# Loop-the-loop question [closed]

Problem: The marble rolls down the track and around a loop-the-loop of radius R. The marble has mass $m$ and radius $r$. What minimum height $h$ must the track have for the marble to make it around the loop-the-loop without falling off?
Express your answer in terms of the variables $R$ and $r$.

I found this solution to be very reasonable:

$$mg = m a_c = m \frac{V^2}R$$ which leads to $$V = \sqrt{g R}$$

The energy at the top of the loop KE = Delta PE

$$\frac12 m V^2 + m g (2R) = m g h \\ \frac12 (g R) + g (2R) = g h \\ \left(\frac12+2\right) R = h$$

so $h = 5/2 R$

However the correct answer is actually $\frac52(R-r)$, I think it's because the radius of the loop is measured from the center of the ball rolling on it, so the they subtracted $R$ from $r$, but how would you derive that? I tried the same steps just using $R-r$ instead of $R$ but I got a different answer.

• Don't you need to take account of the fact that the marble is rolling ? This would suggest to me that the starting height needs to be larger not smaller - since the rolling motion stores additional energy which means you need to start with more potential energy. Apr 14 '14 at 3:56
• As @Floris says, the solution you give is for a slideing block of negligible height. The solution you are supposed to get is for a negligible block of height $2r$ (allowing for the height of it's CoM, and the correct solution must also allow for the angular kinetic energy of the ball (which means it depends on whether the ball is solid or hollow). Apr 14 '14 at 4:07
• Is this a duplicate of physics.stackexchange.com/q/44912 ? Apr 14 '14 at 4:18

Your expression for the velocity looks right; but we have to get a few other things taken care of.

First - the center of the marble doesn't move from 0 to 2R, it moves from r to 2R-r - so the potential energy due to this is smaller than mg(2R) which is what you had in your expression.

On the other hand, you need to take account of the energy of the sphere rolling (which is stated explicitly). The moment of inertia for a solid sphere (the usual case for a "marble") is

$$I=\frac25 m r^2 \\$$

$$E=\frac12I\omega^2=\frac12\frac25 m r^2 (\frac v r)^2 = \frac15mv^2\\$$

Thus your energy equation has to be corrected to

$$\frac12 m v^2 + m g (2R-2r) + \frac15mv^2 = m g h \\$$

Also - note that the marble is moving in a path with radius R-r not radius R; you need to take that into account when you compute the limiting velocity ("fast enough to stick to the track") - you have to put R-r where you have R in your velocity equation:

$$\frac{mv^2}{R-r}=mg\\ v=\sqrt{g(R-r)}$$

Combining these:

$$\frac{7}{10}mg(R-r) +2mg(R-r)=mgh\\ h = \frac{27}{10}(R-r)$$

Note that it is sometimes said that "you can ignore the rotational energy of the marble if it is very small", but that is emphatically not true - the rotational energy for a solid sphere is always 2/5 of the (linear) kinetic energy, regardless of the size of the marble. It can therefore only be ignored for the case of a (frictionless) sliding object.

Finally - since no physics problem is complete without a diagram:

This shows clearly where the R-r term is coming from.

Thus if you ignore the rolling, the diagram explains the "correct" answer (which in a way was your question). If you include rolling, then you need to modify the solution as shown.

• Doesn't the ball's center of mass actually start at a height of h + r as well? Apr 14 '14 at 4:14
• @wgrenard yes but I think that is why you have the 2r term. We are solving for the height of the ramp - the bottom of the marble, not the center. Apr 14 '14 at 4:16
• Ah, of course. I failed to notice you had 2r rather than r on the left. Apr 14 '14 at 4:23

First of all, I would like to say that the answer Floris gave is the correct way to do the problem you've set forward, but I thought it worthwhile to note that the result ${5 \over 2}(R-r)$ is the answer if, rather than rolling a marble down the track, you are sliding a cube. If this problem comes from a textbook, there were some unstated assumptions in providing the solution, namely ignoring the rotational energy.

Thus, with an object sliding (i.e. no rotational energy present) the conservation of energy equation is this: $${1 \over 2} mv^2+mg(2R-r)=mg(h+r)$$Where $$v= \sqrt{g(R-r)}$$ This will simplify to the solution you expect.