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(C) Given the angular velocity $\omega = 	heta^2 + 2	heta$.The angular acceleration $\alpha$ is defined as $\alpha = \frac{d\omega}{dt}$.Using the c

Vedclass · Accepted Answer

$12$

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(C) Given the angular velocity $\omega = 	heta^2 + 2	heta$.The angular acceleration $\alpha$ is defined as $\alpha = \frac{d\omega}{dt}$.Using the chain rule,we can write $\alpha = \frac{d\omega}{d	heta} \cdot \frac{d	heta}{dt}$.Since $\frac{d	heta}{dt} = \omega$,we have $\alpha = \omega \frac{d\omega}{d	heta}$.First,calculate $\frac{d\omega}{d	heta} = \frac{d}{d	heta}(	heta^2 + 2	heta) = 2	heta + 2$.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 	ext{ rad}$:$\alpha = (1^2 + 2(1))(2(1) + 2) = (1 + 2)(2 + 2) = 3 	imes 4 = 12 	ext{ 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 $\theta = 1 \text{ rad}$ is ......... $\text{rad/sec}^2$.

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