Circular Motion - Work: Is Work Done?

[geno921]

Hi, I remember reading that no work is done on an object in a circular orbit experiencing a centripetal force. However, what if that object experiences an external centripetal force greater than that which keeps its orbit in a radius r. That would cause the object to orbit with a smaller radius. Is the work done by that force still 0? Or would it simply be the change in the total kinetic energy of the system (by the work-energy theorem), or simply the constant force times the radius it traveled to its smaller orbit? I don't usually post, but this question baffles me.

 

[rcgldr]

Work only occurs if the speed (magnitude of the velocity) of the object changes, regardless of the path (absent any external fields where force would be affected by position of the object).

The object can remain in a circular path and it's speed can be increased or decreased by a tangental force. The centripetal force does no work but will need to increase or decrease in order to compensate the change in the objects speed in order for the path to remain circular.

Even if the path is not circular, the work done is related to the change in speed and not related to the change in direction.

 

[ashishsinghal]

Is the work done by that force still 0? Or would it simply be the change in the total kinetic energy of the system (by the work-energy theorem), or simply the constant force times the radius it traveled to its smaller orbit?

I hope you know rotational kinetics cause I am explaining it using it. (If you don't know then you can jump to the conclusion).

you are talking about a force acting radially which is more than centripetal force.right?
since it is acting radially, the torque about center of circle will be zero, hence angular momentum (mvr) remains conserved.

Since the force is acting towards the center the radius will decrease and hence velocity will increase. Due to this kinetic energy will increase.

 

[ashishsinghal]

It would not be simply the constant force times the radius it traveled to its smaller orbit because :
as you know mvr is constant. so let vr = c (m is constant)
v=c/r

force = mv^2/r
=mc^2/r^3
hence as r decreases force increases and excess force will decrease. The force you are talking about is not constant

 

[sandy_patch]

Work done is always: F.v (dot product of the vectors). If the object continues to travel in a circular path, the force will always be along the radius and velocity will always be tangential. Thus the two vectors will be mutually perpendicular to each other and their product, 0. Now if the radial force (centripetal) is increased, then the object will have a radial component in its velocity as well, in which case the dot product will be non-zero. However, once the radial forces have equilibrated, the object will begin to move in a circular path (of a different radius than original) and the work done, once again, will be 0.

 

[rcgldr]

force acting radially which is more than centripetal force.

If the force is always perpendicular to the object paths, then no work is done, direction changes but not speed. The easiest way to accomplish this is have the object attached to a string that wraps or unwraps around a pole, the path will be an involute of circle. If instead the string is pullled or released from a fixed hole, then work is done and a variety of paths are possible. See post #3 (hole case) and post #4 (pole case) in this thread:

https://www.physicsforums.com/showthread.php?t=328121

The math for the pole (involute of circle) case is shown in post #32 of this thread (direct link):
https://www.physicsforums.com/showpost.php?p=1435942&postcount=32

 

[geno921]

Sandy_patch, you hit on my idea. I know the work done is 0 in the initial case and the final case taken separately. However, does this imply that the work done in the event of increasing the (centripetal) force, which decreases the radius of orbit, is 0, or is it that since the kinetic energy increases, one can apply the work-energy theorem (I am skeptical because it stipulates rigid bodies). Thank you all for the responses thus far, and for the links rcgldr.

 

[rcgldr]

However, does this imply that the work done in the event of increasing the (centripetal) force, which decreases the radius of orbit.

Again it depends on the direction of the force relative to the path of the object. If the force is kept perpendicular to the direction of the object at all times, such as the involute of circle path, then F and v remain perpendicular and the work done is zero. I'll include the image here of that path, note the string is always perpendicular to the path of the object (the center is a pole that the string winds aorund or unwinds from). The other issue here is what is really "radial" when the path is not a circle, such as a spiral, ellipse, parabola, hyperbola, ... ?

 

[geno921]

But in this spiral path the velocity is not perpendicular to the force. If it were there would not be a spiral path. It is close, but it is not exactly perpendicular and the deviation depends on the magnitude of the centripetal force compared to the current radius of the "orbit?"

 

[rcgldr]

But in this spiral path the velocity is not perpendicular to the force.

It is exactly perpendicular, so the velocity changes direction, but the speed (magnitude of velocity) never changes (until impact with the pole). The equation for the normal line to the involute of circle is the same as the equation of the tangent line of the inner circle (pole). The math is shown in post #32 of this thread (direct link):

https://www.physicsforums.com/showpost.php?p=1435942&postcount=32