# Rolling ball goes through a loop-the-loop...

1. Mar 25, 2016

### JulienB

1. The problem statement, all variables and given/known data

Hi everybody! Here is a classical mechanics problem about a ball (mass m, radius r) rolling down a slope (from height h) and going through a loop-the-loop.
a) What is the minimum height h from which the ball must roll in order to successfully complete the loop-the-loop without losing contact with the ramp? (Consider the radius R of the looping to be much bigger than r)
b) If the ball is rolling from a height h = 6R, what are the horizontal force components at the point Q? (see picture)

2. Relevant equations

Conservation of energy, centripetal acceleration, moment of inertia of a sphere (Isphere = 2/5⋅m⋅r2).

3. The attempt at a solution

a) Okay so I think I should first find the tangential velocity of the sphere as it crosses the highest point of the loop:

m⋅g = m⋅ac = m⋅(vT2/R) ⇒ vT = √(gR)

(Note: I selected R as the radius and not (R - r) because we've been told that R>>r in the problem's statement)

Then I set up an equation of conservation of energy:

PE0 = PE1 + KE1 + Erot1
⇔ m⋅g⋅h = m⋅g⋅2R + ½⋅m⋅vT2 + ½⋅Isphere⋅ω2
⇔ m⋅g⋅h = m⋅g⋅2R + ½⋅m⋅g⋅R + ½⋅(2/5⋅m⋅r2)⋅(vT2/R2)
⇔ g⋅h = g⋅2R + ½⋅g⋅R + 2/10⋅r2⋅g⋅R/R2
h = 5/2⋅R + 1/5⋅r2/R

Does that seem correct to you? (Note: again, I omitted a few "r" because of the indication R>>r)

b) Now h = 6R

I'm always a little confused about how to describe forces when it comes to centripetal and centrifugal forces, but I give it a go:

At the point Q I would say that the horizontal force components are the centrifugal force FZ (that's how we usually denote it in Germany), directing towards the outside of the loop-the-loop, and the normal force (which is also the centripetal force in that case), directing towards the center of the loop-the-loop.

Because there is no horizontal acceleration at that point Q (the velocity of the ball points upwards), I would assume that FZ = FN = m⋅ac = m⋅vT2/R

To find the tangential velocity at that point, I set up my equation of conservation of energy again but solving this time for vT2 and with h = 6R. I get vT2 = (10⋅R3⋅g)/(R2 + r2).

I substitute then that tangential velocity into my first equation and get:

FN = FZ = (10⋅m⋅R3⋅g)/(R2 + r2)

Is that correct? I'm preparing an exam, so I'd like to make sure I am able to solve such problems, and if not hopefully I can correct the mistakes I still make.

Julien.

#### Attached Files:

• ###### Photo 25-03-16 22 24 46.jpeg
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2. Mar 25, 2016

### Nathanael

Your "ω" is supposed to be about the center of mass right? Yet you say ω = vT/R ?
"Centrifugal force" is a fictitious force that arises when viewing things from rotating coordinates. It is best to not think of it's existence in a non rotating frame. The only force acting horizontally is the normal force (as gravity and friction are both vertical at Q).

You say "there is no horizontal acceleration," but this is not true! The velocity starts at Q directed straight up, and soon later it points a bit horizontally. That means there exists a horizontal component of acceleration. The question is how much does it need to accelerate in order to stay on the path of a circle? Well that's exactly the centripetal acceleration. So by newton's ΣFx=max in the horizontal direction, we get FNormal=macentripetal.

See, it is the same answer as you get, but the perspective behind it is hopefully a bit more clear. The centripetal acceleration is not an 'acting-force,' it is simply the acceleration that the 'acting-forces' (gravity, normal forces, friction, EM, etc.) must produce in order to maintain a circular path.

3. Mar 25, 2016

### JulienB

@Nathanael thank you very much, that's precisely the clarification that I was seeking about the centrifugal force. When formulating my answer, I should then only consider the normal force (which in that present case would be labeled centripetal force) if I understand well.

About the ω I indeed confused R with r... It's a bit late where I live, so I will post the corrected answer tomorrow.

Julien.

4. Mar 26, 2016

### JulienB

@Nathanael My new result for a) is h = 2.7R. Is that now correct? I attached the details of the calculation with the post.

Thank you very much.

Julien.

#### Attached Files:

• ###### Photo 26-03-16 13 48 31.jpg
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5. Mar 26, 2016

### Nathanael

Yep, looks good to me.

6. Mar 27, 2016