Ball in Rotating Disk: Kinetic Energy, Work, and Power

[Bestfrog]

Homework Statement

A disk of radius $R$ has a cylindrical hollow that goes through it crossing its center $O$. The disk rotates around its central axis with constant angular speed $\vec{\omega}$. A little ball of mass $m$ and with the same radius of the hollow, is at a distance $R_0$ from $O$. At a certain instant the ball, that was in quiet with respect the disk, is allowed to move. Find, neglecting the friction forces,
1. The radial speed of the ball when it comes out from the disk.
2.The work done by a engine to keep $\vec{\omega}$ constant.
3. The maximum power delivered by the engine.
4. The time to reach the end of the hollow.

The Attempt at a Solution

1. I thought about the potential energy (that is converted in kinetic energy) due to the centrifugal force, so $F=m \omega^2 r$, then $$\int_0^U dU=\int_{R}^{R_0} -m\omega^2 r dr$$ and the final radial speed is $v_r=\omega \sqrt{R^2-R_{0}^2}$. I'm sure (I hope I'm sure :D)
2. There are no dissipative forces, so the total work done is $L=\Delta K$. The final speed is $v^2=v_{r}^2+v_{T}^2$ ($v_T$ is the tangential speed) so $$L=\frac{1}{2}m\omega^2(2R^2-R_{0}^2)$$.
The problem is how to deal with the maximum power delivered! Hints?

 

[TSny]

Parts 1 and 2 look good.

The problem is how to deal with the maximum power delivered! Hints?

Can you find K as a function of r? How is power related to K?

 

[rude man]

Hmm - what bothers me about the above is that there is actually no centrifugal force. Where would it come from? Nothing pushing or pulling on the ball in the r direction.

Rather, the acceleration of the ball in the r direction = 0 = d²r/dt² - ω²r
which with dr/dt|_r=R0 = 0 solves to r = R₀ cosh(ωt)
with boundary condition R = R₀ cosh(ωT)
where T is time spent in the groove.
So, answer to part 4 would be T = (1/ω) cosh^-1(R/R₀)
So dr/dt = ωR₀ sinh(ωt)
and dr/dt(R) = ωR₀ sinh[cosh^-1(R/R0)] as the answer to part 1.
all of which admittedly seems a bit far-out.

Notwithstanding, I can't accept the idea that centrifugal force exists unless someone can point out its source.

 

[TSny]

Your work looks spot-on to me.

The centrifugal force method presupposes working in the rotating frame where there would also be a coriolis force. But the coriolis force does not affect the radial motion in this problem. This approach gives the radial speed as a function of r, which is convenient for the first two questions.

 

[rude man]

Your work looks spot-on to me.

This approach gives the radial speed as a function of r, which is convenient for the first two questions.

As does mine:
dr/dt (r) = ωR₀ sinh[cosh^-1(r/R₀)].
Afraid I didn't understand the rest of your post, sadly.

 

[TSny]

As does mine:
dr/dt (r) = ωR₀ sinh[cosh^-1(r/R₀)].
Afraid I didn't understand the rest of your post, sadly.

Sorry for not being clearer. I was referring to your work as being spot on. You get r as a function of time from which you can answer the various parts of the question. Bestfrog essentially gets the radial speed v_r as a function of r. This can also be used as a basis for answering the different parts of the question.

 

[rude man]

Sorry for not being clearer. I was referring to your work as being spot on. You get r as a function of time from which you can answer the various parts of the question. Bestfrog essentially gets the radial speed v_r as a function of r. This can also be used as a basis for answering the different parts of the question.

Thanks T. You mean you agree with my math? That would imply that his $v_r=\omega \sqrt{R^2-R_{0}^2}$ has to be the same as my dr/dt(R) = ωR₀ sinh[cosh^-1(R/R₀)] as the answer to part 1! I'll put some numbers in & see if they agree.

 

[TSny]

That would imply that his $v_r=\omega \sqrt{R^2-R_{0}^2}$ has to be the same as my dr/dt(R) = ωR₀ sinh[cosh-1(R/R₀)] as the answer to part 1!

To show the equivalence, use the identity sinh²(x) = cosh²(x) - 1.

 

[rude man]

To show the equivalence, use the identity sinh²(x) = cosh²(x) - 1.

Thank you Sir! Will look into this tomorrow. Win-win for everyone it seems!
rudy

 

[Bestfrog]

@TSny yes I forgot to write that I was in the reference frame of the rotating disk.
In the original problem the point 4 can be approached considering the relations $sin(i \alpha)=i(e^{\alpha} -e^{-\alpha})/2$ and $cos(i \alpha)=(e^{\alpha} +e^{-\alpha})/2$.
I don't have experience with complex numbers so I didn't know how to do the point 4 in this way. (I used the relation $dt=\frac{dr}{v_r}=\frac{dr}{\omega \sqrt{r^2 -R_{0}^2}}$, and I used Mathematica to solve this integral)