Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Hollow ball rolling down an incline

  1. Nov 10, 2008 #1
    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. jcsd
  3. Nov 11, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi veronicak5678! :smile:

    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? :wink:
     
  4. Nov 11, 2008 #3
    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
     
  5. Nov 11, 2008 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

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

    what is your equation relating ω and v? :confused:
     
  6. Nov 11, 2008 #5
    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
     
  7. Nov 11, 2008 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi veronicak5678! :smile:

    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. :wink:
     
  8. Nov 11, 2008 #7
    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?
     
  9. Nov 11, 2008 #8

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  10. Nov 11, 2008 #9
    Re: Rotation

    I can't just use the definition of velocity as x / t?
     
  11. Nov 11, 2008 #10

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    No … that only works for constant velocity (zero acceleration).
     
  12. Nov 11, 2008 #11
    Re: Rotation

    Should that be dx/ dt = so dx/xt =√(6gx/10) ?
     
  13. Nov 12, 2008 #12

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Yes!

    Get on with it!! :smile:
     
  14. Nov 12, 2008 #13
    Re: Rotation

    I can't get on with it because I don't know how to solve that!
     
  15. Nov 12, 2008 #14

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Re: Rotation

    You have a separable differential equation:

    [tex]\frac{dx}{dt}=\sqrt{\frac{3gx}{5}} \Rightarrow \frac{dx}{\sqrt{x}}= \sqrt{\frac{3g}{5}} dt[/tex]

    Just integrate both sides of the equation (and don't forget to include a constant of integration!)
     
  16. Nov 12, 2008 #15
    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?
     
  17. Nov 12, 2008 #16

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi veronicak5678! :smile:
    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. :wink:

    (alternatively,if you assume that the acceleration is constant, you can get a from v2 = 6gx/10 and v2 = 2ax :smile:)
     
  18. Nov 12, 2008 #17
    Re: Rotation

    OK. I think I have an answer. Thanks for all your time and patience. Sorry I wasn't understanding you!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook