# Loop-the Loop

1. Nov 30, 2004

### Naeem

Loop-the-Loop

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A small spherical ball of radius r = 1.9 cm rolls without slipping down a ramp and around a loop-the-loop of radius R = 2.6 m. The ball has mass M = 375 g.

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a) How high above the top of the loop must it be released in order that the ball just makes it around the loop?
h = m
5/2*2.6 NO

HELP: Use conservation of energy. Remember that the ball has both translational and rotational kinetic energy. Since it rolls without slipping, the two are related.
HELP: What does it mean to just make it? If you have forgotten, go back and review similar loop-the-loop problems you did without the added complication of rotation. See Lecture 12 for example.

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b) Repeat problem (a) for a disk. Find the ratio of the heights h for the two cases.
hdisk/hsphere =

Can somebody help here!!!!

2. Nov 30, 2004

### ponjavic

Have you tried anything, do you have any ideas?
Have you done what the HELP stated? As in have you reviewed the textbooks?

3. Nov 30, 2004

### marlon

Well in oder to be sure that you don't fall of the loop your centripetal force needs to be equal to the gravity at the top of the loop : mg = mv²/R...and R is the radius of the loop

so the velocity at the top of the loop is v²=gR

At the bottom you can apply conservation of energy :

1/2mv² =mg2R + 1/2mgR where the last term is the potential + kinetic energy at the top...

thus you have that v² = 5gR

and when you start from a height h you have that 1/2mv² = mgh or v²=2gh...

If you set equal these two last expressions for v² you get 2gh=5gR or h=5/2R

the second part is analoguous but you need to use the inertia of motion here because it is no point-particle that moves here...

regards
marlon

4. Nov 30, 2004

### Naeem

for a , i did : h = 5/2R which is 5/2.6 , which I don't think is correct. Am I missing something.

5. Nov 30, 2004

### marlon

h = (5/2)R is correct...
this is the minimal heigth from which you should start out...

regards

6. Nov 30, 2004

### Naeem

I did this:

h = (5/2) which is 2.5 * R
here R = 2.6 then h would be

2.5 * 2.6 which equals 6.5, which the computer says is wrong.

7. Nov 30, 2004

### Staff: Mentor

don't forget rotation

This is incorrect. It would be true for a point particle, but not a rolling ball.

As you found, at the top of the motion $mv^2 = mgR$; use this to find the minimum total KE at that point: translational ($1/2mv^2$) plus rotational ($1/2I\omega^2$).

8. Nov 30, 2004

### ceptimus

Nitpicking a bit, but shouldn't the question state that the ball is not hollow, and that it's made from uniform material? I suppose you have to make that assumption, or you can't calculate the answer accurately?

9. Nov 30, 2004

### ponjavic

Yes a dumb question indeed!

It should also state that the gravity is some constant otherwise you would have to go with the old $$G\frac{m1m2}{d^2}$$
Since friction is not mentioned I'd also like the question to state the aerodynamic properties of the object that is rolling.

Sarcasm indeed, not bulletproof but I'm practicing =)

By the way I wonder, using conservation of energy one assumes there is no friction, but in order for the ball to roll there must be friction, isn't this contradictory?

Last edited: Nov 30, 2004
10. Nov 30, 2004

### Skomatth

The ball is rolling without slipping. This means the point of the ball in contact with the surface is momentarily at rest. Therefore the friction that is producing torque is static. Static friction does no work.

11. Nov 30, 2004

### Staff: Mentor

Good point. Since the problem explicitly considers rotational KE, it is neither a nitpick nor a dumb question.

12. Dec 1, 2004

### ponjavic

Cheap solution :P

13. Dec 1, 2004

### marlon

Correct Doc Al, I gave the solution for a point particle...i did not realize it was a ball...clearly i did not read the question thouroughly...

i apologize for my mistake...

regards
marlon