JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
Moment of inertia of a body about a given axis is \(1.5\, kg\, m^2\) Initially the body is at rest. In order to produce a rotational kinetic energy of \(1200\, J\), the angular acceleration of \(20\, rad/s^2\) must be applied about the axis of rotation for a duration of ......... \(\sec\).
- A \(2\)
- B \(5\)
- C \(2.5\)
- D \(3\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
\(KE = \frac{1}{2}I{\omega ^2} = 1200\left( {given} \right)\) \( \Rightarrow \omega = 40\,rad/s\) \(\,\,\,\,\,\,\,\omega = {\omega _0} + \alpha t\) \(\,\,\,\,\,\,\,40 = 0 + \left( {20} \right)\,t\) \( \Rightarrow t = 2\,\sec .\)
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