What Determines the Minimum Height for a Marble to Complete a Loop-the-Loop?

[MathewsMD]

A solid marble starts from rest and rolls without slipping on the loop-the-loop track in Fig. 10.30. Find the minimum starting height from which the marble will remain on the track through the loop. Assume the marble’s radius is small compared with R.

Solution:

In the question, why is the radius of the circle referred to as R-r instead of just R? Is this common notation since I'm having a little bit of trouble understanding what exactly r is in this case and how this form helps us assess the situation. Also, why must v >= g (R-r)? I think this question stems from my previous one, but if the forces at the top must be at least F_n + mg, why isn't v >= (gr)^0.5 since then v²/r >= g at the top for there to be a normal force still, correct?

Any help would be great! :)

 

[MuIotaTau]

$r$ is the radius of the ball, so the distance from the center of the loop-de-loop to the center of mass of the ball is just the full radius of the loop, $R$, minus the distance the center of mass of the ball is raised up from the edge of the loop, $r$. Make sense? And so the equation they arrive at for the minimum velocity, $v^2 \geq \sqrt{g(R - r)}$, matches what you have, provided that you use $R - r$.

 

[MathewsMD]

$r$ is the radius of the ball, so the distance from the center of the loop-de-loop to the center of mass of the ball is just the full radius of the loop, $R$, minus the distance the center of mass of the ball is raised up from the edge of the loop, $r$. Make sense? And so the equation they arrive at for the minimum velocity, $v^2 \geq \sqrt{g(R - r)}$, matches what you have, provided that you use $R - r$.

Haha okay, wow that does make sense. They stated in the question r is small compared to R so I thought we did not consider it. Thanks for the clarification!

 

[MuIotaTau]

Yeah, sure thing! Actually, given that they said the radius of the ball is small, I would actually be very confused too if I hadn't been working backwards from the solution, so I don't blame you at all.