A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of 25 N m for 40 s, the speed increases to 2100 rpm. The diameter of the disk is m.
Answer
40
Solution
Change in angular speed: = (2100 - 1800) rpm = 300 2 60 = 10 rad/s
Angular acceleration: = t = 10 40 = 4 rad/s2
= I I = 25 /4 = 100 kg m2
For a thin disk about a diameter: I = 14MR2
14 1 R2 = 100 R = 20 m
Diameter = 2R = 40 m
Answer: 40