Energy conservation law problem

[Petrulis]

Homework Statement

There is a ring which radius is R. A little ball moves inside this ring. Ring's plane is perpendicular to the surface of ground. When a ball is moving inside a ring (ring is in quiet), the ball reaches a height which is equal to R/2.
The ring starts to move upright with a fixed acceleration. What is the value of the fixed ring acceleration, if the ball inside the ring reaches the top of the ring?

(I know the answer of this problem (checked at he book) - acceleration is equal to 4g/5 and a ring moves down with this acceleration)

The Attempt at a Solution

What's not clear from the question is where is the ball when the ring acceleration starts. In my opinion, it makes a big difference.

For example, if the ball is at R/2, it has no kinetic energy w/r/t the ring. Since acclereation of the ring creates a relative gravit of g/5, it would still osccilate within the ring at R/2.

If the ball is at the bottom of the ring, then there is kinetic energy when it starts of m*g*R/2. This will get converted to potential energy at a rate of

m*(g-a)*h, or m*g/5*h.​

Since the translational distance within the frame is 2*R, then

m*g*R/2=m*(g-a)*2*R​

Simplify:

g/2=2*g-2*a​

I get:

a= 3*g/4​

What's wrong?

 

[Dick]

I don't see anything wrong. I agree with your analysis. The position of the ball certainly makes a difference.