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Homework Help: Bowling ball problem

  1. Dec 9, 2003 #1
    A bowling ball is thrown such that at the instant it touches the floore it is moving horizontally with a speed of 8m/s and is not rotating. it slides for a time and distance before it begins to roll without slipping. The coefficient of friction between the ball and the floor is .06. What is the final speed of the ball?

    I know how to find the horizontal acceleration but I don't know where to go from there.

    Any hints?
     
  2. jcsd
  3. Dec 9, 2003 #2

    Doc Al

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    Staff: Mentor

  4. Dec 9, 2003 #3
    There's an hard way and an easy way.
    The easy way is to use conservation of angular momentum.

    JMD
     
  5. Dec 9, 2003 #4

    Doc Al

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    Angular momentum is not conserved. (The friction exerts a torque.)
     
  6. Dec 9, 2003 #5
    Angular momentum is conserved about a point, you have to find the correct point.
    and if your looking for the FINAL speed, it's going to be zero.

    JMD
     
    Last edited: Dec 9, 2003
  7. Dec 9, 2003 #6

    HallsofIvy

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    Aren't those two statements contradictory? If the final speed is 0, then the angular momentum is 0 which is not true while it's rolling.

    Actually once it gets to pure roling without sliding, the friction is no longer a factor (that's the whole point of the wheel!).
     
  8. Dec 9, 2003 #7
    ive got a few formulas here:
    alpha = (5 * coeff friction * g)/2R
    omega = (5 * coeff friction * g * t)/2R

    the 8 m/s is beginning translation motion right? but i cant figure out where to begin. ive gone through a few sheets of paper, and i dont think he gave us enough to start with.
     
  9. Dec 9, 2003 #8
    okay, using 8 = R*omega, i was able to find a time. i did 8 = R (5 * coeff friction *g * t)/2r. i ended up with a time of 5.4 seconds. but im stuck now.
     
    Last edited: Dec 9, 2003
  10. Dec 9, 2003 #9
    heres what i did:
    alpha = (5 * mu * g) 2*R
    alpha = a/r, so a/r = (5 * mu * g)/ 2*R. with this i ended up with an a of 1.47.
    i than used this 1.47 in omega = initial omega(which is zero because it was not rotating) + alpha * time.
    i found time using the formula:
    t = 2 * initial velocity/7 * mu * g, which equals 3.89 s.

    than i went with v/r = 0 + 1.47 * t/R
    the r's cancel and i ended up with a final velocity of 5.7 m/s.
    have i done everything correct?
     
  11. Dec 10, 2003 #10
    correct.. You can also use conservation of Angular momentum, about the point where the ball makes contact with the floor.

    When the ball first starts
    [itex]l=mrv_0
    [/itex]
    when the ball starts to roll without sliping
    [itex]l=I \omega [/itex]
    [itex]I = \frac{2}{5}mr^2 + mr^2 [/itex]
    for rolling without slipping you can find
    [itex]\omega = \frac{v}{r} [/itex]

    and you can solve these and find
    [itex]v = \frac{5}{7}v_0
    [/itex]
    Which is 5.7 m/s

    I might have to edit this to get the LaTex to work.

    JMD
     
    Last edited: Dec 10, 2003
  12. Dec 10, 2003 #11
    They are contradictory. I've been ill with the flu and still not thinking clearly all the time. The final speed won't be zero.

    JMD
     
  13. Dec 10, 2003 #12

    Doc Al

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    Excellent. Hadn't thought of that.
     
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