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A thin hollow cylinder open at both ends:
$(i)$ Slides without rotating
$(ii)$ Rolls without slipping, with the same speed
$1:1$
$4:1$
$1:2$
$2:1$
Solution
Given$:$ a thin hollow cylinder open at both ends slides without rotating and then rolls without slipping with the same speed.
To find the ratio of kinetic energies in the two cases Solution$:$
When hollow cylinder slides without rolling, it posses only transnational Kinetic energy $K_{r}=\frac{1}{2} m v^{2}$
When it rolls slipping, it posses both type of Kinetic energy
$K_{N}=\frac{1}{2} m v^{2}\left[1+\frac{K^{2}}{R^{2}}\right]$
Therefore
$\frac{K_{r}}{K_{N}}=\frac{\frac{1}{2} m v^{2}}{\frac{1}{2} m v^{2}\left[1+\frac{K^{2}}{R^{2}}\right]}$
and for hollow cylinder $\frac{K^{2}}{R^{2}}=1$
So the above equation becomes,
$\Longrightarrow \frac{K_{r}}{K_{N}}=\frac{1}{2}$
Hence the ratio of kinetic energies in the two cases is $1: 2$
