Asked by charlie
A 5.35 kg steel block is spun around a motor at 36.5 rpm. The angular acceleration of the block is -2.14 rads/s^2 and it takes 1.79 s for the block to come to a complete stop. What is the angular displacement while the block slows to a stop?
Answers
Answered by
Scott
d = 1/2 a t^2 = 1/2 * 2.14 * 1.79^2
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