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Homework Help: Forces of rolling question

  1. Dec 13, 2003 #1
    Hi all,

    I just had a question on a homework problem I was attempting, but can't find the correct solution for it yet. The answer is in the back of the book, but I wanted to know how it is derived. Anyways, here is the question:

    A hollow sphere of radius .15 with I=.040 rolls without slipping up a surface inclined at 30 degrees. At a certain initial position, the sphere's total kinetic enery is 20J.

    A) How much of the totak K.E. initial is rotation?

    ------------------------------------------------
    Variables:

    Radius of sphere = .15m
    Moment of Inertia = .040 kg * m^2

    ----------------------------------------------

    Attempt:
    I tried this problem with setting the initial kinetic energy of the sphere-incline system to 20J. So I did:
    .5MV^2 + .5I(w)^2 = 20J

    This is the step that I get lost in. Should I try to find the mass of the system first or should I try a whole different approach? Any help would be appreciated! Thanks!
     
  2. jcsd
  3. Dec 13, 2003 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Yes, you'll need the mass of the sphere. (Hint: what's the formula for rotational inertia of a sphere?)

    Also, you'll need to apply the condition for rolling without slipping: how does ω relate to V when there is no slipping?
     
  4. Dec 13, 2003 #3
    Doc Al,

    Thanks for the reply. One of the biggest problems I was having with finding the mass was finding the rotational inertia of a hollow sphere. In my book (Fundamentals of Physics 6th Edition), it only gives the rotational inertia of a solid sphere.

    I know the rotational inertia for a solid sphere is 2/5MR^2. Will that apply to this problem as well? If that applies, then I set:
    2/5MR^2=.040 kg*m^2 and find the mass that way.

    v=w(r) --> w=v/r

    Any hints would be greatly appreciated. Thanks again!
     
  5. Dec 13, 2003 #4
    The moment of inertia for a hollow sphere is 2/3*m*r^2.. find the mass, then with the equation you have (substitute in for v to find w), then plug w back in to the rotational energy equation, and voila
     
  6. Dec 13, 2003 #5
    Hey thanks deltabourne. You gave me the answer to the missing piece of the puzzle! Found the answer to be 8.0J joules, which is correct!
     
    Last edited: Dec 13, 2003
  7. Dec 13, 2003 #6
    Hi all again.

    Now the question wants me the find the TOTAL kinetic energy of the sphere. I keep on trying to figure this out, but I keep getting 10J, while the answer is 6J. I f anyone can point what I am doing wrong, it would be greatly appreciated!

    -------------------------

    I found Vi=2.98m/s
    Height = .75m

    1/2MV^2+1/2I(W)^2 = MGH + 5/6MV^2

    Subtracting 5/6MV^2 to the other side, I keep on getting 10J. What am I doing wrong?

    To make things easier, I attached my work I have done so far as an image:
    http://home.comcast.net/~msharma15/problem_1.jpg
     
    Last edited by a moderator: Apr 20, 2017
  8. Dec 13, 2003 #7

    Doc Al

    User Avatar

    Staff: Mentor

    You'll have to state the problem more clearly. Obviously you are asked to find the total KE at some new point. What point? In your attachment you mention a point 1 m up the incline. Is that the point? Let me pretend that it is. You know the initial KE; when it raises up a height H, the KE will decrease by mgH. For 1 m up the plane, H = 1 sin(30) = 0.5 m.
     
  9. Dec 13, 2003 #8
    Last edited by a moderator: Apr 20, 2017
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