Conservative forces in circular motion?

[disjk47]

TL;DR Summary: The rigid object move around the circle with constant force how it possible the force

 

[disjk47]

The rigid object move around the circle with constant force how it possible the result of the forces are conservative!?

 

[topsquark]

The rigid object move around the circle with constant force how it possible the result of the forces are conservative!?

How do you determine if a force is conservative? What needs to happen?

-Dan

 

[haruspex]

The rigid object move around the circle with constant force how it possible the result of the forces are conservative!?

Why do you think it could not be?

 

[kuruman]

A satellite moves around the Earth in a circular orbit. Is the force on the satellite not conservative?

 

[BvU]

Hello @disjk47 ,

##\qquad ##!
​

Can you tell us a little bit about where you are in the curriculum ?
You know a constant force on a mass towards a central point can maintain a circular trajectory ?
What do you know about conservative forces ?

##\ ##

 

[disjk47]

How do you determine if a force is conservative? What needs to happen?

-Dan

 

[disjk47]

Why do you think it could not be?

The external and constant force enter the system any time it should be the non conservative force

 

[disjk47]

Hello @disjk47 ,

##\qquad ##!
​

Can you tell us a little bit about where you are in the curriculum ?
You know a constant force on a mass towards a central point can maintain a circular trajectory ?
What do you know about conservative forces ?

##\ ##

At the first time it's on (1) point in bottom. The F force enter the mass and it's go around the circle.
I know the external force make system non-conservative

 

[haruspex]

The external and constant force enter the system any time it should be the non conservative force

Constant force as a vector or constant magnitude of force?

 

[disjk47]

Constant force as a vector or constant magnitude of force?

Vector

 

[haruspex]

Vector

Then like local gravity, which is conservative:
https://en.wikipedia.org/wiki/Conservative_force

 

[topsquark]

View attachment 335820

You have a rigid body moving in a circle. I admit that you never used the word "uniform," but that statement "circular motion" usually implies that it is moving with a constant angular speed. (If this is not the case then the problem statement probably should be changed somewhat to reinforce that this is not uniform motion.) That implies that the magnitude of the force is constant and the force vector changes as the object goes on its path. If that is the case then ##\nabla \times \textbf{F} = 0##, and the force is conservative.

If the force is not constant, but the object is still constrained to move in a circle (say with increasing angular speed), then we may not have a conservative force. We would need more information from the problem statement about the nature of the force.

Are you sure the force vector is constant? If that's the case then the object could be constrained to move along a circular path, ie. the path will not be necessarily be closed, but it will eventually settle down to an equilibrium point. In that case things will be a bit stickier and the force may, indeed, not be conservative. Again, we would have no way to tell from the problem statement. (As mentioned above, gravity would fit this requirement, and gravity is a conservative force. But a rocket pack attached to the object would almost certainly not provide a conservative force and could produce a similar motion.)

-Dan

 

[vanhees71]

I think the problem is that the question is not well posed. Is it a particle moving on a circle in the constant gravitational field close to Earth (which is a mathematical pendulum if the circle is vertically oriented)? Then you have in addition reaction forces due to the constraint for the particle moving on the circle. Otherwise the question doesn't make much sense to me.

 

[disjk47]

You have a rigid body moving in a circle. I admit that you never used the word "uniform," but that statement "circular motion" usually implies that it is moving with a constant angular speed. (If this is not the case then the problem statement probably should be changed somewhat to reinforce that this is not uniform motion.) That implies that the magnitude of the force is constant and the force vector changes as the object goes on its path. If that is the case then ##\nabla \times \textbf{F} = 0##, and the force is conservative.

If the force is not constant, but the object is still constrained to move in a circle (say with increasing angular speed), then we may not have a conservative force. We would need more information from the problem statement about the nature of the force.

Are you sure the force vector is constant? If that's the case then the object could be constrained to move along a circular path, ie. the path will not be necessarily be closed, but it will eventually settle down to an equilibrium point. In that case things will be a bit stickier and the force may, indeed, not be conservative. Again, we would have no way to tell from the problem statement. (As mentioned above, gravity would fit this requirement, and gravity is a conservative force. But a rocket pack attached to the object would almost certainly not provide a conservative force and could produce a similar motion.)

-Dan

Thank you dan for your comments
This is an improvise problem for students and F force isn't gravity or magnetic field, it's a vector force.
Question is work of F force after 1 round trip around the circle.
I know if there's a external force in the system it's conservative force but for curiosity calculate conservative force formula quest what? It's equal 0!
My mind can't handle that why?
I am mechatronic engineering not a physics.

