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Rotation v. Translation?

  1. Jul 6, 2010 #1
    1. The problem statement, all variables and given/known data

    I am still - STILL - looking for the rotation-translation equation. You guys have a clue? If a pool ball (radius 57mm, mass 160g) has a forward rotation of 300mm / s and backspin of 6 pi rad / s (which is 1074mm / s at the point of contact)... after one second, what are the ball's coordinates and what are its velocities? After two / three / four seconds? When will it stop?

    (Make up coefficients you don't know, including the rolling / static / dynamic friction coefficients.)


    2. Relevant equations

    Most can be found here - http://archive.ncsa.illinois.edu/Classes/MATH198/townsend/math.html" [Broken] .

    I did find these equations, but I'm not sure if they're right - if they are, can you just tell me they're correct? I'm too tired right now to figure out if these are what I'm looking for.

    eq13.jpg

    eq14.jpg


    3. The attempt at a solution

    I apologize, all of the attempts I made came purely from my head and were not grounded on physical laws. They looked like... something was happening, but I have no way to tell if I was close.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 6, 2010 #2
    i think u have to do it by considering two motions (translation and rotational) separately.
    remember that it will be in a pure rolling state only when v = wr, where v is the translational speed and w the rotational speed and the contact point at the bottom should go backward.
    first examine whether v initial = w initial X r, holds good or not, from given data. i think it will not hold good.
    in that case two cases may arise 1) w initial is more than v initial/r.
    2) v initial is more than w initial X r.
    in case of the first one friction will increase the value of v and friction torque will decrease the value of w such the it comes to a pure rolling state finally.
    for the second case it will be vise-versa.
    now see which case of the two it comes under and write down the eqn. of motion of the form v = u +(or minus) at and w' = w + (or minus) alpha t
    now put the different values of t and find different values of w and v. and remember to find the value of t to reach a state of pure rolling before finding anything. if the values of t, u put is equal to or more than that, u have to take into consideration the pure rolling eqn. also.
     
  4. Jul 6, 2010 #3
    Well, thank you for the help. I can't tell if this is the complete solution or not... why do you have to split it into < / > cases? Since you know the velocity of the ball's center as well as the velocity at the point of contact, it seems like you should be able to come up with something like this:

    Friction[trans.]= [tex]\mu[/tex] * (v-w)
    Friction[rot.]= [tex]\mu[/tex] * (w-v)

    Example:
    V=5, W=0, [tex]\mu[/tex]=.1.
    Friction of translation = .1*(5-0) = .5, so v'=4.5.
    Friction of rotation = .1(0-5) = -.5, so w'=0.5.

    Of course, the original frictional equation doesn't rely on the speed parallel with the surface as much as the normal force, which complicates things. I've been cheating in Flash programming and simply using decay functions like these to simulate friction.
     
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