Mass m sliding without friction inside a rotating tube

[lorenz0]

Homework Statement:

A mass $m$ can slide without friction inside a tube rotated around one of its ends in a horizontal plane with constant angular velocity $\omega$. A second identical mass is hung from the mass by means of an ideal rope passing through the center of rotation as shown in the figure. To determine:
a) the radial distance $R_0$ from the center of rotation at which the first mass must be in order for the system to be in equilibrium;
b) the type of balance, giving adequate reasons for the answer;
c) the opposite lateral constraint reaction from the pipe wall in position $R_0$.
d) supposing that the first mass starts from stationary from the position $R_0$ found, the opposite lateral constraint reaction from the pipe wall when the mass reaches a distance of $2R_0$.

Relevant Equations:

$F_{centr}=m\omega^2 r, F_g=mg$

1) To be in equilibrium, it must be $$\begin{cases}F_{centr}-T=0\\ T-mg=0\end{cases}\Rightarrow F_{centr}=T=mg\Rightarrow m\omega^2 R_0=mg\Rightarrow R_0=\frac{g}{\omega^2}$$

2) It is intuitive that this equilibrium is unstable but I don't know how to formally prove this.

3) In $R_0$ the mass $m$ is in equilibrium so $N_{lat}=0$

4) I have tried by using angular impulse theorem and the work-energy theorem but I haven't got anywhere so far. The result should be $N_{lat}=\sqrt{2}mg$.

 

[PeroK]

For 2) you could think about what happens if you perturb the system by a small amount?

For 4) have you considered looking for an equation of motion for the system when the mass slides outwards?

 

[haruspex]

tried by using angular impulse theorem and the work-energy theorem

Would work be conserved? Would angular momentum be conserved?

 

[lorenz0]

Would work be conserved? Would angular momentum be conserved?

For 2) you could think about what happens if you perturn the system by a small amount?

For 4) have you considered looking for an equation of motion for the system when the mass slides outwards?

it seems to me that in polar coordinates we have $F_{\theta}=N_{lat}=ma_{\theta}=2r' \omega$ but I don't know how to find $r'$

 

[PeroK]

it seems to me that in polar coordinates we have $F_{\theta}=N_{lat}=ma_{\theta}=2r' \omega$ but I don't know how to find $r'$

That doesn't look right. For uniform angular velocity we have:$$a_r = \ddot r - rw^2, \\ a_{\theta} = 2\dot r w$$

I suggest you use the rotating reference frame, using the centrifugal force to analyse the radial motion.

Hint: use energy.

Then use the value for $\dot r$ to calculate $N$.

 

[lorenz0]

That doesn't look right. For uniform angular velocity we have:$$a_r = \ddot r - rw^2, \\ a_{\theta} = 2\dot r w$$

I suggest you use the rotating reference frame, using the centrifugal force to analyse the radial motion.

Hint: use energy.

Then use the value for $\dot r$ to calculate $N$.

shouldn't $a_r=r''-\omega^2 r$? Anyway, since there is no friction inside the tube we should get $a_r=0=r''-\omega^2 r\Rightarrow r(t)=Ae^{-\omega t}+Be^{\omega t}$ and by imposing the initial conditions $r(0)=R_0$ and $r'(0)=0$ we get $r(t)=\frac{R_0}{2}(e^{-\omega t}+e^{\omega t})$ and $r'(t)=\frac{\omega R_0}{2}(e^{-\omega t}+e^{\omega t})$ so $N_{lat}=F_{\theta}=2r'\omega=\omega R_0(e^{-\omega t}+e^{\omega t})$. Now, I just need to find the time necessary $m$ to go from $R_0$ to $2R_0$: any ideas?

 

[PeroK]

What about the influence of the second mass?

Solving for time is messy. That's why I hinted to use energy.

 

[lorenz0]

What about the influence of the second mass?

Solving for time is messy. That's why I hinted to use energy.

I am not sure in that request we are considering the effect of the second body and besides I have no idea how to do that. Anyway, I tried finding $T$ by doing $\int_{0}^{T}\tau dt=m(2R_0)^2 \omega\Rightarrow \int_{0}^{T} r(t)N_{lat}(t)dt=m(2R_0)^2 \omega\Rightarrow \int_{0}^{T} (\frac{R_0}{2}(e^{-\omega t}+e^{\omega t})\cdot \omega R_0 (e^{-\omega t}+e^{\omega t}))dt=m(2R_0)^2 \omega$ which looks messy and should give a solution, but it seems to me not an analytic one. That's the only thing I have been able to come up with up to now.

