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A ball rolls into a spring

  1. Dec 15, 2013 #1
    1. The problem statement, all variables and given/known data

    A problem like this is on my final tomorrow and I can't seem to get past part C...Please Help!

    A .321 kg, .27m radius thin shelled ball rolls (starting at rest) 3.82m from the top down a 35° 10.0m long incline without slipping. After the 3.82 m, the incline becomes frictionless for the rest of the board. After .8m of the frictionless unencumbered movement, the ball reaches a frictionless massless spring with a spring constant of 20 N/m.

    a. How fast is the center of mass of the ball going after 3.82 (linear speed)
    I solved for v using mgh=1/2mv^2 + 1/2(2/3MR^2)(v/r)^2

    so v= 5.076 m/s

    b. How fast is the ball rolling after the 3.82 m (angular speed)
    v/r=5.076/.27 = 18.8 rad/s

    c. How much does the spring compress?
    d. what is the ball's maximum linear speed?
    e) what is the balls angular speed when the spring is fully compressed?

    2. Relevant equations

    I tried using vf^2=vi^2+2(9.8sin35).8
    to get the velocity before it hits the spring


    3. The attempt at a solution

    I'm using 1/2mv^2=1/2kx^2
    and solving for x but I'm not getting the right answer!
     
  2. jcsd
  3. Dec 15, 2013 #2

    haruspex

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    Did you forget that the ball is still descending as the spring compresses?
     
  4. Dec 15, 2013 #3
    so would it be 1/2mv^2=1/2kx^2+1/2mv^2
     
  5. Dec 15, 2013 #4

    haruspex

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    It could only be that if the two v's stand for different things, but i can't think what you have in mind.
    What I meant was that the ball continues to lose PE after contacting the spring.
     
  6. Dec 15, 2013 #5
    I don't know how to set up the equation. A tutor at MIT on InstaEdu couldn't even figure it out. The answer is .211 m but I don't see how my professor got that.
     
  7. Dec 15, 2013 #6

    haruspex

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    Suppose the max compression of the spring is x. Write out the change in PE and the change in KE from contacting spring to max compression.
     
  8. Dec 15, 2013 #7
    I'm thinking but nothing comes to mind!
     
  9. Dec 15, 2013 #8

    haruspex

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    You know the speed at contact with spring. You know the speed at max compression. So you know the change in KE.
    If it compresses the spring a distance x, what is the PE stored in the spring?
    All of the above you calculated and used before. The bit you forgot was:
    If it compresses the spring a distance x, how much more PE did the ball lose while compressing the spring?
     
  10. Dec 15, 2013 #9
    From what you're saying I would think to the equation I posted in #4. I don't understand the last line, I'm stumped.
     
  11. Dec 15, 2013 #10

    haruspex

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    That equation may apply, I've no way of knowing since you still have not clarified what the two 'v's represent. Either way, it isn't the whole story, for the reason given below.
    The whole slope is 10m. After 4.62m it hits a spring. As it compresses the spring it is still going down the slope. if it travels a distance x while compressing the spring, how much gravitational PE does it lose in the process? I don't think I can make it any clearer.
     
  12. Dec 15, 2013 #11
    mg10sin35 - mg5.38sin35 ??
     
  13. Dec 16, 2013 #12

    haruspex

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    " if it travels a distance x"
     
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