# Velocity of ball rolling down ramp

1. Apr 13, 2013

### NathanLeduc1

1. The problem statement, all variables and given/known data
A ball of radius ro rolls on the inside of a track of radius Ro. If the ball starts from rest at the vertical edge of the track, what will be its speed when it reaches the lowest point of the track, rolling without slipping?

Included below is a link to the diagram:
http://s3.amazonaws.com/answer-board-image/0c73c3d19aaa1d4c316cd2e6d2da8e57.jpg

2. Relevant equations
v=Roω
Eo=mgh
E=0.5mv2 + 0.5Iω2 + mgh
I = (2/5)M(ro)2
ω=Δθ/Δt

3. The attempt at a solution
I've tried using conservation of energy to say that ƩEo=ƩE
So:

mgh= 0.5mv2 + 0.5Iω2 + mgh

which, when you plug in the moment of inertia for a sphere and the definition of ω:

mgh= 0.5mv2 + 0.5(2/5)M(ro)2)(Δθ/Δt)2 + mgh

These are my main problems:
1) How do I find a value for h? I understand that from the diagram, h is the distance from the end of the ramp to the ground but how I find a value for that?
2) How do I find time (which is needed to calculate ω)?

The final answer is supposed to be √(10/7)g(Ro-ro)

Thanks for any help!

2. Apr 13, 2013

### cbasst

I have part of a response ready to send, but as I was working out the problem to make sure I was getting the correct answer, I noticed something odd. When I got the final answer, it did not match with the answer you gave. Are you sure that the answer you posted is correct, or is it possible it is mistaken? I know that certain textbooks are notorious for giving wrong answers (I have one like that). I'm pretty sure that the answer should be$$v = \sqrt{\frac{10gR_0}{7}}$$If someone else could verify/refute my answer, that wouldn't be a bad thing. I don't want to continue explaining anything until I know I've got it right.

3. Apr 13, 2013

### Staff: Mentor

Good. But distinguish the two different values of h: h1 & h2 perhaps.

Hint: Apply the condition for rolling without slipping. That will allow you to express ω in terms of v. (Forget about Δθ/Δt.)

I think the diagram is a bit off, so draw your own. Imagine a circular ramp. The initial position is just where the wall is vertical, so a line drawn from the center of the circle to the sphere would be horizontal. Compare the initial and final height of the ball. (Measured from what point?)

You won't need that.

That's fine.

4. Apr 13, 2013

### NathanLeduc1

Okay, if I make the origin of the coordinate plane the center of the circle, then h1 would be 0 and h2 would be Ro-Rocos(45°)

So then:

0=0.5mv2+0.5(0.4Mro2)(v/Ro)+mg(Ro-Rocos45)

-mg(Ro-Rocos45)=0.5mv2+(Mro2v)/5Ro

-mg(Ro-Rocos45)=(5Romv2)/(10Ro)+(2Mro2v)/10Ro

-mg(Ro-Rocos45)=(5Romv2+2Mro2)/10Ro

-mg(Ro-Rocos45)(10Ro) = 5Romv2+2Mro2v

Now what do I do? I can tell that from the answer I shouldn't have an M in the final answer so should that cancel out? Also, should m cancel out?

5. Apr 13, 2013

### Staff: Mentor

Where does the 45° angle come from?

Also: Don't forget the radius of the ball. Measure the height of the ball's center.

6. Apr 13, 2013

### NathanLeduc1

I got the 45 degree angle from the 135 degree angle. If we make the center of the circle the origin of the plane, then the y component of Ro would be Rocos(45). Right?

Also, why do I need the radius of the ball? I have it as part of the moment of inertia formula for the sphere but, other than that, I'm confused why I need it. Does it have something to do with the potential energy?

7. Apr 13, 2013

### Staff: Mentor

That diagram doesn't seem to make sense. Where did you get it? Better to imagine a semicircular track that the ball is rolling down. Note that the starting point is where the wall is vertical, which is not at 135 degrees.

The "trick" is figure how the height of the center of the ball changes as it goes down the track. (Yes, that does affect the potential energy.)

8. Apr 13, 2013

### NathanLeduc1

Okay, well now I'm absolutely lost. If we assume the initial position of the ball to be 0, then the final height would be Rocos(45) because 135-90=45. We're changing quadrants that the ball is in so now the 135° angle is a 45° angle.

If I'm wrong, then I have absolutely no clue where to go next...

9. Apr 13, 2013

### Staff: Mentor

I found this image that might help:

Not perfect, but pretty close to what I imagine the problem to be discussing. Note the initial and final positions of the ball.

Last edited by a moderator: Apr 18, 2017
10. Apr 13, 2013

### NathanLeduc1

Oh, okay, that makes a lot more sense. However, I got a final answer of √(10/7)(gro)

I'm thinking that I did something wrong on the left side of the equation.

The original height of the ball at the beginning would be Ro, right?

If that's the case, I end up with gRo-g(Ro-ro).

11. Apr 13, 2013

### NathanLeduc1

More precisely, here's the work I did:

mgRo=0.5mv2+0.2mro2(v/ro)2+mg(Ro-ro)

gRo=0.5v2+0.2ro2(v/ro)2+g(Ro-ro)

gRo-g(Ro-ro)=0.5v2+0.2v2

gRo-g(Ro-ro)=0.7v2

v=√0.7gro

12. Apr 13, 2013

### NathanLeduc1

Sorry, I mean the final answer I got is v=√(10/7)gro not √0.7gro.

13. Apr 13, 2013

### Staff: Mentor

The initial height, Ro, measured from the lowest point of the track, is correct. That's the initial height of the ball's center.

The second height, Ro-ro, is not correct. Rethink that.

14. Apr 13, 2013

### NathanLeduc1

Oh, okay. So the second height is ro and I get the right answer... my question is how does that make sense?

If ro is the radius of the ball but the ball is already at the bottom of the ramp, then how could the radius of the ball be part of potential energy? In order to reduce the potential energy, you'd have to reduce the radius of the ball.

I understand how the math works out now but I'm confused as to the theory behind the math.

15. Apr 13, 2013

### Staff: Mentor

What's the height of the center of the ball initially?

What's the height of the center of the ball when it is at the bottom?

What matters to gravitational PE is how the ball's center of mass changes height.

16. Apr 13, 2013

### NathanLeduc1

Ah, I understand now. So it all comes down to what frame of reference we're looking at it from...

Many thanks for the help.

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