 

[PeroK]

I am not sure in that request we are considering the effect of the second body and besides I have no idea how to do that. Anyway, I tried finding $T$ by doing $\int_{0}^{T}\tau dt=m(2R_0)^2 \omega\Rightarrow \int_{0}^{T} r(t)N_{lat}(t)dt=m(2R_0)^2 \omega\Rightarrow \int_{0}^{T} (\frac{R_0}{2}(e^{-\omega t}+e^{\omega t})\cdot \omega R_0 (e^{-\omega t}+e^{\omega t}))dt=m(2R_0)^2 \omega$ which looks messy and should give a solution, but it seems to me not an analytic one. That's the only thing I have been able to come up with up to now.

Effectively you have system of mass $2m$. The first mass $m$ is pulled in the direction of increasing $r$ by the centrifugal force. The second mass is pulled in the opposite direction by the gravitational force.

That simplifies things conceptually- and agrees with your equilibrium from part 1).

It also helps with part 2).

For part 4) you can integrate force with respect to $r$ to get the increase in KE.

 

[lorenz0]

Effectively you have system of mass $2m$. The first mass $m$ is pulled in the direction of increasing $r$ by the centrifugal force. The second mass is pulled in the opposite direction by the gravitational force.

That simplifies things conceptually- and agrees with your equilibrium from part 1).

It also helps with part 2).

For part 4) you can integrate force with respect to $r$ to get the increase in KE.

Sorry, but I don't understand... I don't have the expression of a force which I can integrate with respect to distance and it seems to me that at most I could use $\int_{0}^{\theta}rN_{lat}=\frac{1}{2}m(\omega 2R_0)^2-\frac{1}{2}m(\omega R_0)^2$ but I don't know the angle... so I am stuck.

 

[haruspex]

Now, I just need to find the time necessary $m$ to go from $R_0$ to $2R_0$: any ideas?

No, you don't need to get r as a function of time. You just need $\dot r$ as a function of r and plug that into your equation for F.
The trick is not to worry unduly about the initial conditions. We are to assume an infinitesimal nudge away from equilibrium.

That said, I don't get the square root in the answer.

 

[lorenz0]

No, you don't need to get r as a function of time. You just need $\dot r$ as a function of r and plug that into your equation for F.
The trick is not to worry unduly about the initial conditions. We are to assume an infinitesimal nudge away from equilibrium.

That said, I don't get the square root in the answer.

Do you mean $N_{lat}=F_{\theta}=m\cdot 2r'\omega\overset{r'=\omega r}{=}m\cdot 2\omega^2 r$?

 

[haruspex]

Do you mean $N_{lat}=F_{\theta}=m\cdot 2r'\omega\overset{r'=\omega r}{=}m\cdot 2\omega^2 r$?

Yes.

 

[lorenz0]

Yes.

by integrating $m\cdot 2\omega^2 dr$ from $R_0$ to $2R_0$ I find $2mg$. I still don't get why we should integrate $F_{\theta}$ over a variation of radial length.

 

[haruspex]

by integrating $m\cdot 2\omega^2 dr$ from $R_0$ to $2R_0$ I find $2mg$.

That's what I get as the answer, but I'm not sure why you are doing that integration.

I still don't get why we should integrate Fθ over a variation of radial length.

No, I don't see how it can be solved just using energy, if that's what you are referring to.

 

[PeroK]

I got the book answer but I won't have time to post until.later today.

 

[haruspex]

That's what I get as the answer,

Ha! Schoolboy error on my part. The suspended mass is acceleration too.

 

[lorenz0]

That's what I get as the answer, but I'm not sure why you are doing that integration.

No, I don't see how it can be solved just using energy, if that's what you are referring to.

Isn't the difference in kinetic energy the work done by a force over a certain distance? My doubt is that $N_{lat}$ is not acting from $R_0$ to $2R_0$ i.e. not radially but tangentially. How did you get that value otherwise?

 

[lorenz0]

Ha! Schoolboy error on my part. The suspended mass is acceleration too.

Yes, and since they are connected by an ideal rope the one hanging has the same upward acceleration that the the mass inside the tube has radially. This is what does not allow one (it seems to me) to solve this problem like the classic "bead on a rotating wire" problem.

 

[haruspex]

Yes, and since they are connected by an ideal rope the one hanging has the same upward acceleration that the the mass inside the tube has radially. This is what does not allow one (it seems to me) to solve this problem like the classic "bead on a rotating wire" problem.

