Rotational Motion

JEE Main level, single correct, medium

176 of 280

Question 176 of 280

Moment of inertia of a body about a given axis is 1.5kgm2.1.5kg{m}^{2}. Initially the body is at rest. In order to produce a rotational kinetic energy of 1200J,1200J, the angular acceleration of 20rad/s220rad/{s}^{2} must be applied about the axis for a duration of:

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Answer: (B) 2s2s

We know that,
Rotational kinetic energy, K=12Iω2K=\dfrac{1}{2}I{\omega }^{2}
⇒12Iω2=1200\Rightarrow \dfrac{1}{2}I{\omega }^{2}=1200
⇒12×1.5×ω2=1200\Rightarrow \dfrac{1}{2}\times 1.5\times {\omega }^{2}=1200
⇒ω=40rads−1\Rightarrow \omega =40rad{s}^{-1}
We also know that, ω=αt\omega =\alpha t
⇒t=4020=2s\Rightarrow t=\dfrac{40}{20}=2s

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