# Ball Down a Ramp! Help appreciated

1. Mar 17, 2013

### Nfinley1

1. The problem statement, all variables and given/known data
A ball of mass 2.60 kg and radius 0.120 m is released from rest on a plane inclined at an angle θ = 36.0° with respect to the horizontal. How fast is the ball moving (in m/s) after it has rolled a distance d=1.90 m? Assume that the ball rolls without slipping, and that its moment of inertia about its center of mass is 1.40×10-2 kg·m2.

2. Relevant equations
Given: I = 1.40×10-2 kg·m2.
Linear Acceleration = r*ω
KE = 0.5Iω^2
PE = MGH
μ=Tanθ
3. The attempt at a solution
Using the hypotenuse and the angle, I solved that the Height from the ground should be 1.12 m
ΔKE = 1/2*I*ω(final)^2 - 1/2*I*ω(initial)^2 Since it's not moving initially
ΔKE = 1/2*I*ω^2
Then Since I wanted to get linear acceleration I solved a=rω for omega and got ω=a/r

I plugged that in to get an equation of ΔKE = 1/2*I*(a/r)^2. Although I've just realized my question prompts for a velocity not an acceleration, so I guess I can go back to
ΔKE = 1/2*I*ω^2

Then with ΔKE + ΔPE + ΔEThermal = 0

ΔPE = MGH final - MGH initial so 0 - MGH

1/2*I*ω^2 - MGH + ΔEThermal = 0

I know that μ should be equal to Tanθ after solving MGsinθ = μ *MGcosθ

Here's where I've kinda blanked out... I can't seem to figure out what to put for friction. As I know that exists because the ball is not slipping down the ramp

Help would be very much appreciated. Am I even headed in the correct direction?

Edit: And I'm realizing now that this path leads me to an Imaginary number... so I've missed it somewhere.

Last edited: Mar 17, 2013
2. Mar 17, 2013

### rcgldr

You need to include linear KE. In this case, since the ball starts at rest:

ΔKE = 1/2*I*ω^2 + 1/2*m*v^2

It will be easier if you convert this equation to a function of v rather than ω.

If you continue with this energy approach, you can solve for the friction force afterwards.

If the ball was uniform, the moment of inertia for the ball would be (2/5)*m*r^2 = 0.014976 * kg * m^2, not 0.014000 * kg * m^2, but you should use the given 0.014000 * kg * m^2 value for this problem.

Last edited: Mar 17, 2013
3. Mar 17, 2013

### haruspex

You don't need to worry about friction. There are no losses due to friction in rolling contact (no thermal energy produced).
I agree with all the above, except that I don't think the question asks for the friction. But I don't see any reference to an impulse, or a string, or a spring:
Was there something else in the OP, now edited out?

4. Mar 17, 2013

### Nfinley1

Thanks for Helping

So then, now we have

1/2*Iω^2 + 1/2mv^2 - mgh + Δ Thermal Energy = 0

For change in thermal should I be using μ * N ? so Tanθ*Mgcosθ ? To get the energy from friction in this case.

This likely sounds dumb, but can you help me do this?

Or should I be solving for V with the linear kinetic energy equation?

I jumped into this course co-requisite with calculus, and while I can mostly do everything no problem, sometimes I just need somebody to call me an idiot and point me into the correct direction.

5. Mar 17, 2013

### Nfinley1

Was there something else in the OP, now edited out?[/QUOTE]

Nope... that part sort of confused me, but I looked passed it as well.

I think the friction is only necessarily because without it the object would slip rather than roll.
Unless that's not correct?

6. Mar 17, 2013

### rcgldr

Somehow an edit I made to another post in another thread ended up in my last post of this thread, and then I couldn't get back into physics forums for a while to correct this. Sorry for the confusion.

You're supposed to assume static friction, not sliding friction and no other losses such as rolling resistance, so there is no change in thermal energy.

Stick with the energy equation you have now. You can convert 1/2*I*ω^2 into c*m*v^2, where c is some constant, in this equation:

ΔKE = 1/2*I*ω^2 + 1/2*m*v^2 = c*m*v^2 + 1/2*m*v^2 = m*g*h

so that it's a just a function of velocity and proceed from there. It would have been nicer if the ball was uniform, since in that case the radius of the ball wouldn't matter.

Last edited: Mar 17, 2013
7. Mar 17, 2013

### Nfinley1

How would I go about defining the constant then?

8. Mar 17, 2013

### haruspex

rcgldr's c is just I/(2mr2) = 1.40×10-2 / (2*2.6*.122). The idea is just to make the equation a littler simpler. The important part is to replace ω by v/r so that v is the only unknown.

9. Mar 17, 2013

### Nfinley1

So then I'd use (1/4X10^2)/(2*2.6*.122) in the place of C?
I see why that's important, i'm just working on how we did it? And what do we do with it now.

Thanks for being patient with me! :)

Edit: I just used what you guys said to get down to the correct answer!

I can't thank you guys enough! I learned a lot!

Last edited: Mar 17, 2013