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

(75°f − 32) × 5/9 = 23.889°c

measuring things 1 litre of water