1. Nov 19, 2007

wowolala

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. Nov 20, 2007

azatkgz

$$\omega_b=15 rad/s$$

$$\omega_r=(20-5t)rad/s$$

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

$$v'=\omega'R=(\omega_b+\omega_r)R$$

3. Nov 20, 2007

wowolala

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

4. Nov 20, 2007

azatkgz

Ok.

$$\omega_b=20 rad/s$$

$$\omega_r=(15-5t) rad/s$$

as I said that

$$\omega'R=(\omega_r+\omega_b)R$$

we just eliminate R

$$\omega'=\omega_r+\omega_b$$

Original formula is $$d\theta=\omega dt$$
so we use this formula

$$d\theta=\omega'dt=(20+15-5t)dt$$

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

$$N=\frac{82.5}{2\pi}=13$$ times

5. Nov 21, 2007

wowolala

thx so much

THank YOU, YOu are so smart....