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Complicated disk rolling problem (Angular Momentum and it's ilk)

  1. Nov 1, 2008 #1
    1. The problem statement, all variables and given/known data
    A disk has mass 3kg and outer radius 50 cm with a radial mass distribution (which may not be uniform) so that its moment of inertia is (2/7)mR^2. The disk is given a hard kick (impulse) along a horizontal surface at time t0. The kicking force acts along a horizontal line through the disk's center, so the disk aquires a linear velocity 3.5 m/s but no initial angular velocity. The coefficiant of friction between disk and surface is .06

    The kinetic friction force between the surface and the disk shows down its linear motion while at the same time making the disk spin on its axis at an accelerating rate. Eventually, the disk's rotation catches up with its linear motion, and the disk begins to roll at time Trolling without slipping on the surface.

    1) How long after the kick does it take for the ball to roll without slipping? The acceleration of gravity is 9.8 m/s^2

    2) Once the disk rolls without slipping, what is its linear speed?

    3) How far does the ball slide until it begins to roll without slipping?

    4) Through what angle does the disk rotate while sliding before it begins to roll without slipping?

    5) What is the ratio of final kinetc energy (when pure rolling occurs) to the initial kinetic energy?


    2. Relevant equations
    No idea, terribly sorry, please see below.


    3. The attempt at a solution

    My professor had to go to Denmark or something for the second lecture of the week, so instead of finishing rotational motion, we get hit with statics. So, this problem would probably have been easier to understand with that lecture. However, I only have time to do stuff over the weekend due to commuting/schedule. So, can anyone please tell me how the heck I'm supposed to even approach this problem? And why does it matter if something is slipping or not? Is there some sort of "non slipping condition" formula?

    Thank you in advance. This is a 5 part problem, and I just can't ignore it/wait for the next lecture to hopefully get something answered.


    Edit: Here is a picture of the problem, if it helps at all: http://img340.imageshack.us/my.php?image=discrollmz0.png
     
    Last edited: Nov 1, 2008
  2. jcsd
  3. Nov 1, 2008 #2

    LowlyPion

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    Start with what you know.

    Put yourself in the frame of reference of the disk. There is a tangential force of friction u*m*g that would be going into angular momentum and increasing angular speed.

    From what you are given can't you figure the angular acceleration? You have the moment of inertia. You have the force supplied.

    Won't it stop slipping when the linear speed is the same as the rotational speed? And isn't there a conservation of momentum involved here? The increase in angular momentum is coming from the linear momentum?
     
  4. Nov 1, 2008 #3
    I don't have a clue about angular momentum, sorry, the lecture hasn't gotten that far, which is one thing that makes this problem hard.

    And also, the slipping thing confuses me, so I really can't answer that question. I know there's a law of angular conservation of momentum. I do not know how linear and angular momentum relate to each other.

    See, if I knew more about angular momentum and some of it's formulas, I could probably figure out how to start this monster problem. But without that knowledge, it makes it harder. Sorry. <.<
     
  5. Nov 1, 2008 #4

    LowlyPion

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    Torque = F * r = I * a

    where I is moment of inertia and a is angular acceleration. This gives you angular acceleration immediately.

    Here are some equations

    http://hyperphysics.phy-astr.gsu.edu/Hbase/rotq.html#drot

    Notice how similar they are to the linear motion equations.
     
  6. Nov 1, 2008 #5

    LowlyPion

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    They are conserved. What one gains the other loses.
     
  7. Nov 1, 2008 #6
    Alright, that's a start, thanks, I'll post again if I run into trouble further along the line. Thanks for your help.
     
