Rotational Motion

JEE Main level, single correct, medium

140 of 280

Question 140 of 280

A particle is rotating in a circular path and at any instant its motion can be described as θ=5t440−t33\theta = \dfrac{5t^4}{40} - \dfrac{t^3}{3}. The angular acceleration of the particle after 1010 seconds is _______ rad/s2^2.

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Answer: (C) 130130

Given θ=5t440−t33=t48−t33\theta = \dfrac{5t^4}{40} - \dfrac{t^3}{3} = \dfrac{t^4}{8} - \dfrac{t^3}{3}

Angular velocity ω=dθdt=4t38−3t23=t32−t2\omega = \dfrac{d\theta}{dt} = \dfrac{4t^3}{8} - \dfrac{3t^2}{3} = \dfrac{t^3}{2} - t^2

Angular acceleration α=dωdt=3t22−2t\alpha = \dfrac{d\omega}{dt} = \dfrac{3t^2}{2} - 2t

At t=10t = 10 s, α=3(10)22−2(10)\alpha = \dfrac{3(10)^2}{2} - 2(10)

α=150−20=130\alpha = 150 - 20 = 130 rad/s2^2

Answer: 130130

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