Frictionless ring sliding on circle in constant acceleration

[Feodalherren]

Homework Statement

A box is sliding with constant acceleration a to the right. Inside the box there is a quarter of a circle upon which a frictionless ring can slide. Find the angle theta in terms of the other given variables.

Picture in solution

Homework Equations

F=ma, etc. basic stuff

The Attempt at a Solution

Pretty sure I did something wrong as I didn't even use "r".

 

[TomHart]

Looks right to me.

 

[Feodalherren]

Hi Tom,

Thanks for the quick response. They just put that "r" on there to throw me off huh?

 

[TomHart]

They just put that "r" on there to throw me off huh?

I'm not sure if it was to throw you off - possibly. But for a given acceleration, the angle has to be the same, regardless of the radius.

 

[Feodalherren]

But only in case there is no friction, such as here, correct? Because my intuition tells me it shouldn't be the same but I feel like that is because there would be far more friction if r was larger. Just want to check my logic here.

 

[TomHart]

But only in case there is no friction, such as here, correct? Because my intuition tells me it shouldn't be the same but I feel like that is because there would be far more friction if r was larger. Just want to check my logic here.

The angle is independent of radius when there is no friction. What would happen if there was friction, I'm not sure.

 

[PeroK]

But only in case there is no friction, such as here, correct? Because my intuition tells me it shouldn't be the same but I feel like that is because there would be far more friction if r was larger. Just want to check my logic here.

If you use the reference frame of the inside of the box, then there is a fictitious force to the left (proportional to the mass of the ring). This will slide the ring up a slope unless and until the gravity down the slope equals the fictitious force up the slope. As both of these forces are independent or $r$, then so is the angle.

Friction would also be proportional to the mass and change with the angle, and again be independent of $r$.

It shouldn't be too hard to add a coefficient of friction and see what happens.

 

[Feodalherren]

If you use the reference frame of the inside of the box, then there is a fictitious force to the left (proportional to the mass of the ring). This will slide the ring up a slope unless and until the gravity down the slope equals the fictitious force up the slope. As both of these forces are independent or $r$, then so is the angle.

Friction would also be proportional to the mass and change with the angle, and again be independent of $r$.

It shouldn't be too hard to add a coefficient of friction and see what happens.

Hmm. The way I see it, the distance that the ring must travel to reach the same angle increases with theta, that ought to mean that more energy is lost to friction but then again it's not a closed system.

 

[PeroK]

Hmm. The way I see it, the distance that the ring must travel to reach the same angle increases with theta, that ought to mean that more energy is lost to friction but then again it's not a closed system.

Yes, exactly, it's not a closed system, so the external force needed to keep the constant acceleration will vary depending on what's going on inside the box (at least initially).

If you found the original problem easy enough, you ought to try adding friction. There's no pressure, though!

 

[Feodalherren]

Yes, exactly, it's not a closed system, so the external force needed to keep the constant acceleration will vary depending on what's going on inside the box (at least initially).

If you found the original problem easy enough, you ought to try adding friction. There's no pressure, though!

I think I might try that this weekend. It sounds fairly trivial, wouldn't it just be the the coefficient of friction multiplied by the normal force acting in a direction tangential to point of contact between the ring and the slider facing the down and to the right?

 

[haruspex]

I think I might try that this weekend. It sounds fairly trivial, wouldn't it just be the the coefficient of friction multiplied by the normal force acting in a direction tangential to point of contact between the ring and the slider facing the down and to the right?

How do you know which way friction will act?

 

[Feodalherren]

Friction always acts counter to the direction of movement, no?

 

[haruspex]

Friction always acts counter to the direction of movement, no?

It acts to oppose relative lateral motion of the surfaces in contact. That may be actual motion (kinetic) or potential motion (static).
So which way is the potential motion?

 

[PeroK]

Friction always acts counter to the direction of movement, no?

If you look at the original problem, you have calculated the angle at which the forces are balanced. But, at this point the ring will still be moving up the circle. The ring, therefore, will oscillate about this equilibrium point.

If you have friction, you can still calculate the first equlibrium point. Again, at this point, the ring will be moving upwards. Then, you might think about what happens to the ring after it moves past the equilibrium point.

Alternatively, you could think about what would happen if the ring was held at the top of the circle and then released.

I must admit, I wasn't thinking as deeply as @haruspex when I suggested adding friction. I was only thinking about the first equilibrium point.

Perhaps we've got you in a bit deep now!

 

[Feodalherren]

I certainly recall doing harmonic motion in my systems and vibrations class but I can't say that I remember much of anything about how to calculate equilibriums. I might be able to draw it up in simulink hah :).

 

[PeroK]

I certainly recall doing harmonic motion in my systems and vibrations class but I can't say that I remember much of anything about how to calculate equilibriums. I might be able to draw it up in simulink hah :).

I wouldn't try to calculate the actual motion by solving equations of motion, but you could try to understand it. In the first case (without friction) there is only a single equilibrium point. The case with friction may not be so simple.

In any case, I think you are still a long way from this, as you haven't shown the force equations when you add friction. That's the first step and, to be honest, that was the exercise that I thought was worth doing.

 

[Feodalherren]

I wouldn't try to calculate the actual motion by solving equations of motion, but you could try to understand it. In the first case (without friction) there is only a single equilibrium point. The case with friction may not be so simple.

In any case, I think you are still a long way from this, as you haven't shown the force equations when you add friction. That's the first step and, to be honest, that was the exercise that I thought was worth doing.

Yeah I catch your drift. I think it's a little bit above my skill level right now. I'm going to work my way up to harder problems but I haven't done physics in a while so I don't want to overwhelm myself immediately. I'm sure the book will work its way up to harder problems in due time.