  8. Nov 1, 2008 #7

    djeitnstine

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    Things you need to know are F = [tex]I \omega^{2}[/tex]

    [tex]\omega[/tex] is the angular velocity of the disc in radians per second

    [tex]\omega[/tex] =[tex]\frac{\Delta \theta}{t}[/tex]
     
  9. Nov 2, 2008 #8
    Ok, I tried working out the problem for the first part, and it's saying the answer is wrong, and I"m not quite sure where in my math I'm screwing up, so here's my work:

    Torque = r x F

    r= .50 m (outer radius)

    F= u*M*G
    u= .06 (coeficiant of friction)
    M= 3 KG
    G= 9.8 m/s^2

    So, F= 1.764 N
    and Torque = .882 N-m

    T = I * a -> a= T/I

    I = (2/7)M*r^2 = .2142...

    a = 4.116

    With Angular Accerleation, I can find time now

    w = wi +at
    w= 3.5 (linear speed, since the angular needs to match the linear speed)
    wi (initial)= 0
    a= 4.116

    t= (w-wi)/a

    t=.85 seconds

    Can someone please see where I made the mistake in my work? Thanks.
     
  10. Nov 2, 2008 #9

    LowlyPion

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    I'm with you up until there.

    What you are needing to do now is determine when the forward momentum has slowed and the angular momentum has increased so that the w*r of the angular momentum is the same as the speed of the linear momentum v.

    Won't that be when

    I*w + m*v = m*vo
     
  11. Nov 2, 2008 #10
    That makes sense...

    Thoough now I'm a tad confused, as I see two unknowns (w and v) and only one equation to solve them.

    Where's the second equation? Or am I missing something here?
     
  12. Nov 2, 2008 #11

    LowlyPion

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    What you're missing is that w = v/r
     
  13. Nov 2, 2008 #12
    Ahhhh, yes, that makes it so I can solve it with algebra now. Thank you, I'll post if I have trouble, though hopefully I won't run into any more.
     
  14. Nov 2, 2008 #13
    Sorry to bug you again, but for some reason something's not quite right.

    Ok, I calculated the final linear velocity as 21, and the final angular velocity as 42. I plugged it back in to the first equation, and everything matches up.

    So, I tried plugging in 42 into the w= w0 +at equation, and it says it's the wrong answer. Did I skip a step?

    I now have 10.2 seconds as my answer.
     
  15. Nov 2, 2008 #14

    LowlyPion

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    Whoa. How so big?

    You start at 3.5 m/s.

    And it slows down.

    I * w + m * v = m * vo = 2/7*m*r2 * v/r + m * v

    2/7 r * v + v = vo

    v = 3.5/1.14
     
  16. Nov 2, 2008 #15

    LowlyPion

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    The best double check is always to review the original problem to make sure that the answer seems reasonable.

    Good luck.
     
  17. Nov 2, 2008 #16
    Ok, I re-ran those numbers, got the w = 6.14

    However, when I plug it into the w= w0+at equation, the homework STILL says it's wrong.

    I have t = 1.49

    Sorry to keep bugging you, but this problem just seems to be made to throw me off. >.>

    Incidentally, is the 3.07 for the final v here, that would be the linear speed of the disk once it stops slipping, correct?

    Also, this is my last attempt for the homework here, so I need to make sure that I get it right <.<
     
    Last edited: Nov 2, 2008
  18. Nov 2, 2008 #17
    Ok, as a small update, I assumed that the linear velocity found via the equation you gave me was the answer for number 2.

    Upon inputing it, the homework system said it was wrong. So, does this mean that I calculated the linear velocity wrong?
     
  19. Nov 2, 2008 #18

    LowlyPion

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    That looks like what I get. And v/r = 3.07/.5 = 6.14 and that gets to 1.49.

    But hold on. If that's wrong maybe I should have worked it as rotational kinetic energy and linear kinetic energy. Let me look at that a little.
     
  20. Nov 2, 2008 #19

    LowlyPion

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    I suppose after looking at it that the work done in creating angular kinetic energy was what I should have been thinking. So I apologize for misleading - both of us really.

    In which case, the rotational kinetic energy and the linear kinetic energy are the two quantities related by the work of the friction.

    Hence the relationship:

    1/2*I*w2 + 1/2*m*v2 = 1/2*m*(3.5)2

    Eliminating 1/2*m

    2/7*r2*(v/r)2 +v2 = 3.52

    That gets to v2 = 7*(3.5)2/9

    Which gets to V = 3.087
     
  21. Nov 3, 2008 #20
    Hmmmmmmmmmmm....

    Quick question, would the answer to the second part of my problem be the V you just calculated?
     
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