Question

Part complete a 1.15-kg grinding wheel 22.0 cm in diameter is spinning counterclockwise at a rate of 20.0 revolutions per second. when the power to the grinder is turned off, the grinding wheel slows with constant angular acceleration and takes 80.0 s to come to a rest. (a) what was the angular acceleration (in rad/s2) of the grinding wheel as it came to rest if we take a counterclockwise rotation as positive?

Answer 1

initial angular speed is given as 20 rev/second

so it is given as

`w = 2 \pi f`

`w = 2 \pi 20 = 40\pi`

now it comes to rest after 80 s

now we can use kinematics to find angular deceleration

`w = w_o + \alpha * t`

`0 = 40\pi + \alpha * 80`

`\alpha = -1.57 rad/s^2`

so angular deceleration is 1.57 rad/s^2