The angular displacement of a body performing | vedclass

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(D) The angular displacement is given by $	heta = 5 \sin \frac{\pi t}{6}$.Angular velocity $\omega$ is the rate of change of angular displacement,def

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zero

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(D) The angular displacement is given by $	heta = 5 \sin \frac{\pi t}{6}$.Angular velocity $\omega$ is the rate of change of angular displacement,defined as $\omega = \frac{d	heta}{dt}$.Differentiating $	heta$ with respect to $t$:$\omega = \frac{d}{dt} \left( 5 \sin \frac{\pi t}{6} ight) = 5 \cdot \cos \left( \frac{\pi t}{6} ight) \cdot \frac{\pi}{6} = \frac{5\pi}{6} \cos \left( \frac{\pi t}{6} ight)$.At $t = 3 \ s$:$\omega = \frac{5\pi}{6} \cos \left( \frac{\pi 	imes 3}{6} ight) = \frac{5\pi}{6} \cos \left( \frac{\pi}{2} ight)$.Since $\cos \frac{\pi}{2} = 0$,we get $\omega = \frac{5\pi}{6} 	imes 0 = 0 \ rad/s$.

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

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