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?