a) The length of the string for a complete rotation of the yo-yo is

`d_0 =2*pi*R =2*pi*0.005 =0.0314 m`

For a total length of the string `d = 30 cm=0.3 m` the number of revolutions the yo-yo need to make is

`N =d/d_0 =0.3/0.0314 =9.55 revolutions`

b)

The linear acceleration...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

a) The length of the string for a complete rotation of the yo-yo is

`d_0 =2*pi*R =2*pi*0.005 =0.0314 m`

For a total length of the string `d = 30 cm=0.3 m` the number of revolutions the yo-yo need to make is

`N =d/d_0 =0.3/0.0314 =9.55 revolutions`

b)

The linear acceleration of the yo-yo comes from

`d =v_0*t + (a*t^2)/2` with initial speed `v_0 =0 m/s`

Thus `a =(2*d)/t^2 =(2*0.3)/25^2=9.6*10^-4 m/s^2`

This linear acceleration is tangent to the yo-yo disk. The angular acceleration `epsilon` (perpendicular to the yo-yo disk plane) comes from

`a =epsilon*R rArr epsilon = a/R =(9.6*10^-4)/(5*10^-3) =0.192 (rad)/s^2`

(The above relation is analog `v = omega*R` )

c)

The angular velocity as a function of time equation is

`omega=omega_0 +epsilon*t` (analog `v=v_0 +a*t` )

For `omega_0 =0 (rad)/s` we have

`omega =0.192*25 =4.8 (rad)/s`