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Pure Rolling

A thin ring o...

Question

A thin ring of mass $1 k g$ and radius $1 m$ is rolling at a speed of $1 m s^{- 1}$ . Its kinetic energy is :

2 J

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1 J

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0.5 J

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Z e r o

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Solution

The correct option is A $1 J$
The kinetic energy due to translation motion is $m v^{2} / 2 = 1 \times 1^{2} / 2 = 0.5 J$
The Moment of Intertia is $I = m r^{2} = 1$
The angular velocity is $ω = v / r = 1 s e c^{- 1}$
The kinetic energy due to rotational motion is $I ω^{2} / 2 = 1 \times 1^{2} / 2 = 0.5 J$
Total K.E. is $0.5 + 0.5 = 1 J$

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