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(C) Given the angular velocity $\omega = 	heta^2 + 2	heta$.The angular acceleration $\alpha$ is given by the relation $\alpha = \omega \frac{d\omega

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) Given the angular velocity $\omega = 	heta^2 + 2	heta$.The angular acceleration $\alpha$ is given by the relation $\alpha = \omega \frac{d\omega}{d	heta}$.First,differentiate $\omega$ with respect to $	heta$:$\frac{d\omega}{d	heta} = \frac{d}{d	heta}(	heta^2 + 2	heta) = 2	heta + 2$.Now,substitute $\omega$ and $\frac{d\omega}{d	heta}$ into the expression for $\alpha$:$\alpha = (	heta^2 + 2	heta)(2	heta + 2)$.At $	heta = 1 \ rad$:$\alpha = (1^2 + 2(1))(2(1) + 2) = (1 + 2)(2 + 2) = 3 	imes 4 = 12 \ rad/sec^2$.

If the angular velocity of a disc depends on the angle rotated $\theta$ as $\omega = \theta^2 + 2\theta$,then its angular acceleration $\alpha$ at $\theta = 1 \ rad$ is ......... $rad/sec^2$.

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