How do you calculate the terminal velocity of these two balls?

[algebra topology]

TL;DR Summary: Almost as difficult as the Physics Olympics semi-finals or finals

The coefficient of friction of the two arcs is the same.The radius of the ball is not taken into account.The roll of the ball is ignored.Can it be calculated or estimated quantitatively?The difficulty is estimated to be above the Physics Olympics semi-finals.

 

[Juanda]

I'm not sure I'm understanding your post but I'll try.

Assuming the bodies remain attached to the track, the velocity of a body at each point of the track will be a function of the height of said point because of the conservation of energy.
For example, for the body on the left:
$$\frac{1}{2}mv_A^2+mgh_A=\frac{1}{2}mv_B^2+mgh_B$$
Where A and B are two different points on the track.

If there is dynamic friction between the body and the track that consumes the energy then you'll have to subtract the amount of energy consumed to get there.
$$\frac{1}{2}mv_A^2+mgh_A=\frac{1}{2}mv_B^2+mgh_B-W_{f}$$

I just realized the energy consumed during the displacement $W_f$ could be a little harder to calculate than I initially anticipated. It depends on the normal force $N$ which I believe is not constant during the trajectory. Are you sure the problem states there is sliding friction? The fact this is a high school problem and the statement mentions a ball I'd initially think this is the typical simplification of a small ball that can be considered a punctual mass with no dynamic friction going on.

 

[algebra topology]

I'm not sure I'm understanding your post but I'll try.

Assuming the bodies remain attached to the track, the velocity of a body at each point of the track will be a function of the height of said point because of the conservation of energy.
For example, for the body on the left:
$$\frac{1}{2}mv_A^2+mgh_A=\frac{1}{2}mv_B^2+mgh_B$$
Where A and B are two different points on the track.

If there is dynamic friction between the body and the track that consumes the energy then you'll have to subtract the amount of energy consumed to get there.
$$\frac{1}{2}mv_A^2+mgh_A=\frac{1}{2}mv_B^2+mgh_B-W_{f}$$

I just realized the energy consumed during the displacement $W_f$ could be a little harder to calculate than I initially anticipated. It depends on the normal force $N$ which I believe is not constant during the trajectory. Are you sure the problem states there is sliding friction? The fact this is a high school problem and the statement mentions a ball I'd initially think this is the typical simplification of a small ball that can be considered a punctual mass with no dynamic friction going on.

Thanks.I am sure that there is sliding friction.So the difficulty of the problem is to solve the sliding friction.

 

[kuruman]

I would like to see a full statement of the problem. Is this a bead with a hole and a wire through it that is constrained to stay on the curved paths? If not, it is entirely possible for the mass to leave the track and follow a ballistic parabolic trajectory while on the left curve.

 

[algebra topology]

I would like to see a full statement of the problem. Is this a bead with a hole and a wire through it that is constrained to stay on the curved paths? If not, it is entirely possible for the mass to leave the track and follow a ballistic parabolic trajectory while on the left curve.

This is a problem in space.But it is confined to a plane.The wire stands for a rough transversal semicircle.Because the speed is not big, so do not consider the speed is too big and out of orbit.The support force divided by the friction force is considered constant μ.

 

[Juanda]

Is this really from high school? I definitely didn't work on problems like this in high school .

Anyways, the full statement of the problem as Kuruman said would be extremely useful. If the original statement includes pictures you could add them as well to help us understand you.

 

[kuruman]

This is a problem in space.But it is confined to a plane.The wire stands for a rough transversal semicircle.Because the speed is not big, so do not consider the speed is too big and out of orbit.The support force divided by the friction force is considered constant μ.

So we have a "rough" wire on which the bead slides, no gravity, and the centripetal force is provided by the wire. Is that your understanding? If "yes", then it seems to me that you can use Newton's second law to derive a simple differential equation for the linear acceleration along the path that is easy to solve.

 

[algebra topology]

So we have a "rough" wire on which the bead slides, no gravity, and the centripetal force is provided by the wire. Is that your understanding? If "yes", then it seems to me that you can use Newton's second law to derive a simple differential equation for the linear acceleration along the path that is easy to solve.

I mean just consider gravity, elasticity and friction.

 

[Juanda]

I mean just consider gravity, elasticity and friction.

