Physics - Rotation Question with Solution | TestHub

Let the initial mass of the rolled carpet be $M$ and its radius be $R$. The center of mass of the cylindrical roll is at a height $R$ from the floor.

Initial potential energy of the carpet, $U_i=MgR$

When the carpet is unrolled to a radius $R/2$, the mass of the remaining rolled part becomes $M^{\prime }$. Since the mass of the roll is proportional to its cross-sectional area (assuming uniform density and constant length):

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

The center of mass of this remaining rolled part is at a height of $R/2$ from the floor. The unrolled part of the carpet lies flat on the floor, so its center of mass is at a height of approximately zero.

Final potential energy of the carpet, $U_f=M^{\prime }g\left(\frac{R}{2}\right)+(M-M^{\prime })g(0)$

$U_f=\left(\frac{M}{4}\right)g\left(\frac{R}{2}\right)=\frac{MgR}{8}$

Change in potential energy, $\Delta U=U_i-U_f=MgR-\frac{MgR}{8}=\frac{7}{8}MgR$

Answer: $\frac{7}{8}\mathrm{MgR}$