Question

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 ?

Answer 1

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

By definition we know that

$KE + KR = PE$

Where,

KE =Kinetic Energy

KR = Rotational Kinetic Energy

PE = Potential Energy

In this way

$\frac{1}{2} mv^2 +\frac{1}{2} I\omega^2 = mgh$

Where,

m = mass

v= Velocity

I = Moment of Inertia

$\omega =$ Angular velocity

g = Gravity

h = Height

We know as well that $\omega = v/r$ for velocity (v) and Radius (r)

Therefore replacing we have

$\frac{1}{2} mv^2 +\frac{1}{2} I\omega^2 = mgh$

$[tex]h= \frac{1}{2} \frac{v^2}{g} +\frac{1}{2} \frac{I}{mg}(\frac{v}{r})^2$[/tex]

$h= \frac{1}{2}v^2 ( \frac{1}{g} +\frac{I}{mg}\frac{1}{r^2} )$

$h= \frac{1}{2}0.75^2 ( \frac{1}{9.8} +\frac{2.9*10^{-5}}{(0.056)(9.8)}\frac{1}{(0.0064)^2} )$

$h = 0.3915m$

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