A ball is spun with angular acceleration α = 6t2 - 2t where t is in second and α is in rads ⁻². At t = 0, the ball has angular velocity of 10 rads ⁻¹ and angular position of 4 rad. The most appropriate expression for the angular position of the ball is: (a) 3/2t⁴ - t² + 10t (b) t⁴/2 - t³/3 + 10t + 4 (c) 2t⁴/3 - t³/6 + 10t + 12 (d) 2t⁴ - t³/2 + 5t + 4

Answer: (b). ω = ∫ (6t² - 2t) dt = 2t³ - t² + C; ω(0) = 10 ⇒ C = 10. θ = ∫ (2t³ - t² + 10) dt = t⁴/2 - t³/3 + 10t + C'; θ(0) = 4 ⇒ C' = 4 So the answer is (b) t⁴/2 - t³/3 + 10t + 4.

A ball is spun with angular acceleration α = 6t2 - 2t where t is in… | JEE Main 2022 (28 Jun, Shift 2)

Q31JEE Main 2022 · 28 JUN · SHIFT 2Medium

Rotational Kinematics & Dynamics

A ball is spun with angular acceleration $\alpha$ = 6t2 $-$ 2t where t is in second and $\alpha$ is in rads ${}^{-2}$ . At t = 0, the ball has angular velocity of 10 rads ${}^{-1}$ and angular position of 4 rad. The most appropriate expression for the angular position of the ball is:

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Answer: (b)

$\omega = \int (6t^2 - 2t)\,dt = 2t^3 - t^2 + C$ ; $\omega(0) = 10 \Rightarrow C = 10$ .
$\theta = \int (2t^3 - t^2 + 10)\,dt = \frac{t^4}{2} - \frac{t^3}{3} + 10t + C'$ ; $\theta(0) = 4 \Rightarrow C' = 4$
So the answer is (b) $\frac{t^4}{2} - \frac{t^3}{3} + 10t + 4$ .

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