Question

You tie the loose end of a 0.1 kg yo-yo string to your finger and then release the yo-yo so that it spins down toward the ground (the yo-yo is released from rest and the end of the string tied to your finger remains motionless). After the yo-yo falls a distance of 0.9 m, it has a translational speed of 4 m/s and an angular speed of 180 rad/s. What is the moment of inertia of the yo-yo

Answer 1

The answer is "
`5.06 * 10^(-6) \ kg \ m^2`"

Step-by-step explanation:

`\to E_1=0..............(i)\\\\\to E_2= (mV^2)/(2) +(Iw^2)/(2) - mgh.............(ii)\\\\ \Delta E=0\\\\\to mgh= (mV^2)/(2) +(Iw^2)/(2) \\\\ \to 2 \ mgh= mV^2 +Iw^2\\\\ \to 2 \ mgh- mV^2 =Iw^2\\\\ \to m(2gh- V^2) =Iw^2\\\\ \to I= (m(2gh- V^2))/(w^2)`

`= 5.06 * 10^(-6) \ kg \ m^2`