Elasticity? Where is the spring?
You also mentioned an orbit at some point... I don't see how that applies here.
I'm definitely getting lost in this problem.

I believe it's what @kuruman said at #7 but with the addition of gravity as OP said on #8. A bead with a hole through the middle so it is constrained to follow the path set by the rough and rigid wire which is set in a constant gravitational field. Basically, the wire is held on both ends.

I think it'd be solved with just conservation of energy without using differential equations.
The expression for normal acceleration is as follows:
$$a_n=\omega^2r=\frac{v^2}{r}$$
So it's possible to get the normal force that will give an expression for the friction force acting on the body. Also, the displacement is known as well so it's possible to obtain the energy consumed as the bead moves through the wire.
EDIT: Nope, I again believe calculus is necessary to calculate the energy dissipated by friction. The expression for the normal force probably can be a useful link to express the functions in the necessary terms for the integration.

 

[Juanda]

Then it seems to me that you can use Newton's second law to derive a simple differential equation for the linear acceleration along the path that is easy to solve.

I think I got the differential equations that describe the problem. I'd like to post them to verify if they're OK but I'm not sure if that follows the policy regarding people asking how homework is done.
Do I just post them or should I wait for OP to try to come up with a solution by himself?

By the way, I got a not-very-good-looking differential equation. I'd be extremely surprised if these things are being solved in high schools abroad.

 

[algebra topology]

I think I got the differential equations that describe the problem. I'd like to post them to verify if they're OK but I'm not sure if that follows the policy regarding people asking how homework is done.
Do I just post them or should I wait for OP to try to come up with a solution by himself?

By the way, I got a not-very-good-looking differential equation. I'd be extremely surprised if these things are being solved in high schools abroad.

In our semi-finals Physics Olympiad here, differential equation are a common occurrence.

 

[Juanda]

Out of curiosity, where is "here"? I guess the country would suffice.

Anyway, still I believe you need to show how you approached the problem first.

Oh, and remember to use LaTex if you're going to post some math. It makes it easier for anyone to read what you write. You can see the guide here. I also often use this to have a preview of what I'm writing. Then you just need to put it between $$ or ## and it'll show it in the LaTex format once you post the reply.

 

[erobz]

I think I got the differential equations that describe the problem. I'd like to post them to verify if they're OK but I'm not sure if that follows the policy regarding people asking how homework is done.
Do I just post them or should I wait for OP to try to come up with a solution by himself?

By the way, I got a not-very-good-looking differential equation. I'd be extremely surprised if these things are being solved in high schools abroad.

Send them privately to verify. Full solutions aren't allowed until after the OP came up with a correct one (I believe).

 

[haruspex]

I also often use this to have a preview of what I'm writing.

Alternatively you can just click the magnifying glass button at the right hand end of the toolbar above the edit box. If it doesn’t work, do a page refresh and try again.
If it was a lot of typing, do a select all and copy to paste buffer before the refresh. The refresh should not lose your edits, but it did happen to me once.

 

[Juanda]

Alternatively you can just click the magnifying glass button at the right hand end of the toolbar above the edit box. If it doesn’t work, do a page refresh and try again.

WOW. I didn't know about that feature! That's very useful. Thanks for the heads up.

 

[haruspex]

In our semi-finals Physics Olympiad here, differential equation are a common occurrence.

Then please post your attempt at the equations.

 

[MatinSAR]

First of all, I apologize for interfering. By terminal velocity, Does the question ask us to find velocity at the point where there's no net force acting on the ball?

 

[Juanda]

First of all, I apologize for interfering. By terminal velocity, Does the question ask us to find velocity at the point where there's no net force acting on the ball?

I noticed that as well. There are a few words (terminal velocity does not apply here, naming elasticity but there is no spring on sight, orbits, etc.) that simply make no sense in this problem.
We asked for the full statement of the original problem but with no luck so far. I attempted to describe the original problem the way I see it in post #9 but we don't really know if that's what's really going on.
At least that's the problem I worked on when deriving the equations that describe the problem.

 

[kuruman]

I interpreted terminal velocity to mean velocity at the end of each arc.

 

[haruspex]

I mean just consider gravity, elasticity and friction.

Are you sure we are to consider gravity? It seems unlikely.

 

[kuruman]

Does the question ask us to find velocity at the point where there's no net force acting on the ball?

