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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.

  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.

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


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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
    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
    Which is 5.7 m/s

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

    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.

  13. Dec 10, 2003 #12

    Doc Al

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    Excellent. Hadn't thought of that.
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