A simple pendulum of length 2 m is given a horizontal push through… | MHT CET 2023 (9 May, Shift 2)

Q12MHT CET 2023 · 9 MAY · SHIFT 2Medium

Vertical Circle: Speed & Tension

A simple pendulum of length $2\text{ m}$ is given a horizontal push through angular displacement of $60^\circ$ . If the mass of bob is 200 gram, the angular velocity of the bob will be (Take Acceleration due to gravity $= 10\text{ m/s}^2$ ) $\left(\sin 30^\circ = \cos 60^\circ = 0.5, \cos 30^\circ = \sin 60^\circ = \sqrt{3}/2\right)$

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Answer: (c)

Bob rises through $60^\circ$ : $h = L(1 - \cos 60^\circ) = 1$ m
Speed at the lowest point: $v = \sqrt{2gh} = \sqrt{20}$ m/s
$\omega = \frac{v}{L} = \frac{\sqrt{20}}{2} = \sqrt{5} \approx 2.24$ rad/s
This equals $2\sqrt{1.25}$ rad/s, which is not among the options.

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Vertical Circle: Speed & Tension question from MHT CET 2023 (9 May, Shift 2)

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