If the object stays on the circular arc at all times, its velocity (if not the speed) is changing direction at all times which means that it is accelerating at all times. Therefore the net force is non-zero at all times.

 

[algebra topology]

I interpreted terminal velocity to mean velocity at the end of each arc.

The terminal velocity is the instantaneous velocity to the end of another circular orbit

 

[algebra topology]

Out of curiosity, where is "here"? I guess the country would suffice.

Anyway, still I believe you need to show how you approached the problem first.

Oh, and remember to use LaTex if you're going to post some math. It makes it easier for anyone to read what you write. You can see the guide here. I also often use this to have a preview of what I'm writing. Then you just need to put it between $$ or ## and it'll show it in the LaTex format once you post the reply.

China.

 

[Juanda]

The terminal velocity is the instantaneous velocity to the end of another circular orbit

China.

Okay, so it seems this is a language barrier thing. I'm not English either so I feel you.
Did you maybe mean "final velocity" instead of "terminal velocity"?
Anyway, we're still expecting your attempt at the problem.

 

[algebra topology]

Okay, so it seems this is a language barrier thing. I'm not English either so I feel you.
Did you maybe mean "final velocity" instead of "terminal velocity"?
Anyway, we're still expecting your attempt at the problem.

Yes.

 

[algebra topology]

Okay, so it seems this is a language barrier thing. I'm not English either so I feel you.
Did you maybe mean "final velocity" instead of "terminal velocity"?
Anyway, we're still expecting your attempt at the problem.

$v\frac{\mathrm{d} v}{\mathrm{d} \theta } + gRcos \theta + \mu gRcos \theta = \mu v^2$On the right, I get a complex differential equation.

 

[algebra topology]

$$ v\frac{\mathrm{d} v}{\mathrm{d} \theta } + gRcos \theta + \mu gRcos \theta = \mu v^2 $$

 

[haruspex]

$$ v\frac{\mathrm{d} v}{\mathrm{d} \theta } + gRcos \theta + \mu gRcos \theta = \mu v^2 $$

As I posted, unless it states that you should include gravity then I strongly suggest you should not. With gravity, the ODE looks insoluble to me; without gravity it is straightforward.

 

[Juanda]

$$ v\frac{\mathrm{d} v}{\mathrm{d} \theta } + gRcos \theta + \mu gRcos \theta = \mu v^2 $$

It looks very different to what I got. But I'm not an expert on this. I'll share my attempt.

I solved the left half of the problem. The equations that describe the other half should be equivalent. It's just a matter of getting the velocity of the bead at the end of the left loop and using it as the initial condition for the next loop.

Here is the diagram I used.

Using Newton we know that:
$$\sum \vec{F}=m\ddot{\vec{r}} \rightarrow m\vec{g}+\vec{F_f}+\vec{N}=m\ddot{\vec{r}} \tag {1}$$
(I prefer to express them all being positive and allow trigonometry to give me the right signs)

Since we're dealing with a circular path, it makes sense to use polar coordinates. Therefore, I'll express the cartesian coordinates $x$ and $y$ in terms of $r$ and $\theta$. Take into account that $r$ is constant so its derivative with respect to time will be 0.
$$x=r\cos\theta \tag{2}$$
$$\dot{x}=-r\dot{\theta}\sin\theta \tag{3}$$
$$\ddot{x}=-r(\ddot{\theta}\sin\theta+\dot{\theta}^2\cos\theta) \tag{4}$$
$$y=r\sin\theta \tag{5}$$
$$\dot{y}=r\dot{\theta}\cos\theta \tag{6}$$
$$\ddot{y}=r(\ddot{\theta}\cos\theta-\dot{\theta}^2\sin\theta) \tag{7}$$

I'll also need to express the horizontal and vertical components of the forces. The weight is clear, it's always pointing down and with the same magnitude. For the other forces, we will need $\theta$.
$$N_x=N\cos(\theta+\pi)=-N\cos\theta \tag{8}$$
$$N_y=N\sin(\theta+\pi)=-N\sin\theta \tag{9}$$
$$F_{fx}=F_f\cos(\theta +\pi /2)=-F_f\sin(\theta)=-\mu N\sin\theta \tag{10}$$
$$F_{fy}=F_f\sin(\theta +\pi /2)=F_f\cos(\theta)=\mu N\cos\theta \tag{11}$$

