# Loop-the-Loop with rotational motion

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?
---
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...

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

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

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

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

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

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