A hollow 0.358 kg sphere rolls without slipping down an inclined plane that makes an angle of 41.0o with the horizontal direction. The sphere is released from rest a distance 0.734 m from the lower end of the plane.
a. How fast is the hollow sphere moving as it reaches the end of the plane?
b. At the bottom of the incline, what fraction of the total kinetic energy of the hollow sphere is rotational kinetic energy?
This is what i got but i'm not sure its correct. Any help would be much appreciated!
a)
cons. of energy
1/2mv^2+1/2Iw^2=mgLsinθ
where
I=2/3mr^2
w=v/r
then
1/2mv^2+1/3mv^2=mgLsinθ
5/6mv^2=mgLsinθ
v=(6gLsinθ/5)^1/2
v=(6*9.8*0.734*sin41/5)^1/2 = 2.38 m/s
b)
K=1/2mv^2+1/3mv^2
K=5/6mv^2
Krotational=1/3mv^2
Krot/K=2/5=0.4
1 answer
looks ok to me, I didn't do numbers.
1 answer