A thin hollow cylinder,open at both ends,slid | vedclass

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(B) Case $1$: When the hollow cylinder slides without rotating,it possesses only translational kinetic energy.$K_{slide} = \frac{1}{2}mv^2$Case $2$: W

Vedclass · Accepted Answer

$1 : 2$

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(B) Case $1$: When the hollow cylinder slides without rotating,it possesses only translational kinetic energy.$K_{slide} = \frac{1}{2}mv^2$Case $2$: When it rolls without slipping,it possesses both translational and rotational kinetic energy.The total kinetic energy is given by $K_{roll} = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$,where $k$ is the radius of gyration.For a thin hollow cylinder,the moment of inertia $I = mR^2$,so $mk^2 = mR^2$,which implies $\frac{k^2}{R^2} = 1$.Substituting this into the formula: $K_{roll} = \frac{1}{2}mv^2(1 + 1) = mv^2$.Now,the ratio of kinetic energies is:$\frac{K_{slide}}{K_{roll}} = \frac{\frac{1}{2}mv^2}{mv^2} = \frac{1}{2}$.Thus,the ratio is $1 : 2$.

$A$ thin hollow cylinder,open at both ends,slides without rotating and then rolls without slipping with the same speed. The ratio of kinetic energies in the two cases will be:

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