Question

A power drill spins up to 1800 rpm in half of a second (0.5s). If the drill initially starts with angular position 6 = 0, wha
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Answer #1

acceleration of the drill

a = w/t = ( 1800* 2* 3.14 / 60)^2 / 0.5

a = 70898.12 rad/s^2

angular displacement in 0.5 s

x = 0.5* 70898.12* 0.5^2

x = 8873.64 rad

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