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

The correct option is 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: $P.E_i=MgR$

$\because M=d\times \pi R^2$ , when $R$ becomes $\frac{R}{2}$ then,

Mass of the carpet of radius $\frac{R}{2}=\frac{M}{4}$

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 $\Delta (P.E)=MgR-\frac{MgR}{8}=\frac{7}{8}MgR$

The change in kinetic energy at this instant is given by,

$\Delta K.E=\frac{1}{2}\frac{M}{4}{v}^2+\frac{1}{2}I{\omega }^2$ [initially at rest]

Since, $I=\frac{M}{4}\frac{{\left(\frac{R}{2}\right)}^2}{2}$ and $\omega =\frac{v}{R/2}$

Substituting the values in the above equation, we get

$\Delta K.E=\frac{1}{2}\frac{M}{4}{v}^2+\frac{1}{2}\frac{M}{4}\frac{{\left(\frac{R}{2}\right)}^2}{2}\frac{v^2}{{\left(\frac{R}{2}\right)}^2}$

$\Rightarrow K.E_f=\frac{3}{16}M{v}^2$

Now, from conservation of mechanical energy principle,
$\Delta P.E=\Delta K.E$

Substituting the values, we get

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

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


Hence, option (c) is correct answer.