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

[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.

 

[haruspex]

Since N points inwards I thought I should give it a negative sign.

You show N pointing inwards, which means that is the positive direction for N. If your equations give a negative value then the implication is that the normal force acts outwards. Clearly you should get a positive value, but your eqn 14 can't do that.

what'd be that additional equation?

The effect of friction.

 

[kuruman]

I also think that you miss an important problem here. If you write Newton's second law in the radial direction, you get $$N+mg\sin\!\theta=\frac{mv^2}{R}\implies N=\frac{mv^2}{R}-mg\sin\!\theta.$$As you can see the magnitude of the normal force is a function of both the angle $\theta$ and the speed. An additional complication is that the speed always decreases because of friction which, of course, depends on $N$. I don't see how $v(\theta)$ can be determined independently of $N$.

I am still waiting for a full statement of the problem or a web link, but I am not holding my breath.

 

[haruspex]

I also think that you miss an important problem here. If you write Newton's second law in the radial direction, you get $$N+mg\sin\!\theta=\frac{mv^2}{R}\implies N=\frac{mv^2}{R}-mg\sin\!\theta.$$As you can see the magnitude of the normal force is a function of both the angle $\theta$ and the speed. An additional complication is that the speed always decreases because of friction which, of course, depends on $N$. I don't see how $v(\theta)$ can be determined independently of $N$.

I am still waiting for a full statement of the problem or a web link, but I am not holding my breath.

As noted in post #32, there is no gravity.

 

[kuruman]

I just spotted this in post #5:

So there is definitely no need to consider gravity.

I considered the "in space" equivalent to "no gravity" and that is why I tried to guide (oh so gently) the OP in that direction in post #7. However, in post #8 OP said explicitly that there is gravity. I still think that the exact statement of the problem (or an appropriate link) should be posted here before spending any more time on this.

 

[nasu]

I still think that the exact statement of the problem (or an appropriate link) should be posted here before spending any more time on this.

This happens too often. The OP does not bother to provide a full description and the thread goes all over the place.

 

[kuruman]

This happens too often. The OP does not bother to provide a full description and the thread goes all over the place.

Yup.

 

[kuruman]

Anyone still interested in this problem may consider a formulation with streamlined solution posted here.

 

[anuttarasammyak]

I take it easy as sketched.

 

[kuruman]

I take it easy as sketched.

View attachment 333676

$N=mg~\sin\theta~$ "as sketched" only if the mass is at rest at each point on the track. Otherwise, it is what is necessary to provide the observed acceleration.

 

[anuttarasammyak]

N=mg sin⁡θ "as sketched" only if the mass is at rest at each point on the track. Otherwise, it is what is necessary to provide the observed acceleration.

In order that we can neglect centrifugal force in N, the condition
g >>\frac{v^2}{R}
should be necessary.

[EDIT] Now I perceived under this condition the ball cannot overcome the top where potential energy mgR < 1/2 mv_i^2 initial kinetic energy. So this is not the case we are looking for.

[EDIT2] I further recognize that the ball leaves the slope in my ball and bamp model. The ball just go straight up with initial velocity leaving the slope, stops and falls. How the ball is constrained on the designed trajectory ? Friction takes place between ball surface and WHAT ? I am afraid there is no generic answer to this problem but are specific answers according to these conditions given.

 

[kuruman]

In order that we can neglect centrifugal force in N, the condition
g &gt;&gt;\frac{v^2}{R}
should be necessary.

So how would you write Newton's second law? If the mass is to follow the semicircular paths, its direction is changing continuously. If you ignore the centripetal (centrifugal if you prefer) force in N, then the direction of the velocity cannot change. Maybe I don't understand what you mean so please write Newton's second law for the part of the motion when the mass is where you show in your diagram.