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Sphere rolling up an incline then back down

  1. Dec 18, 2004 #1
    Need Help Here!!

    A sphere or mass, m and radius r rolls along a horizontal surface with a constant velocity, Vi approaches an incline with (angle theta). ie bottom angle :smile: If it rolls without slipping,
    a) what is the maximum distance,x it will travel on the incline?
    b) If it begins to roll back down, find the time it takes to get to horizontal surface.
    c)What will be its final velocity. ie. at time it gets to horizontal surface.

    My solution:

    I want to write energy equations and use conservation of energy to solve it.

    K.E. = 1/2M(Vi^2) + 1/2I(w^2) where I = 2/5 r^2
    P.E = mgh(max) - kNx where k is coeficient of friction and N is normal force

    using trig, h(max) = xsin(theta) so x = h(max)/sin(theta)
    so
    P.E = mgh (max) - [kNh(max)] /sin(theta) ; N = mgcos(theta)
    P.E = mgh(max) - [kmgcos(theta)h(max)]/ sin(theta)

    h(max) is the vertical distance travelled ie. less than h and
    x is distance travelled on incline i.e less than d

    Are my equations right? and if so,the way to go now is to substitute w= v/r, set K.E = P.E and solve for x?
     
    Last edited: Dec 19, 2004
  2. jcsd
  3. Dec 18, 2004 #2

    Doc Al

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    Staff: Mentor

    Good. That's how to solve part a).

    I = 2/5 M r^2
    The friction does no work, since it rolls without slipping.

    Right.
    Correct your expression for PE and you're good to go.
    Right.
     
  4. Dec 18, 2004 #3
    Hey Doc Al,
    Thanks. I think i figured out the 3rd part.

    I first found the acceleration.
    since w = v/r ; v = wr
    a = w(dot -on top of it--hehe) r
    w(dot) = a / r

    I w(dot) = Fr where F is friction
    ma = mg sin (theta) -F

    If you plug expression for F into it
    a = 5/7g sin(theta)

    Now, knowing acceleration, i just separate variables and integrate to find velocity

    and i got it to be 5/7gsin(theta) t

    If this is right, i'll need to figure out the time it takes to get down. do i just make t the subject? that seems trivial...any help?
     
  5. Dec 19, 2004 #4

    Doc Al

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    Staff: Mentor

    Looks good to me.

    Right.

    It's as easy as you think it is. :smile: [itex]V = 5/7 g sin\theta t[/itex], since you know the final speed [itex]V = V_i[/itex], just solve for t.
     
  6. Dec 19, 2004 #5
    okay, but when i solve for t, my expression will involve v not vi. Should i then make vi the subject of my x expression from a) and plug vi into my t? (the vi expression will involve v meaning my final t expression will involve v)

    I'm also not really sure why you say v = vi at the end. Is it because the ball goes back to rolling on the horizontal surface? Thanks. I'm almost there:smile:
     
  7. Dec 19, 2004 #6

    Doc Al

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    Staff: Mentor

    Yes. Since the ball started up the incline with a linear speed of Vi, that's what it will end up with when it rolls back down.
     
  8. Dec 19, 2004 #7
    so t = 7v/5gsin (theta) but v = vi so
    t = 7vi/5sin(theta)?

    It just seems wierd that we're suddenly replacing the velocity with which it hits the ground with its initial velocity (even though i understand that vf = vi)

    Is there something i'm missing? or am i just thinking too hard?
     
  9. Dec 19, 2004 #8

    Doc Al

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    Staff: Mentor

    Right (but don't leave out the g).

    It may seem weird, but vi is the only information you are given, so your answer had better be in terms of it!

    If you want to, why not figure the time it takes to go down the incline using the distance x that you already calculated? (You'd better get the same answer!)
     
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