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?
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 = ?.
To find the angular displacement of the steel block while it slows to a stop, we can use the equation:
angular displacement = angular velocity * time + (1/2) * angular acceleration * time^2
First, let's convert the given angular velocity from rpm to radians per second (rad/s). One revolution is equal to 2π radians.
angular velocity = 44.0 rpm * (2π rad/1 rev) * (1 min/60 s)
angular velocity = 44.0 * 2π * (1/60) rad/s
Now, we can substitute the values into the equation:
angular displacement = (44.0 * 2π * (1/60)) * 3.54 + (1/2) * (-1.3) * (3.54)^2
Calculating the above expression will give us the angular displacement.