Rotational Motion

JEE Main level, single correct, medium

136 of 280

Question 136 of 280

A pulley of radius 2 m2 \mathrm{~m} is rotated about its axis by a force F=(20t−5t2)\mathrm{F}=\left(20 \mathrm{t}-5 \mathrm{t}^2\right) Newton (where t\mathrm{t} is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation made by the pulley before its direction of motion if reversed, is :

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Answer: (A) more than 3 but less than 6

r×F=l×α\mathrm{r} \times \mathrm{F}=\mathrm{l} \times \alpha
2(20t−5t2)=10α⇒α=4t−t22\left(20 \mathrm{t}-5 \mathrm{t}^2\right)=10 \alpha \Rightarrow \alpha=4 \mathrm{t}-\mathrm{t}^2
dωdt=4t−t2\frac{\mathrm{d} \omega}{\mathrm{dt}}=4 \mathrm{t}-\mathrm{t}^2
 dω=(4t2−t2)dt\mathrm{~d} \omega=\left(4 \mathrm{t}^2-\mathrm{t}^2\right) \mathrm{dt}
ω=2t2−t33\omega=2 \mathrm{t}^2-\frac{\mathrm{t}^3}{3} (on integration)
ω=0⇒t=6 s\omega=0 \Rightarrow \mathrm{t}=6 \mathrm{~s}

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