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) \[\frac{3}{16}M{{\upsilon }^{2}}=\frac{7}{8}MgR\] \[\upsilon =\sqrt{\frac{14}{3}gR}\]
You need to login to perform this action.
You will be redirected in
3 sec