Question

A solid ball with radius of 1.8 cm is released from the height of hs=85.9 cm on a non-slip surface. After reaching its lowest point the ball begins to rise again, on a frictionless surface. How high does the ball rise on that side? Express your answer in cm.

Answer 1

We need to find the final velocity when it reaches the lowest point. On the descent, the solid ball has both rotational and translational kinetic energy.
The initial energy = potential energy = m g h
Where m is the mass of the ball
m g h = (1/2) I ² + (1/2) m v²
The moment of inertia of the sphere I = (2/5) m r².
= v / r
Where r is the radius of the sphere.
m g h = (1/2) (2/5) m r² (v/r)² + (1/2) m v²
m g h = (7/10) m v ²
v² = 10 g h / 7  
v = 3.47 m/s.
The ball rises on the frictionless surface
On the frictionless surface, the ball has no rotational motion. Thus the conservation of mechanical energy
(1/2) m v² = m g h_f
Where h_f is the maximum height reached.
h_f = v² / 2 g
h_f = (3.47 )² / 2 x 9.81 m/s²
h_f =  0.614 m
h_f = 61.37 cm.