JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A pulley of radius \(1.5\,m\) is rotated about its axis by a force \(F =\left(12 t -3 t ^{2}\right)\,N\) applied tangentially (while \(t\) is measured in seconds). If moment of inertia of the pulley about its axis of rotation is \(4.5\,kg\,m ^{2}\), the number of rotations made by the pulley before its direction of motion is reversed, will be \(\frac{K}{\pi}\). The value of \(K\) is \(.....\)
- A \(18 \)
- B \(9\)
- C \(3\)
- D \(6\)
Answer & Solution
Correct Answer
(A) \(18 \)
Step-by-step Solution
Detailed explanation
\(\tau= I \alpha \Rightarrow\left(12 t -3 t ^{2}\right) 1.5=4.5 \alpha\) \(\alpha=4 t - t ^{2}\) \(\frac{ d \omega}{ dt }=4 t - t ^{2} \Rightarrow \omega=\int_{0}^{ t }\left(4 t - t ^{2}\right) dt\) \(\omega=2 t ^{2}- t ^{3} / 3\) For…
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