A particle is rotating in a circular path and at any instant its motion can be described as $ heta

Question

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

Accepted Answer

Sol. [Key Concept: Angular acceleration $\alpha = \frac{d^{2}	heta}{{dt}^{2}}$; evaluate at $t = 10$.]$$	heta = \frac{5t^{4}}{40} - \frac{t^{3}}{3} $$$$\frac{d	heta}{dt} = \frac{5}{40}\left( 4t^{3} ight) - \frac{3t^{2}}{3} = \frac{t^{3}}{2} - t^{2} $$$$\alpha = \frac{d^{2}	heta}{{dt}^{2}} = \frac{3t^{2}}{2} - 2t $$at $t = 10$$$\alpha = \frac{3}{2}(10)^{2} - 2 	imes 10 = 130rad/\sec^{2} $$

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