\mathrm{\Delta U}=-\left(\frac{3}{4}\mathrm{MgR}+\frac{1}{4}\mathrm{Mg}\frac{\mathrm{R}}{2}\right)=-\frac{7}{8}\mathrm{MgR}$

Finally substituting $\mathrm{\Delta U}\text{ and }\mathrm{\Delta K}$ in the equation (i), we have

$\Rightarrow \frac{3{\mathrm{Mv}}^2}{16}+\left(-\frac{7}{8}\mathrm{MgR}\right)=0$

$\Rightarrow \mathrm{v}=\sqrt{\frac{14}{3}\mathrm{gR}}=\sqrt{\frac{14}{3}\times 10\times 1.05}=7 \mathrm{m}/\mathrm{s}$