A particle is rotating in a circular path and | vedclass

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(C) The angular position is given by $	heta = \frac{5t^4}{40} - \frac{t^3}{3} = \frac{t^4}{8} - \frac{t^3}{3}$.Angular velocity $\omega$ is the rate

Vedclass · Accepted Answer

$130$

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(C) The angular position is given by $	heta = \frac{5t^4}{40} - \frac{t^3}{3} = \frac{t^4}{8} - \frac{t^3}{3}$.Angular velocity $\omega$ is the rate of change of angular position: $\omega = \frac{d	heta}{dt} = \frac{d}{dt}(\frac{t^4}{8} - \frac{t^3}{3}) = \frac{4t^3}{8} - \frac{3t^2}{3} = 0.5t^3 - t^2$.Angular acceleration $\alpha$ is the rate of change of angular velocity: $\alpha = \frac{d\omega}{dt} = \frac{d}{dt}(0.5t^3 - t^2) = 1.5t^2 - 2t$.At $t = 10 	ext{ s}$, substituting the value of $t$ in the expression for $\alpha$:$\alpha = 1.5(10)^2 - 2(10) = 1.5(100) - 20 = 150 - 20 = 130 	ext{ rad/s}^2$.

$A$ particle is rotating in a circular path and at any instant its motion can be described as $\theta = \frac{5t^4}{40} - \frac{t^3}{3}$. The angular acceleration of the particle after $10 \text{ s}$ is . . . . . . $\text{rad/s}^2$.

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