Rotational Motion

JEE Main level, numerical answer, medium

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Question 158 of 280

A pulley of radius 1.5m1.5m is rotated about its axis by a force F=(12t−3t2)NF=\left(12t-3{t}^{2}\right)N applied tangentially (while tt is measured in seconds). If moment of inertia of the pulley about its axis of rotation is 4.5kgm24.5kg{m}^{2}, the number of rotations made by the pulley before its direction of motion is reversed, will be Kπ\dfrac{K}{\pi }. The value of KK is _____ .

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Answer: 18

Torque of pulley about its axis is τ=Iα\tau =I\alpha, where, II is moment of inertia and α\alpha is angular acceleration.
Torque of a force about its axis is τ=FR\tau =FR, here, RR is radius of pulley.
Then, (12t−3t2)1.5=4.5α\left(12t-3{t}^{2}\right)1.5=4.5\alpha
⇒α=4t−t2\Rightarrow \alpha =4t-{t}^{2}
Angular acceleration in terms of angular velocity is
⇒α=dωdt=4t−t2\Rightarrow \alpha =\dfrac{d\omega }{dt}=4t-{t}^{2}
⇒ω=∫0t(4t−t2)dt\Rightarrow \omega ={\int }_{0}^{t}\left(4t-{t}^{2}\right)dt
⇒ω=2t2−t33\Rightarrow \omega =2{t}^{2}-\dfrac{{t}^{3}}{3}For ω=0,2t2−t33=0⇒t2(2−t3)=0\omega =0,2{t}^{2}-\dfrac{{t}^{3}}{3}=0\Rightarrow {t}^{2}\left(2-\dfrac{t}{3}\right)=0
⇒t=0or6s\Rightarrow t=0\text{or}6s
Now, using ω=dθdt\omega =\dfrac{d\theta }{dt}, we have
dθdt=2t2−t33⇒θ=∫06(2t2−t33)dt\dfrac{d\theta }{dt}=2{t}^{2}-\dfrac{{t}^{3}}{3} \Rightarrow \theta ={\int }_{0}^{6}\left(2{t}^{2}-\dfrac{{t}^{3}}{3}\right)dt
=[2t33−t412]06={\left[\dfrac{2{t}^{3}}{3}-\dfrac{{t}^{4}}{12}\right]}_{0}^{6}
=63(23−612)=63(8−612)={6}^{3}\left(\dfrac{2}{3}-\dfrac{6}{12}\right)={6}^{3}\left(\dfrac{8-6}{12}\right)=636=36=\dfrac{{6}^{3}}{6}=36
Number of revolutions =362π=18π=\dfrac{36}{2\pi }=\dfrac{18}{\pi }
∴K=18∴K=18

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