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Rotation and Rolling

  1. Nov 24, 2003 #1
    Ok theres this ball rolling down a ramp. The ball has some radius r and some mass m. At the bottom of the ramp there is a loop. The loop has radius R.

    The question is, if the ball is 1/4 of the way completed the loop, what is the horizontal component of the net force on the ball. And it is also given that the ball starts up the ramp at height 6R.

    Ok so ive been messing around, and what getting me is that when the ball starts the go up in the loop, gravity will do some work. The problem is, I do not know how to calculate the work done by gravity.

    Perhaps I am doing the question wrong because I am analysing the rotation of the ball in the loop, not the actual rotation of the ball itself.

    Basically what I am doing is using energy and heres my equations

    mg(6R) + Wgrav = 1/2mv^2 + 1/2I(omega)^2 + mgR
    and
    v = R(omega)

    so we make the substitution and solve for omega

    then we go back and solve for v

    the we do

    a = v^2/R

    and finally

    F=ma

    but I may be doing this completely wrong so any help will be great. Here is the original question with diagram:




    A small solid marble of mass 12 g and radius 3.8 cm will roll without slipping along the loop-the-loop track shown in Fig. 12-33 if it is released from rest somewhere on the straight section of track. (a) From what initial height h above the bottom of the track must the marble be released if it is to be on the verge of leaving the track at the top of the loop? (The radius of the loop-the-loop is 4.0 m; Note that it is much greater than the radius of the marble.) (b) If the marble is released from height 6R above the bottom of the track, what is the horizontal component of the force acting on it at point Q?


    The answer with these number is 0.840N.
     

    Attached Files:

  2. jcsd
  3. Nov 24, 2003 #2

    Hurkyl

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    I see terms in both sides for gravitational potential energy... that is the work done by gravity!

    More precisely, the work done by gravity is the difference in gravitational potential energy, so you've already accounted for the work done by gravity because you've included the GPE terms... you don't need to add an additional work due to gravity term.
     
  4. Nov 24, 2003 #3
    Alright but I still dont get the right answer.

    Should i maybe analyse the ball instead of the loop?


    Im really lost
     
  5. Nov 24, 2003 #4

    Hurkyl

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    Aha, here's your problem:

    Those are different omegas! In the first equation, omega is the angular velocity with which the ball is rotating, and in the second equation, omega is the angular velocity with which the ball progresses around the center of the loop!


    I think what you want here is:

    v = r ω

    Where this ω is the same as the one in your first equation and r is the radius of the ball. (I'm using the fact that the radius of the loop is much larger than the radius of the ball, thus allowing me to assume the loop is approximately flat)
     
    Last edited: Nov 24, 2003
  6. Nov 24, 2003 #5
    Still no good

    more lost then ever

    :\

    Perhaps someone else has a suggestion?
     
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