Yo-yo man releases a yo-yo from rest and allows it to drop, as he keeps the top end of the string stationary. the mass of the yo-yo is 0.056 kg, its moment of inertia is 2.9×10−5kg⋅m2 and the radius, r, of the axle the string wraps around is 0.0064 m.

through what height must the yo-yo fall for its linear speed to be 0.75 m/s ?

To solve this problem it is necessary to use the conservation equations of both kinetic, rotational and potential energy.

By definition we know that

Where,

KE =Kinetic Energy

KR = Rotational Kinetic Energy

PE = Potential Energy

In this way

Where,

m = mass

v= Velocity

I = Moment of Inertia

Angular velocity

g = Gravity

h = Height

We know as well that for velocity (v) and Radius (r)

Therefore replacing we have

" alt="h= \frac{1}{2} \frac{v^2}{g} +\frac{1}{2} \frac{I}{mg}(\frac{v}{r})^2" />" />

Therefore the height must be 0.3915 for the yo-yo fall has a linear speed of 0.75m/s

d is the correct answer

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!