What would be the minimum height for a complete lap?

Question

An empty ball, of  mass $m$ and  moment of inertia $I = frac{2}{3}m.r^{2}$, is rolling across the path shown below:

there is friction $f_{r}$ from  A to C  .

$r$ is the radius of the ball, and $R$ is the radius of the circular part within the path.

what would be the minimal height $h_{0}$ (in function of $R$) , so that the ball can make a complete lap in the loop (the circular part)?

after applying Newton's 2nd law and some rotational dynamics I found the translational acceleration $a_{T}$ and the friction $ f_{r}$:

$$a_{T}  =  frac{g.sin(alpha)}{frac{2}{3}r^{2}-1}~~$$

$$f_{r} = frac{2m.g.sin(alpha).r^{2}}{2r^{2}-3}$$

I know that the ball must roll on a distance of $~~ 2pi R~~$  ( the perimeter of the loop). but I can't figure out how to link between the acceleration and this distance?

Nb: of course the ball is rolling without sliding.

sammy gerbil · Answer

Look at the situation at the top of the loop when the ball just makes it round the loop. Here the minimum force on the ball from the track is zero - anything less means that the ball has left the track. The net force on the ball is $F=mg$ and its (centripetal) acceleration is $a=v^2/R$ where $v$ is the speed of the CM of the ball. So here you apply $F=ma$.

Then, equating the total mechanical energy of the ball at the top of the loop (linear and translational KE plus PE) with its initial PE at the top of the incline (A) will give you the answer - after you substitute the value of $v$ found above.

I notice there is a sharp turn at B. This implies that there is a collision between the ball and the track at this point. Assuming this collision is inelastic (so that the ball does not bounce up from the track) the vertical component of momentum (and therefore also speed) will be lost at this point. You will need to take account of this in your calculation of required height.

What would be the minimum height for a complete lap?

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