A flywheel rotates with a uniform angular acceleration. Its angular velocity increases from $20\pi\ rads^{- 1}$ to $40\ \pi\ rads^{- 1}$ in 10 s. how many rotations did it make in this period?

Question

Dr. Abubacker · Accepted Answer

(d) As $\omega_{2} = \omega_{1} + \alpha t\ \ 	herefore 40\pi = 20\pi + \alpha 	imes 10$
or $\alpha = 2\pi\ rads^{- 1}$
From, $\omega_{2}^{2} - \omega_{1}^{2} = 2\alpha	heta$
$(40\pi)^{2} - (20\pi)^{2} = 2 	imes 2\pi	heta,\  \Rightarrow 	heta = \frac{1200\pi^{2}}{4\pi} = 300\pi$
Number of rotations completed
$= \frac{	heta}{2\pi} = \frac{300}{2\pi} = 150$

Q. A flywheel rotates with a uniform angular acceleration. Its angular velocity increases from $20\pi\ rads^{- 1}$ to $40\ \pi\ rads^{- 1}$ in 10 s. how many rotations did it make in this period?

Explanation

Reviewed by Dr. Abubacker

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