It only changes the radial acceleration equation a little. Do you now get the book answer using the method in post #13?

 

[haruspex]

Isn't the difference in kinetic energy the work done by a force over a certain distance? My doubt is that $N_{lat}$ is not acting from $R_0$ to $2R_0$ i.e. not radially but tangentially. How did you get that value otherwise?

How did I get an answer without using energy?
I just wrote down the force equations and solved the differential equation from the radial acceleration to get the velocity as a function of position. Then I plugged that into the tangential force equation. Isn't that what you did in post #13?

 

[lorenz0]

How did I get an answer without using energy?
I just wrote down the force equations and solved the differential equation from the radial acceleration to get the velocity as a function of position. Then I plugged that into the tangential force equation. Isn't that what you did in post #13?

$F_r=m(r''-\omega^2 r), F_{\theta}=2r'\omega$. Now, $F_{\theta}=N_{lat}$ and $F_r=m\omega^2 r-T$. Now, $T-mg=ma$ and $a=r''-\omega^2 r$ so it should be $r''=\frac{T}{m}-g$. If now I am to solve the radial differential equation for $r''$ how do I eliminate $T$ first? Since it is an internal force between the two masses perhaps by considering the system consisting of both masses and then I would have $F_r=2m(r''-\omega^2 r), F_{\theta}=N_{lat}=2r'\omega$. Now, the question is: what is $F_r$? If I put myself in the rotating system I would say $F_{rad}=m\omega^2 r$ which would give the differential equation $r''=\frac{3}{2}\omega^2 r$ to solve with the initial conditions $r(0)=R_0, r'(0)=0$. Is this correct?

 

[kuruman]

$F_r=m(r''-\omega^2 r), F_{\theta}=2r'\omega$. Now, $F_{\theta}=N_{lat}$ and $F_r=m\omega^2 r-T$. Now, $T-mg=ma$ and $a=r''-\omega^2 r$ so it should be $r''=\frac{T}{m}-g$. If now I am to solve the radial differential equation for $r''$ how do I eliminate $T$ first?

You have to eliminate the tension. If you eliminate the tension, you get the appropriate differential equation to solve for $r(t)$. It's similar to what you have in post #6, but includes $g$ and gives the value you found in part (a) for $R_0$ when you set the radial acceleration equal to zero. Personally, I think that solving the diff eq is more straightforward than using energy considerations. Hint: What is the difference between $r''$ and $a$ in the equations quoted above? Answer this and you will stop going around in circles.

 

[PeroK]

Taking radial motion in the rotating reference frame we have:$$F = mrw^2 - mg$$ Integrating this from $r_0$ to $2r_0$ gives the kinetic energy:$$\frac 3 2 mw^2r_0^2 - mgr_0 = \frac 1 2 m\frac {g^2}{w^2}$$ Which we equate to$$\frac 1 2 (2m) \dot r^2$$ Giving $$\dot r = \frac 1 {\sqrt 2} \frac g w$$ Finally, using $$ N = 2m \dot r w$$ gives the book answer.

 

[PeroK]

PS I posted the above before I realized there was no longer any doubt about the book answer. Given the effort spent by the OP on this thread perhaps a complete outline solution is useful at this stage? If not I guess it should be deleted!

 

[haruspex]

gives the kinetic energy:

Now I understand your KE approach: in the rotating frame there is no tangential KE and the normal force from the tube is balanced by the fictitious force from the rotation.
Of course, essentially the same algebra works, but either not calling it energy or thinking of it as the 'radial component' of the kinetic energy, in the lab frame.

 

[lorenz0]

Taking radial motion in the rotating reference frame we have:$$F = mrw^2 - mg$$ Integrating this from $r_0$ to $2r_0$ gives the kinetic energy:$$\frac 3 2 mw^2r_0^2 - mgr_0 = \frac 1 2 m\frac {g^2}{w^2}$$ Which we equate to$$\frac 1 2 (2m) \dot r^2$$ Giving $$\dot r = \frac 1 {\sqrt 2} \frac g w$$ Finally, using $$ N = 2m \dot r w$$ gives the book answer.

Now I understand how to take into account the tension due to the second mass. Thanks!

 

[PeroK]

Now I understand how to take into account the tension due to the second mass. Thanks!

I thought this was an excellent problem, as there were many possible approaches and it could be simplified conceptually.

It worth close study, I think, to really understand that you can trust Newton's laws with fictitious forces.