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The angular velocity of a particle moving along a circle of radius 50 cm is increased in 5 minutes from 100 revolutions per minute to 400 revolutions per minute. Find angular acceleration.

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$\frac{π}{60} rad s^{- 2}$

$\frac{2 π}{45} rad s^{- 2}$

$\frac{2 π}{15} rad s^{- 2}$

$\frac{π}{30} rad s^{- 2}$

answer is A.

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Detailed Solution

Given: Initial angular velocity, $ω_{1} = \frac{200 π}{60} r a d s^{- 1}$
Final angular velocity, $ω_{2} = \frac{800 π}{60} {rads}^{- 1}$

Time, t = 5 min = 300 sec

Angular acceleration,

$\begin{array}{rcl} α & = & \frac{ω_{2} - ω_{1}}{t} \\ = & \frac{(\frac{800 π}{60} - \frac{200 π}{60})}{300} \\ = & \frac{π}{30} rad / \sec^{2} \end{array}$

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