# Loop-the-Loop with rotational motion

1. Mar 10, 2006

### nahya

A small spherical ball of radius r = 1.5 cm rolls without slipping down a ramp and around a loop-the-loop of radius R = 2.6 m. The ball has mass M = 358 g.

How high above the top of the loop must it be released in order that the ball just makes it around the loop?
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i just calculated the height above zero potential energy (ground) and subtract 2R from it after my calculations.
the initial potential energy, then, is p = mgH.
at the top of the loop, the potential energy is all converted to translational and rotational KE, so...
mgH = 1/2(mv^2 + Iw^2).
the minimum speed needed to barely complete the loop is Sqrt(gR), which, in this case, is ~5.05 m/s.
I = 2/5mR^2 and w = v/R, so Iw^2 = 2/5mv^2
gH = 1/2(v^2 + 2/5v^2) = 1/2v^2(1 + 2/5) = 0.7v^2
so i get H = 1.8216, which is less than the radius of the loop!
this can't be right...
what am i doing wrong?
the equation for the minimum vel is correct, right?
then i should be doing this right...

2. Mar 10, 2006

### BerryBoy

You equation is right, before you put numbers into your equations, use algebra. This gives a nice equation that cancels down to:
h = (19/10) R
But the delta 'h' is the change in height from where the sphere is released to the top of the loop. You don't need to subtract 2R.

Regards,
Sam

3. Mar 10, 2006

### nahya

hmm... h = 1.9 * 2.6 = 4.94 m above 2R, but it says that it's incorrect.
bleh TT

4. Mar 10, 2006

### Hootenanny

Staff Emeritus
At the top of the loop there will still be some potential energy.
$$E_p = 2mgR$$

5. Mar 10, 2006

### BerryBoy

Sorry, I got the moment of inertia for a sphere wrong. It should be:
$$\Delta h=\frac{7}{10} R$$.

Regards,
Sam

Last edited: Mar 10, 2006
6. Mar 10, 2006

### BerryBoy

$$mg\Delta h= \frac{1}{2} (mv^2 + I \omega^2)$$
$$v= \sqrt{gR} \ \ I = \frac{2}{5} mr^2$$

These are the formulae I used

Last edited: Mar 10, 2006