Rotational Motion

JEE Main level, numerical answer, medium

154 of 280

Question 154 of 280

For rolling spherical shell, the ratio of rotational kinetic energy and total kinetic energy is x5\dfrac{x}{5}. The value of xx is _____.

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Answer: 2

The rotational kinetic energy is given by
Krot=Iω22=13MR2ω2{K}_{rot}=\dfrac{I{\omega }^{2}}{2}=\dfrac{1}{3}M{R}^{2}{\omega }^{2}.
The total kinetic energy is given by
Ktot=12mv2+13MR2ω2=mR2ω22+13MR2ω2=56MR2ω2{K}_{tot}=\dfrac{1}{2}m{v}^{2}+\dfrac{1}{3}M{R}^{2}{\omega }^{2} =\dfrac{m{R}^{2}{\omega }^{2}}{2}+\dfrac{1}{3}M{R}^{2}{\omega }^{2} =\dfrac{5}{6}M{R}^{2}{\omega }^{2}
Thus, the ratio is
KrotKtot=13MR2ω256MR2ω2=25\dfrac{{K}_{rot}}{{K}_{tot}}=\dfrac{\dfrac{1}{3}M{R}^{2}{\omega }^{2}}{\dfrac{5}{6}M{R}^{2}{\omega }^{2}}=\dfrac{2}{5}
Comparing with the given value,
x5=25⇒x=2\dfrac{x}{5}=\dfrac{2}{5} \Rightarrow x=2

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