A 3.65 kg steel block is spun around a motor at 44.0 rpm. The angular acceleration of the block is -1.3 rads/s2 and it takes 3.54 s for the block to come to a complete stop. What is the angular displacement while the block slows to a stop?

Answer 1

Vo = 44rev/min. * 6.28rad/rev * 1min/60s. = 4.61 rad/s.

V^2 = Vo^2 + 2a*D.
0 = 4.61^2 - 2.6D, D = ?.