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

`(1)/(2) mv^2 +(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

`(1)/(2) mv^2 +(1)/(2) I\omega^2 = mgh`

`[tex]h= (1)/(2) (v^2)/(g) +(1)/(2) (I)/(mg)((v)/(r))^2`[/tex]

`h= (1)/(2)v^2 ( (1)/(g) +(I)/(mg)(1)/(r^2) )`

`h= (1)/(2)0.75^2 ( (1)/(9.8) +(2.9*10^(-5))/((0.056)(9.8))(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