Question

Tim attaches the string of a 0.10 kg yo-yo to his finger and releases the yo-yo from rest. As the yo-yo spins down, Tim keeps his finger motionless which allows the yo-yo to be treated the same as a freely rolling object. After the yo-yo drops a distance of 0.90 m, it has a translational speed of 4.0 m/s and an angular speed of 180 rad/s. What is the moment of inertia of the yo-yo? Air resistance and friction are negligible. A B C D E 4.40e-6 kgm 9.97e-7 kg.m 5.06e-6 kg•m? 6.00e-6 kg.m 4.51e-5 kg•m?

Answer 1

C) 5.06e-6 kg.m^2   is the correct answer.

m = 0.1 kg

h = 0.9 m

v = 4.0 m/s
w = 180 rad/s

let I is the moment of inertia of the yo-yo

Apply conservation of energy

gain in kinetic energy = loss of potential energy

(1/2)*m*v^2 + (1/2)*I*w^2 = m*g*h

(1/2)*0.1*4^2 + (1/2)*I*180^2 = 0.1*9.8*0.9

0.8 + 16200*I = 0.882

I = (0.882 - 0.8)/16200

= 5.06*10^-6 kg.m^2