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Homework Help: A question about rotational speed

  1. Nov 19, 2007 #1
    A casino roulette wheel is set spinning with an initial angular speed of 15 rad/s. The ball is set spinning in the opposite direction with a constant angular speed of 20 rad/s as the
    "00" passed by. If friction makes the wheel slow down with an angular acceleration of 5 rad/s^2, How many times does the ball pass by the "00" after 3 seconds? ( Ignore the slowing down of the ball due to friction. )


    thx so much... could somebody help me to solve this quesion?
     
  2. jcsd
  3. Nov 20, 2007 #2
    [tex]\omega_b=15 rad/s[/tex]

    [tex]\omega_r=(20-5t)rad/s[/tex]

    Now imagine,you're sitting on the roulette.You'll see that roulette is not moving.And ball is moving with velocity

    [tex]v'=\omega'R=(\omega_b+\omega_r)R[/tex]
     
  4. Nov 20, 2007 #3
    thx , i still don't quite understand.


    you wrote the angular speed for the ball is 15 rad/s, but the question says the wheel is 15 rad/s , is something wrong.

    in your last step, how can we find the R, since R is unknown.

    finally, please tell me how many times the ball passes by "00", now, i am so confused..


    thx
     
  5. Nov 20, 2007 #4
    Ok.

    [tex]\omega_b=20 rad/s[/tex]

    [tex]\omega_r=(15-5t) rad/s[/tex]

    as I said that

    [tex]\omega'R=(\omega_r+\omega_b)R[/tex]

    we just eliminate R

    [tex]\omega'=\omega_r+\omega_b[/tex]


    Original formula is [tex]d\theta=\omega dt[/tex]
    so we use this formula

    [tex]d\theta=\omega'dt=(20+15-5t)dt[/tex]

    after integrating
    [tex]\theta=20t+15t-\frac{5t^2}{2}=20\times 3+15\times 3-\frac{5\times 3^2}{2}=82.5 rad[/tex]
    [tex]2\pi[/tex] is one cycle,so it passes

    [tex]N=\frac{82.5}{2\pi}=13[/tex] times
     
  6. Nov 21, 2007 #5
    thx so much

    THank YOU, YOu are so smart....
     
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