# Hollow ball rolling down an incline

1. Nov 10, 2008

### veronicak5678

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

A ball has a diameter of 25.4 cm and a mass of 566 g. It can be approximated as a hollow spherical shell. Starting at rest, how long will it take to roll 10 m down an an incline of 30 degrees if it doesn't slip?

2. Relevant equations
Energy conservation
I = 2/3 mr^2

3. The attempt at a solution
I manage dto find the final velocity ( I think it is 7 m/s), but not sure where to go from here. The only equation I have to solve for time involves angular acceleration ( α = Δω / Δt) which I also need to know time for. I also don't know the number of rotations this took.

2. Nov 11, 2008

### tiny-tim

Hi veronicak5678!

If you find the velocity for a general distance, x, you should be able to get a differential equation.

Can you show us the equations you did get?

3. Nov 11, 2008

### veronicak5678

Re: Rotation

I used cons. of energy :

1/2 I* ω^2 + 1/2 m*vfinal ^2 = m*g*yintial

vfinal ^2 = g*yinitial

9.8*5m = 49 m^2/s^2

vf = 7 m/s
I don't know differential equations, but can't I just use v = x/t?

7 m/s = 10.0 m /t

t = 1.43 s

4. Nov 11, 2008

### tiny-tim

Hi veronicak5678!
I'm not following anything after this …

what is your equation relating ω and v?

5. Nov 11, 2008

### veronicak5678

Re: Rotation

1/2 I* ω^2 + 1/2 m*vfinal ^2 = m*g*yintial
1/2(2/3 m*r^2)(v^2/r^2) + 1/2 (m*vfinal ^2 = m*g*yi

cancel out the masses and the radii

1/2 (2/3 vfinal ^2) + 1/2 vfinal ^2 = g*yinitial
I think I made a mistake here!
It should be 1/3 (vfinal )^2 + 1/2 vfinal ^2 = gyinitial
5/6 vf^2 = gyinitial
vfinal ^2 = 6/5 (gyinitial)
9.8*5m*(6/5)
vf = 7.67 m/s

6. Nov 11, 2008

### tiny-tim

Hi veronicak5678!

Yes, that's fine, except …

i] just write v, not vfinal (and specify v(0) = 0)

ii] use x for the distance along the slope (or some other letter, if you'd rather, and write mgx/2 instead of mgy)

And now you have v = √(6gx/10), which you should be able to solve.

7. Nov 11, 2008

### veronicak5678

Re: Rotation

Where did √(6gx/10) come from? I need to find time. Is it OK to use
7.67 m/s = 10.0 m / t?

8. Nov 11, 2008

### tiny-tim

You had 5/6 v2 = gy, which is the same as gx/2,

so v2 = 6gx/2*5 = 6gx/10

so dx/xt =√(6gx/10), which you should be able to solve.

9. Nov 11, 2008

### veronicak5678

Re: Rotation

I can't just use the definition of velocity as x / t?

10. Nov 11, 2008

### tiny-tim

No … that only works for constant velocity (zero acceleration).

11. Nov 11, 2008

### veronicak5678

Re: Rotation

Should that be dx/ dt = so dx/xt =√(6gx/10) ?

12. Nov 12, 2008

### tiny-tim

Yes!

Get on with it!!

13. Nov 12, 2008

### veronicak5678

Re: Rotation

I can't get on with it because I don't know how to solve that!

14. Nov 12, 2008

### gabbagabbahey

Re: Rotation

You have a separable differential equation:

$$\frac{dx}{dt}=\sqrt{\frac{3gx}{5}} \Rightarrow \frac{dx}{\sqrt{x}}= \sqrt{\frac{3g}{5}} dt$$

Just integrate both sides of the equation (and don't forget to include a constant of integration!)

15. Nov 12, 2008

### veronicak5678

Re: Rotation

I've never done differential equations. I know the left side will be 2 rootx. I don't know the other. Also, since we don't need to know this for my class, I thought there must be some other way. How can i find time from this ? Should I change g to its value?

16. Nov 12, 2008

### tiny-tim

Hi veronicak5678!
ah … well, in that case, instead of using conservation of energy, you should have taken moments about the instantaneous point of contact, and used torque = rate of change of angular momentum, which will give you the acceleration directly.

(alternatively,if you assume that the acceleration is constant, you can get a from v2 = 6gx/10 and v2 = 2ax )

17. Nov 12, 2008

### veronicak5678

Re: Rotation

OK. I think I have an answer. Thanks for all your time and patience. Sorry I wasn't understanding you!