Physics Help Sphere on Incline Rotations!?

Question

A solid sphere of uniform density starts from rest and rolls without slipping a distance of d = 2.8 m down a q = 29° incline. The sphere has a mass M = 5.8 kg and a radius R = 0.28 m
a) Of the total kinetic energy of the sphere, what fraction is translational?
b) What is the translational kinetic energy of the sphere when it reaches the bottom of the incline?
c) What is the translational speed of the sphere as it reaches the bottom of the ramp?
d) Suppose now that there is no frictional force between the sphere and the incline. Now, what is the translational kinetic energy of the sphere at the bottom of the incline?

I don't know where to start. Please help. Thanks!! :)

Answer 1

For a solid sphere
I=(2/5)mR^2

Ke = (1/2)mV^2
Ker= (1/2) I w^2=
Ker= (1/2) (2/5)mR^2 (V/R)^2
Ker= (1/2) (2/5) m V^2

(a)
Ke/(Ke+Ker) = (1/2) m V^2/[(1/2)mV^2 + (1/2) (2/5) m V^2]=
Ke/(Ke+Ker)= 1 /[1+ (2/5) ]=1/(7/5)
Ke/(Ke+Ker) =5/7

(b)
Total potential energy
Pe=mgh
(note h= length of the incline times the sin(29°))
Ke=(1/2)mV^2
Total kinetic energy
Ket=Ke+Ker=Pe
Ket=Pe
Ke=5/7 Ket so Ke=5/7Pe so
(c)
V=sqrt(2Ke/m)
V=sqrt(2(5/7)Pe/m)
V=sqrt( (10/7)gh)

(d) What do you think? Will it roll?