 

[disjk47]

I think the problem is that the question is not well posed. Is it a particle moving on a circle in the constant gravitational field close to Earth (which is a mathematical pendulum if the circle is vertically oriented)? Then you have in addition reaction forces due to the constraint for the particle moving on the circle. Otherwise the question doesn't make much sense to me.

This is an improvise problem for students and F force isn't gravity or magnetic field, it's a vector force.
Question is work of F force after 1 round trip around the circle.
I know if there's a external force in the system it's conservative force but for curiosity calculate conservative force formula quest what? It's equal 0!
My mind can't handle that why?

 

[disjk47]

I think the problem is that the question is not well posed. Is it a particle moving on a circle in the constant gravitational field close to Earth (which is a mathematical pendulum if the circle is vertically oriented)? Then you have in addition reaction forces due to the constraint for the particle moving on the circle. Otherwise the question doesn't make much sense to me.

This is an improvise problem for students and F force isn't gravity or magnetic field, it's a vector force.
Question is work of F force after 1 round trip around the circle.
I know if there's a external force in the system it's conservative force but for curiosity calculate conservative force formula quest what? It's equal 0!
My mind can't handle that why?

 

[vanhees71]

A force is of course always a vector. In non-relativistic physics you usually have pair-interaction forces (closed systems) or external forces like the gravitational field for a small body around a very large one (including planets around the Sun in our solar system, where the mass of the Sun is so much heavier than any of the planets, including Jupiter).

The most simple case thus is a single particle moving in some force field, ##\vec{F}(\vec{x})##. It's conservative if there's a scalar potential,
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}).$$
So what is the problem you try to solve?

 

[kuruman]

I think you have confused yourself with your work shown in post #7.
First you assume that you have a force field in cylindrical coordinates that is constrained in the xy-plane and set ##F_z=0.## That's OK.
Then you take the curl using an incorrect expression. The correct expression is $$
\vec{\nabla}\times\vec F=
\begin{vmatrix}
\dfrac{1}{\rho}~\hat{\rho} & \hat{\phi} & \dfrac{1}{\rho}~\hat{z} \\
\dfrac{\partial}{\partial \rho} & \dfrac{\partial}{\partial \phi} & \dfrac{\partial}{\partial z} \\
F_{\rho} & \rho~F_{\phi} & 0
\end{vmatrix} $$Note the difference in the top row with what you have. However, that is not the problem because fortuitously you get the appropriate expression for the z-component of the curl for ##\rho \neq 0 ## $$\frac{\partial}{\partial \rho}(\rho~F_{\phi})-\frac{\partial F_{\rho}}{\partial \phi}=0.
$$Then you set ##F_{\rho}=0.## That's not OK. You cannot have an object follow a curved path if the radial component of the force acting on it is zero. A ball tied to the end of a string will go around in a circle as long as the string is there. If you cut the string and set the radial force equal to zero, the ball will move in a straight line.

 

[vanhees71]

For a conservative force, you must of couse have ##\vec{\nabla} \times \vec{F}=0##. That's not a problem a priori. The problem is that the problem is not clearly posed.

E.g., if you have a central force, there are always solutions for motion of a particle along a circle. All trajectories are always in plane due to angular-momentum conservation. So put the angular momentum in ##3##-direction. Then
$$\vec{x}=R \begin{pmatrix}\cos \phi \\ \sin \phi \\0 \end{pmatrix}=R \vec{e}_r,$$
where ##\phi=\phi(t)## and
$$\dot{\vec{x}}=R \dot{\phi} \begin{pmatrix}-\sin \phi \\ \cos \phi \\ 0 \end{pmatrix}=R \dot \phi \vec{e}_{\theta}.$$
Now ##\vec{e}_r^2=\vec{e}_{\phi}^2=1## and ##\vec{e}_r \cdot \vec{e}_{\theta}=0## and ##\vec{e}_r \times \vec{e}_{\theta}=\vec{e}_3##. From this you get
$$\vec{L}=m \vec{x} \times \dot{\vec{x}}=R^2 \dot{\phi} \vec{e}_3=L \vec{e}_3,$$
and ##L=\text{const}## from which ##\dot{\phi}=\omega=\text{const}## and thus
$$m \ddot{\vec{x}}=-m R \omega^2 \vec{e}_r=F(R) \vec{e}_r,$$
from which you get ##F(R)=\text{const}##. That's of course the most simple case of a uniform circular motion in a central-force field. There the question is, whether it's a stable orbit or not, which you can check with linear-response theory to small deviations from circular motion.

The central force is easily seen to be derivable from a potential,
$$\vec{F}(\vec{x})=F(r) \vec{e}_r =-V'(r) \vec{e}_r.$$
You only need ##F(r)## being integrable.