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disjk47
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TL;DR Summary: The rigid object move around the circle with constant force how it possible the force
How do you determine if a force is conservative? What needs to happen?disjk47 said: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?disjk47 said:The rigid object move around the circle with constant force how it possible the result of the forces are conservative!?
topsquark said:How do you determine if a force is conservative? What needs to happen?
-Dan
The external and constant force enter the system any time it should be the non conservative forceharuspex said:Why do you think it could not be?
At the first time it's on (1) point in bottom. The F force enter the mass and it's go around the circle.BvU said: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 ?
##\ ##
Constant force as a vector or constant magnitude of force?disjk47 said:The external and constant force enter the system any time it should be the non conservative force
Vectorharuspex said:Constant force as a vector or constant magnitude of force?
Then like local gravity, which is conservative:disjk47 said:Vector
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.disjk47 said:
Thank you dan for your commentstopsquark said: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
This is an improvise problem for students and F force isn't gravity or magnetic field, it's a vector force.vanhees71 said: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.vanhees71 said: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.
Conservative forces in circular motion are forces that do not dissipate energy and can be described by a potential energy function. These forces are always perpendicular to the direction of motion and do not change the total mechanical energy of the system.
Some examples of conservative forces in circular motion include gravity, magnetic forces, and electric forces. These forces follow the law of conservation of energy and do not cause any energy loss in a system.
Conservative forces affect circular motion by changing the direction of the motion without changing the total mechanical energy of the system. This means that the speed of the object may change, but the total energy remains constant.
Conservative forces and potential energy are closely related. The potential energy of a system is defined by the work done by conservative forces. As the object moves in a circular path, the potential energy is converted into kinetic energy and vice versa.
Non-conservative forces, such as friction or air resistance, do not follow the law of conservation of energy and can dissipate energy in a system. These forces can change the total mechanical energy of the system and are not described by a potential energy function.