a question about rotational speed

[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?

[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

[wowolala]

\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


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

[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

[wowolala]

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

THank YOU, YOu are so smart....