question_answer

A carpet of mass M is rolled along its length so as to form a cylinder of radius R and is kept on a rough floor. When a negligibly small push is given to the cylindrical carpet, it starts unrolling itself without sliding on the floor. Calculate horizontal velocity of cylindrical part of the carpet when its radius reduces to \[R/2\].

Answer:

                If \[\rho \] is cross sectional density of the cylindrical carpet, then in the initial Position, its mass, \[M=\pi {{R}^{2}}\rho \], and in the final position, when radius reduces to R/2, its mass, \[m=\pi {{\left( R/2 \right)}^{2}}\rho =\frac{\pi {{R}^{2}}\rho }{4}=\frac{M}{4}\] Decrease in P.E. of carpet = \[MgR-mgr=MgR-\frac{M}{4}g.\frac{R}{2}=\frac{7}{8}MgR\]                                               ... (i) Increase in total kinetic energy of carpet = \[{{K}_{t}}+{{K}_{r}}=\frac{1}{2}m{{\upsilon }^{2}}+\frac{1}{2}I{{\omega }^{2}}\] \[=\frac{1}{2}m{{\upsilon }^{2}}+\frac{1}{2}\left( \frac{1}{2}m{{r}^{2}} \right){{\omega }^{2}}=\frac{1}{2}m{{\upsilon }^{2}}+\frac{1}{4}m{{\upsilon }^{2}}=\frac{3}{4}m{{\upsilon }^{2}}=\frac{3}{4}\left( \frac{M}{4} \right){{\upsilon }^{2}}=\frac{3}{16}M{{\upsilon }^{2}}\]                                        ... (ii) Assuming that there is no stray loss of energy, we get from (i) and (ii) \[\frac{3}{16}M{{\upsilon }^{2}}=\frac{7}{8}MgR\] \[\upsilon =\sqrt{\frac{14}{3}gR}\]