Finally, applying Newton to the horizontal and vertical directions results in:
$$\sum F_x=m\ddot{x}\rightarrow N_x+F_{fx}=m\ddot{x}$$
$$-N\cos\theta-\mu N\sin\theta=-mr(\ddot{\theta}\sin\theta+\dot{\theta}^2\cos\theta) \tag{12}$$
$$\sum F_y=m\ddot{y}\rightarrow -mg +N_y+F_{fy}=m\ddot{y}$$
$$-mg-N\sin\theta+\mu N\cos\theta=mr(\ddot{\theta}\cos\theta-\dot{\theta}^2\sin\theta) \tag{13}$$

I believe from $(12)$ and $(13)$ it should be possible to obtain the expressions for $N$ and $\theta$.

I believe we can take advantage of the fact that, since we know it's a circular path, the centripetal acceleration is defined.
$$a_n = \dot{\theta}^2r \rightarrow m\dot{\theta}^2r = - N - mg \cos(\theta-\pi /2) \tag{14}$$
Although I have some doubts about $14$. I guess one possibility would be to solve it both ways and see if the methods are equivalent or not. I can't do that though because those differential equations are too hard for me now.

 

[kuruman]

The extra complication is that in the equations above $N$ is not constant but depends on $\theta$ and $v^2$ as a free body diagram would show.

 

[Juanda]

The extra complication is that in the equations above $N$ is not constant but depends on $\theta$ and $v^2$ as a free body diagram would show.

Do you think the procedure from #29 would give the right result?

Do you have some insight about what I mentioned related to equation $14$?

I believe from $(12)$ and $(13)$ it should be possible to obtain the expressions for $N$ and $\theta$.

I believe we can take advantage of the fact that, since we know it's a circular path, the centripetal acceleration is defined.
$$a_n = \dot{\theta}^2r \rightarrow m\dot{\theta}^2r = - N - mg \cos(\theta-\pi /2) \tag{14}$$
Although I have some doubts about $14$. I guess one possibility would be to solve it both ways and see if the methods are equivalent or not. I can't do that though because those differential equations are too hard for me now.

 

[haruspex]

Do you think the procedure from #29 would give the right result?

Do you have some insight about what I mentioned related to equation $14$?

I just spotted this in post #5:

This is a problem in space.

So there is definitely no need to consider gravity.

 

[Juanda]

I just spotted this in post #5:

So there is definitely no need to consider gravity.

OP has been giving confusing bits describing the problem during the thread. I tried to provide a clearer description in post #9 but I got no answer from him about it. Still, if there really is no gravity as you just said it's just a matter of setting $g=0$.
I believe the procedure I posted should be OK but I don't know for sure. I'm especially doubtful about the comments on equation $14$ as I said in post #31.

 

[haruspex]

OP has been giving confusing bits describing the problem during the thread. I tried to provide a clearer description in post #9 but I got no answer from him about it. Still, if there really is no gravity as you just said it's just a matter of setting $g=0$.
I believe the procedure I posted should be OK but I don't know for sure. I'm especially doubtful about the comments on equation $14$ as I said in post #31.

Your method in #29 is long-winded. You can write equation 14 straightaway (except, g=0 and you have a sign error; according to your diagram N should be positive). You just need another equation to relate N to $\ddot\theta$.

 

[Juanda]

Your method in #29 is long-winded.

There might be a shortcut but I wasn't sure about it so I tried to be thorough.

You can write equation 14 straightaway (except, g=0 and you have a sign error; according to your diagram N should be positive). You just need another equation to relate N to $\ddot\theta$.

Do you mean it'd be like this (with $g=0$)?
$$a_n = \dot{\theta}^2r \rightarrow m\dot{\theta}^2r = N $$
Since $N$ points inwards I thought I should give it a negative sign. It is for these kinds of possible confusions that I try to avoid these "shortcuts" if I have no way to check for the solution to see if the result makes physical sense. I know it's possible to derive that expression using vectors and then the signs appear by themselves but I don't remember how to do that now.

By the way, do you know what'd be that additional equation?
In the long version of the problem, the solution comes from equations $12$ and $13$ with the possibility of using $14$ (when expressed correctly) if it makes the system of differential equations easier to solve. But I don't know what would be the equation "$15$" that you mentioned.