posted by Alie on .
A spherically symmetric object with radius of .7m and mass of 1.6kg rolls without slipping accross a horizontal surface with velocity of 1.7m/s. It then rolls up an invline with an angle of 28degrees and comes to rest a distance d of 2.3m up the incline, before rolling back down. Find the moment of inertia of this object about an axis through its center of mass.
Isn't I just= mr^2?
No, not for a solid sphere.
You can find it from the experiment.
Conservation of energy:
final energy= energy at top of incline
1/2mv^2 + 1/2 I w^2= mgh= 1.6*9.8*2.3sin28
but w=v*r, so
1/2 mv^2+1/2 I v^2 r^2= 1.6*9.8*2.3sin28
Solve for I.