The angular displacement of body performing circular motion is given by θ = 5sin (π t)/(6). The angular velocity of the body at t = 3 second will be [sin π/2 = 1, cos π/2 = 0] (a) 5 rad/s (b) 1 rad/s (c) 2.5 rad/s (d) zero rad/s

Answer: (d). ω = (dθ)/(dt) = 5 · π/6cos (π t)/(6). At t = 3 s: cos π/2 = 0, so ω = 0. So the answer is (d) zero.

The angular displacement of body performing circular motion is given… | MHT CET 2021 (22 Sep, Shift 1)

Q30MHT CET 2021 · 22 SEP · SHIFT 1Easy

Angular Variables

The angular displacement of body performing circular motion is given by $\theta = 5\sin\frac{\pi t}{6}$ . The angular velocity of the body at $t = 3$ second will be $\left[\sin\frac{\pi}{2} = 1, \cos\frac{\pi}{2} = 0\right]$

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Answer: (d)

$\omega = \frac{d\theta}{dt} = 5 \cdot \frac{\pi}{6}\cos\frac{\pi t}{6}$ .
At $t = 3$ s: $\cos\frac{\pi}{2} = 0$ , so $\omega = 0$ .
So the answer is (d) zero.

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