Suppose a cube of mass $m$ enters a loop-de-loop of radius $r$, travelling halfway around and reaching the top. At the top, it has velocity $v$, and the second half of the loop is frictionless.

How can I determine whether or not it will continue around the loop, or fall off?

According to the notes I took during class I can check to see if $v^2/r < g$, but I don't understand why that works. Can someone explain this?

How would I determine if the cube stays on the loop if the second half of the loop isn't frictionless?

  • 1
    $\begingroup$ @Mark Eichenlaub: This isn't homework - I'm studying for a final exam. $\endgroup$ – Cam Dec 15 '10 at 10:19
  • $\begingroup$ In my mind, those are pretty much the same as far as the functionality of the site. Some other users may disagree so I'll wait to hear what they have to say. My opinion is that the "homework" tag is there to say to answerers "don't just give away the answer because then the questioner won't have to reason it through." For final exam review questions or homework questions, the goal is to give hints towards the answer and have you work towards the rest of it as much as possible. Anyway, I'll start a conversation about it on meta. $\endgroup$ – Mark Eichenlaub Dec 15 '10 at 10:23
  • $\begingroup$ Discussion on the homework tag here: meta.physics.stackexchange.com/questions/234/… $\endgroup$ – Mark Eichenlaub Dec 15 '10 at 10:27
  • $\begingroup$ @Mark: I agree with you about the use of the homework tag, but this question is asking for explanation of a general concept, it's not a "please solve this problem for me" kind of thing. So I think it'd be okay to give a complete explanation in this case. (I've soft-deleted my answer until we have some time to discuss the issue, though, just to be on the safe side) $\endgroup$ – David Z Dec 15 '10 at 10:30
  • $\begingroup$ @Mark: Yeah I wasn't sure about that, which is why I changed the tag back. You're probably right and your reasoning makes sense. $\endgroup$ – Cam Dec 15 '10 at 10:30

If the second half of the loop is frictionless, this is just a simple application of Newton's second law. In order to stay on the loop, the cube needs to move in a circle of radius $r$, but no smaller, which means it has to be subject to a net inward force of $mv^2/r$, but no larger. Gravity always provides a force of $mg$, so if $g > v^2/r$, the inward force will be too large for the block to stay on the loop.

If there is friction on the second half of the loop, you could use the same idea in principle, but it becomes a lot messier because you need to check the force at every point on the second half, not just the top. Consider a point at angle $\theta$ around the loop, where $\theta = 0$ corresponds to the very top. If you draw a free-body diagram, you'll see that there are two forces acting perpendicular to the loop (gravity and the normal force) and two components acting parallel to the loop (gravity and friction). Write out Newton's second law in each case and you get

$$mg\cos\theta + F_N(\theta) = m r\dot\theta^2$$

(perpendicular) and

$$mg\sin\theta - \mu F_N(\theta) = mr\ddot\theta$$

(parallel). Notice that I've written the centripetal acceleration ($r\dot\theta^2$) and tangential acceleration ($r\ddot\theta$) in polar coordinates for convenience.

This is a system of coupled differential equations that you would need to solve to find the normal force at each angle. If $F_N(\theta) < 0$ for any $\theta$, the block will fall off at that point. Of course, those are rather complicated equations, and I'm pretty sure you could only solve them numerically.

You could also try using the Euler-Lagrange equation and Lagrange multipliers for the case with friction. I don't have time to write that up now but I'll come back to this as soon as I can and fill it in.

| cite | improve this answer | |
  • $\begingroup$ Thanks! For the friction case, I'm new to polar coordinates - can you explain your notation for the centripetal acceleration? $\endgroup$ – Cam Dec 15 '10 at 10:58
  • $\begingroup$ @Cam: Have a look at Wikipedia, or any of the many other web pages that describe polar coordinates. But basically, you can write $v = r\dot\theta$ (where $\dot\theta$ is the angular velocity, $\mathrm{d}\theta/\mathrm{d}t$) and substitute into $v^2/r$ to find that the centripetal acceleration is $r\dot\theta^2$. $\endgroup$ – David Z Dec 15 '10 at 11:02
  • $\begingroup$ Ah, that makes sense. Thanks - this was very helpful. $\endgroup$ – Cam Dec 15 '10 at 11:04

Try finding the normal force between the loop and the block. Ask yourself what normal forces are physically possible. Is there some condition on the normal force?

If the second half of the loop is not frictionless, you would have a fairly complicated calculation on your hands to find the full motion. You'd need to find the energy dissipated due to friction, which in turn depends on the normal force. However, my intuition is saying that if the block gets up to the top of the loop and isn't going to fall off at the top, it isn't going to fall off anywhere in the second half of the loop, no matter how high the friction coefficient. You might want to check that with some math if you're interested.

| cite | improve this answer | |
  • $\begingroup$ Okay... For the frictionless case, does this make sense? If the cube is in contact with the loop at the top of the loop, it will remain that way until the end, so we just need to check that the cube is in contact at the top. At the top of the loop, the total centripetal force will be $mv^2/r$, so if the normal force is 0, we have $mg=mv^2/r$, and the result follows. $\endgroup$ – Cam Dec 15 '10 at 10:28
  • $\begingroup$ @Cam That sounds good to me! $\endgroup$ – Mark Eichenlaub Dec 15 '10 at 10:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.