Rotational Motion

JEE Main level, numerical answer, medium

205 of 280

Question 205 of 280

A solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about axis of rotation of the sphere to the total energy of moving sphere is π:22\pi :22 then, the value of its angular speed will be_________rad{s}^{-1}.

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Answer: 4

The angular momentum is given by
L=ICMω=2MR2ω5L={I}_{CM}\omega =\dfrac{2M{R}^{2}\omega }{5}
The total kinetic energy is
E=12(Mv2+2MR2ω25)E=\dfrac{1}{2}(M{v}^{2}+\dfrac{2M{R}^{2}{\omega }^{2}}{5})
The ratio of angular momentum and kinetic energy is
LE=ICMω12Mv2+12⋅25MR2ω2=25MR2ω710MR2ω2\dfrac{L}{E}=\dfrac{{I}_{CM}\omega }{\dfrac{1}{2}M{v}^{2}+\dfrac{1}{2}\cdot \dfrac{2}{5}M{R}^{2}{\omega }^{2}}=\dfrac{\dfrac{2}{5}M{R}^{2}\omega }{\dfrac{7}{10}M{R}^{2}{\omega }^{2}}
⇒π22=47⋅1ω\Rightarrow \dfrac{\pi }{22}=\dfrac{4}{7}\cdot \dfrac{1}{\omega }⇒ω=4rads−1\Rightarrow \omega =4rad{s}^{-1}

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