Question

To spin up a toy gyroscope you wrap a string around the 2 mm diameter shaft of the gyro 20 times (20 turns around the shaft). You actually put the end of the string through a hole in the shaft to anchor it. When you pull the other end of the string rapidly away from the gyro, the string automatically comes loose after it spins the gyro through 20 revolutions. If you pull with a constant force of 10 N on the string and reach an angular velocity of 3000 rpm's, how much work did you apply to the gyro? O 88 J 5.6 O 22J 1.4

Answer 1

c = circumference of the shaft = $\pi$ d = 3.14 x 2 x 10^-3 = 6.28 x 10^-3 m

n = number of loops = 20

d = distance for which the force was applied = n c = 20 (6.28 x 10^-3) = 0.1256 m

F = force = 10

W = work done = F d = 10 x 0.1256 = 1.256 =1.4