A ball is spun with angular acceleration $\alpha=6 \mathrm{t}^{2}-2 \mathrm{t}$ where t is in second and $\alpha$ is in rads $^{-2}$. At $\mathrm{t}=0$, 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:

(B) $\frac{t^{4}}{2}-\frac{t^{3}}{3}+10 t+4$

(A) $\frac{3}{2} t^{4}-t^{2}+10 t$

(C) $\frac{2 t^{4}}{3}-\frac{t^{3}}{6}+10 t+12$

(D) $2 t^{4}-\frac{t^{3}}{2}+5 t+4$

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Kinematics

A ball is spun with angular acceleration

α = 6 t^{2} - 2 t

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) $\frac{3}{2} t^{4} - t^{2} + 10 t$
(B) $\frac{t ^{4}}{2} - \frac{t ^{3}}{3} + 10 t + 4$ ✓ Correct
(C) $\frac{2 t ^{4}}{3} - \frac{t ^{3}}{6} + 10 t + 12$
(D) $2 t^{4} - \frac{t ^{3}}{2} + 5 t + 4$

Answer: (B)

Solution

Given

α = \frac{d ω}{d t} = 6 t^{2} - 2 t

. Integrating with initial condition

ω (0) = 10

ω = 10 + 2 t^{3} - t^{2}

. Since

ω = \frac{d θ}{d t}

, integrating again with

θ (0) = 4

θ = 4 + 10 t + \frac{t ^{4}}{2} - \frac{t ^{3}}{3}

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Single correct option

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