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

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$π / 15 {ms}^{- 2}$

$π / 60 {ms}^{- 2}$

$60 {ms}^{- 2}$

$π / 30 {ms}^{- 2}$

answer is D.

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Detailed Solution

$\begin{matrix} α = (ω_{2} - ω_{1}) / t = (400 - 100) / 5 = 60 rev \min^{- 2} \\ = \frac{60 \times 2 π}{(60)^{2}} = \frac{2 π}{60} {rads}^{- 2} \\ a_{t} = αr = \frac{2 π}{60} \times \frac{50}{100} = \frac{π}{60} {ms}^{- 2} \end{matrix}$

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