# Car moving at a constant velocity and many other questions on KE

1. Sep 3, 2006

### leright

car moving at a constant velocity....and many other questions on KE

This is NOT hw...just a semantics inquiry.

Say you have a car moving at a constant velocity. Is the engine doing work on the object since it is exerting a force on the car in the direction of the motion? I am thinking not, since the frictional force balances the force the engine is exerting. I can calculate the work the engine does, which is the force the engine exerts in the direction of the motion over a distance (integral of the force dotted with a differential length, which becomes FD since force is constant) and I get a positive number. Now, if I calculate the work the friction force does, which is negative (the friction force is in the opposite direction of the length, so the dot product is negative but the magnitude is the same since the car is moving at a constant velocity and the forces balance) and the same magnitude as the work the motor does. This results in a net work of zero. This is consistent with the work-KE theorem. No work done means no change in kinetic energy, and there is no change in kinetic energy of the car as it is moving at constant speed. A lecturer (AT UNIVERSITY EVEN) explained work by pushing a piece of chalk on the table work is done, and since the force is constant, the work is simple the pushing force times the distance. But really, the only work done was the KE. Therefore, the only time work was done was the tiny moment of time (and tiny distance) when the pushing force was greater than the static friction force...but then when the friction became larger (kinetic friction) the pushing force was balanced and then no work is done, and there is no change in KE.

Now, this brings me to my next question. I know KE is relative. Now say I have a car moving at a constant velocity and I am standing still. Then I move to a constant velocity relative to the car. The KE of the car is going to change (say decrease, since I move in the same direction as the car) and my kinetic energy is going to change. The kinetic energy of the car decreases and the kinetic energy of me also increases. However, energy needs to be put into me that is equal to my kinetic energy, right? So my kinetic energy balances the energy put into me, so energy is conserved. But, the kinetic energy of the car DECREASES. There is no longer a balance in energy. Kinetic energy is not conserved, it seems! Can someone explain this? What am I not accounting for?

Now, can someone explain rotational kinetic energy? Total KE is the translational KE + rotational KE, right? Well, people seem to say that, for instance, centripetal forces cannot do work on objects because they do not change the kinetic energy (and the force is perpendicular to the movement, so, by definition of work, no work is done), but now, it seems they CAN do work on objects, because they change the rotational kinetic energy, and total KE is the rotational KE + translational KE, and if we have a change in KE we do work. So, do centripetal forces do work or do they not do work?

I suppose that circular motion occurs because a centripetal force, in one frame, causes a displacement in the direction of the centripetal force and work is done, which causes a change in rotational KE. Now, In the next frame, that same centripetal force (not translated to the next frame) is perpendicular to the motion. But the centripetal force is in fact in the same direction as the motion, so it DOES do work.

Also, people say that B-fields don't do work. They do do work even though they cause a centripetal force as they change rotational KE of an object, and the object gains a component of displacement in the direction of the centripetal force. However, is the centripetal force causing this displacement component in the same frame, or in the following frame. I suppose if it is the following frame then the centripetal force does no work. HOWEVER, THE CENTRIPETAL FORCE HAS TO DO SOME INITIAL WORK TO PROVIDE AN ANGULAR VELOCITY.

:yuck:

2. Sep 3, 2006

### Galileo

Okay, one thing at a time...

The solution is simple: Don't switch inertial systems! Ofcourse the kinetic energy of the car is 1/2mv^2, where v is measured wrt to inertial frame. As seen from the car itself for example, the kinetic energy is zero. Newton's laws (and conservation principles) aren't meant to hold when you switch systems.

Rotational kinetic energy is also simply kinetic energy. When you have a rigid body (extended, not like a point particle) rotating you can break up the kinetic energy of the object (which is just the sum of all the 1/2mv^2 of the particles that make up the body) into two parts. One is attributable to translation, the other due to rotation. It's a very convenient way to keep track of things, but not a fundamental new kind of energy or something.
A centripetal force does no work, it simply maintains circular motion of the particle (changes direction, not the speed)

Magnetic forces never do work. Like a centripetal force, a magnetic force is always at right angles to the velocity of the particle. The particle may get a small component in the direction of the centripetal force, but the centripetal force continuously changes direction as to stay perpendicular to the velocity, which is exactly how circular motion is brought about.

Last edited: Sep 3, 2006
3. Sep 3, 2006

### leright

let's say we have a point mass moving at a constant velocity. Then, we have a centripetal force act on the particle which causes it to move on a circular path of radius r, say. All of the translational energy is converted into rotational kinetic energy, but the total kinetic energy is the same. Therefore, the centripetal force does no work.

ok, got it. Thanks.

And also, magnetic forces can do work. what about a mass being lifted by an electromagnet, or two wires carrying current in the same direction?

4. Sep 3, 2006

### Galileo

Whether you consider a point mass moving in a circle as rotational kinetic energy or not is really just how you look at it. Kinetic energy is 1/2mv^2. If you consider it rotational kinetic energy it's 1/2Iw^2, where I is the moment of inertia and w the angular velocity. In the case of a single point moving in a circle of radius R about the origin: I=mR^2 and w=v/R, so the rotational kinetic energy is 1/2(mR^2)(v/R)^2=1/2mv^2, the same ofcourse.

The magnetic force on a charged particle in magnetic field is $q(\vec v \times \vec B)$ and is thus at right angles with the velocity. Therefore magnetic forces can never do work.

In the case of an electromagnet lifting a piece of metal (like the big ones they have in a car scrap junkyard) the work is really done by the agancy supplying the electromagnetic with power. Someone has to keep the current through the electromagnet going, so an electric potential has to be maintained. It's sometimes hard to figure out what agency is doing the work, but it will never be just the magnetic force.

5. Sep 3, 2006

### leright

So, what about a ferromagnetic material lifting some metal? As the magnet lifts the metal, you would think the emf induced in the iron by faraday's law would diminish the eddy currents in the magnet, and over time the magnet would lose its magnetic dipole moment....however, this is not the case (or is it?)

If the magnetic dipole moment doesn't diminish, then where does the energy come from then?

6. Sep 3, 2006

### leright

so why do people call things magnetic energy, and why is there a magnetic energy density if magnetic fields can never do work? Why am I learning about magnetic work in my thermal physics class?

So, really, magnetic fields don't have energy, because they cannot do work, but can only take other forms of energy and convert then into energies that can do work?