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: Plz can any one solve it ><

  1. Jan 4, 2010 #1
    Hi everybody how r u all ? In fact I had a home work and I tried to solve it many times but I couldn't can any one help me to find out the answer >>>
    He's the problem
    A solid sphere of mass 0.7 kg is at rest at the top of a ramp . It rolls to the bottom without slipping .The apper end of the ramp is 10 m higher than the lower end .What is the sphere's total kinetic energy .When it reaches the bottom ?
    Last edited by a moderator: Jan 4, 2010
  2. jcsd
  3. Jan 4, 2010 #2
    you need to work out the potential energy of the ball at the top of the ramp as
    "energy can't be created or destroyed only changed from one from to another"
    the ball will have all the potential energy as kenetic energy at the bottom, assuming that friction is neglible, since you have not given any coefficients of friction.
    Petential energy= mgh
    g=acceleration due to gravity
    h= height
  4. Jan 4, 2010 #3
    Thanksssssss my dear ^_^ but I think it's more complicated than this I think we have to find angular speed and moment of inertia ...in my opinion we should use mgh = .5 * mv^2 +.5 I w^2 don't you agree with me ?
    Last edited: Jan 4, 2010
  5. Jan 4, 2010 #4


    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi Mr.Right! Welcome to PF! :smile:

    (have an omega: ω and try using the X2 tag just above the Reply box :wink:)
    Yes, that's correct …

    now find the relation between v and ω, and solve. :smile:
  6. Jan 4, 2010 #5
    Hi tiny-tim thanks for your comment can u plz solve it for me in a simple way I mean can u tell me what's the relation between v and ω and I'll appreciate ur
    help ^_-
  7. Jan 4, 2010 #6


    User Avatar
    Science Advisor
    Homework Helper

    Hint: the question says "without slipping", so the instantaneous speed of the "bottom" of the sphere must be zero. :wink:
  8. Jan 4, 2010 #7
    oh ic OK what about moment of inertia how can we find it ?
  9. Jan 4, 2010 #8
    I'm sorry tiny for all these Qs but I was absent at that lesson >_< so i need u r help
  10. Jan 4, 2010 #9


    User Avatar
    Science Advisor
    Homework Helper

    he he :biggrin:

    you look it up (or, preferably, learn all the usual ones) …

    see eg http://en.wikipedia.org/wiki/List_of_moments_of_inertia" [Broken] :wink:
    Last edited by a moderator: May 4, 2017
  11. Jan 4, 2010 #10
    Thanks top man ^_^
  12. Jan 4, 2010 #11
    Though you've turned it into a great exercise in rotational mechanics, you should read the wording of the question carefully to see what is expected in this case.

    Since you aren't given the dimensions of the sphere you can't give the answer in terms of linear K.E. and rotational K.E. at all!

    Look at the text of the problem closely, it only asks for the total K.E. at the bottom. In this problem, you should only regard, "Without slipping" as meaning, "Without loss of energy."

    Not to mention that the sphere's mass distribution might not be uniform (Making [tex]I_{sphere}=\frac{2}{5}MR^2[/tex] not hold in this case)

    As for how to find the moment of inertia for an arbitrary object... That's quite a hassle in and of itself, it's a pure math exercise:

    The moment of inertia of an object about its center of mass is defined as:

    [tex]I_{CM}=\int r^2 dm[/tex] where [tex]r[/tex] is the magnitude of the radius vector from the center of mass to the point in question. The integral sweeps the whole of the object's mass, giving each a 'weight' according to how far away from the center of mass it is.

    You can see right away that if all of your mass is located very far away from the center of mass, then you moment of inertia about your center of mass is huge! If, however, all of that mass were brought closer to the center of mass, then the corresponding moment of inertia would diminish greatly.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook