Yo-Yo conservation of energy

[postfan]

Ok, thanks for all your help!

 

[Simon Bridge]

Good to see someone else was helping while I was away :)

Just checking...

A yo-yo has mass 0.8 kg and rotational inertia 0.12 kg m2 measured about an axis perpendicular to the screen and passing through its center of mass. A light thin string is wrapped around the central part of the yo-yo that has radius 0.03 m. The string is attached to the ceiling as shown. The yo-yo is released from rest and begins to descend as the string unwraps.

Use conservation of energy to find the linear speed of the yo-yo by the time it descended a distance 0.8 m. Use g = 9.8 m/s2.

energy lost falling distance h is $mgh = (0.8\text{kg})(9.8\text{N/kg})(0.8\text{m}) = 6.2720\text{J}$
... notice that working and the answer includes the units?

this energy goes into kinetic energy by $$mgh=\left ( \frac{m}{2}+\frac{I}{2r^2}\right )v^2 = \frac{mr^2+I}{2r^2}v^2\Rightarrow v^2=\frac{2mghr^2}{mr^2+I}$$ ... the trick is to pick the correct r.

The yoyo is basically rolling down the string on it's inner radius.
(Aside: This is what I mean by "show your reasoning" ... if you just memorize equations and do a bit of algebra, you will keep getting things wrong.)

When it's center of mass has a speed v, it is turning with angular velocity ... ??

Using r=0.03m, I get v=0.30581m/s. which is 0.3m/s (1sf)

If your program says that is not correct, you should check the input format: these computer things can be fussy about getting things exactly the way they expect
- does the sig fig matter for eg?

 

[pgardn]

Sorry post fan.

I got stuck on the radius of the yo yo needing to be on the very outside like a disk with a string attached.
The yo yo is indeed rotating from the distance given in the problem. If they had given the radius of the yo yo from middle to the outside the moment of inertia for the yo yo would have been too large.

My bad, as they say on the urban courts of physics