# Acceleration without force in rotational motion?

This has really been bugging me. I hope someone can point out the flaw in my logic.

1. Force is required to change velocity.
2. A revolving object in space is continually changing its velocity by virtue of this revolution.
3. Therefore this revolving object is forever experiencing a force.
4. A force requires the expenditure of some energy.
5. Therefore a revolving object requires a constant input of energy to keep rotating.

Point 5 is obviously wrong by experience, but why is it wrong?

• Your item 4 is wrong.
– hft
Feb 28 at 21:04
• A rotating object changes its angular velocity due to a torque.
– J.G.
Feb 28 at 22:02
• I am confused by the votes to close. This isn't a personal theory. He is asking what is wrong with his thinking because he isn't getting mainstream results. Mar 1 at 3:23
• @mmesser314 I voted to close this question because it includes a false premise as part of its setup. How is anyone supposed to answer the question other than to point out that OP's premise is wrong?... which I already did in a comment. To me this is a comment, not an answer. This is just a bad/confused question.
– hft
Mar 1 at 23:58
• Mar 2 at 0:28

Point 4 sounds perfectly reasonable, but it turns out to be wrong upon closer examination!

Force does not require an expenditure of energy. Only force directed along the path of a moving object requires expenditure of energy.

To phrase that more mathematically:

Energy Expenditure = $$\int \vec{F} \cdot d\vec{x}$$, where $$\vec{x}$$ is the coordinate along the path of motion.

For a rotating or circularly revolving object, force and motion are perpendicular and therefore energy expenditure is 0.

• Another intuitive way to see why point 4 is flawed might be the coffee cup on my desk. The cup is most certainly applying a force to said desk, and has been doing so for quite a while, but it has not, to my knowledge, tapped into any energy source in order to do so. Mar 1 at 9:26
• Your last sentence assumes the object has no angular acceleration. For a rotating object that has angular acceleration the rotational kinetic energy increases. For rotation of a rigid body in a plane, the force/distance relationship is expressed as torque/angular displacement. The textbook Physics by Halliday and Resnick shows how force/distance relationship can be expressed as torque/angular displacement. Mar 1 at 16:24
• @JohnDarby it's pretty clear from context what I'm referring to, your comment just feels like a classic physics stack exchange "well actually..." Mar 1 at 21:57
• @oerkelens - When I was in high school, I was confused by this point. If I hold my arms up, they get tired. They are tapping into an energy source. On the other hand, a coffee cup is not. It can take a while to see the difference. E.G. Suppose I support a coffee cup with a helicopter... Mar 2 at 0:41
• Er wait a sec. I thought motion was relative. Isn't there always an inertial reference frame where dx is 0? And conversely, many where it's not? Mar 2 at 17:55

The work done by a force is the force times the distance along which the force acts. This means that it is the force times the distance traveled times the cosine of the angle between them. In constant rotation, the force needed is centripetal force, which is radially inward toward the axis of rotation. Also, the motion itself is tangential to the axis of motion, since the object remains at the same distance from the axis. Forces providing centripetal force thus do no work and rotation requires no energy input.

One of the nifty things about physics is that we can predict a lot of things about an isolated object in space, without any information about the detailed structure of that object.

In particular, astronomers can and have calculate highly accurately the path of Mercury and Venus (as a whole) long before knowing their rate of rotation.

1. Force is required to change velocity

Yes!

1. A rotating object in space is continually changing its velocity by virtue of this rotation

Any particular piece of this rotating object is continually changing its velocity.

However, the object as a whole is not changing its velocity by virtue of this rotation. We can calculate a special point -- the center of mass -- that "represents" the entire object, and moves "the same" whether the object is rotating or not.

1. Therefore this rotating object is forever experiencing a force

An object isolated in space is not experiencing a force. Nothing is "touching" it.

However, any particular piece of this rotating object is experiencing a force. Other pieces of that object that touch that piece generally push or pull on that piece. With a and a bit of math, you can show that the total mechanical energy going into any particular piece of this isolated rotating object at any particular short period of time is exactly balanced by mechanical energy coming from some other touching pieces, so the net energy to or from the outside world is always zero.

1. A force requires the expenditure of some energy

As Señor O pointed out previously, "4 sounds perfectly reasonable, but it turns out to be wrong upon closer examination!"

There's at least 3 common situations off the top of my head where there is definitely a mechanical force for long periods of time that require no external energy source during that time:

• Zero motion: The spring inside my clicky pen is compressed, pushing out against other parts of the pin, and it can continue to push with that force while sitting on my desk (or floating in space) indefinitely.
• motion at right angles to the force: a hockey puck pushing down on the ice while coasting across an ice rink, the pull of the string on each weight of a bola, the pull of a rotating space tether on the end masses, etc.
• Oscillation: A ringing bell, a clock spring, etc. can oscillate back and forth in isolation for a surprisingly long amount of time. There is no external energy input during that time to the whole object. However, different parts of the object push on each other and trade kinetic energy and potential energy back and forth.

The 4th point is not generally true. Let me give you an example for a field which exerts a force on the particle but doesn't do any work. Consider the magnetic Lorentz force $$\vec{F} = q(\vec{v} \times \vec{B})$$. where $$\vec{B}$$ is the external magnetic field which is present in the region. Now consider the work done by this force. $$W = \int \vec{F} \cdot \vec{dl}$$. This could be further be written as $$W = \int q(\vec{v} \times \vec{B}) \vec{dl}$$. We can also write $$\vec{dl} = \vec{v} dt$$ which after substitution would be $$W = \int q(\vec{v} \times \vec{B}) \cdot \vec{v}dt$$ We can immediately notice that the term $$(\vec{v} \times \vec{B}) \cdot (\vec{v} dt) = 0$$

The 4th point is not necessarily true. This is because $$\vec{v} \times \vec{B}$$ is always perpendicular to $$\vec{v}$$. Therefore $$W = 0$$ for magnetic fields.

Therefore, you can see that although there is force on the charged particle due to the magnetic field, it doesn't necessarily mean that the the field is doing work on the particle. It's the $$\vec{F} \cdot d\vec{l}$$ that matters when we are concerned about the work done by something.