One last thing to note. We are being rather cavalier about the normal force exerted by the pulley on the string – all we can easily tell is that the total force acting up on the string must equal T1 + T2, the force that the string pulls down on the pulley with. Similarly, since the center of mass of the pulley does not move, we have something like Tp − T1 − T2 − M g = 0. In other words, there are other questions I could always ask about pictures like this one, and by now you should have a good idea how to answer them.

5.7: Rotational Work and Energy

ω

Center of Mass

v

r dm

Figure 71: A blob of mass rotates about an axis through the center of mass, with an angular velocity as shown.

We have already laid the groundwork for studying work and energy in rotating systems. Let us consider the kinetic energy of an object rotating around its center of mass as portrayed in figure 71. The center of mass is at rest in this figure, so this is a center of mass inertial coordinate system.

It is easy for us to write down the kinetic energy of the little chunk of mass dm drawn into the figure at a distance r from the axis of rotation. It is just:

To find the total, we integrate over all of the mass of the blob:

which works because ω is the same for all chunks dm in the blob and is hence a constant that can be taken out of the integral, leaving us with the integral for I.

If we combine this with the theorem proved at the end of the last chapter we at last can precisely describe the kinetic energy of a rotating baseball in rest frame of the ground:

That is, the kinetic energy in the lab is the kinetic energy of the (mass moving as if it is all at the) center of mass plus the kinetic energy in the center of mass frame, 12 Iω2. We’ll have a bit more to say about this when we prove the parallel axis theorem later.

5.7.1: Work Done on a Rigid Object

We have already done rotational work. Indeed we began with rotational work in order to obtain Newton’s Second Law for one dimensional rotations above! However, there is much to be gained by considering the total work done by an arbitrary force acting on an arbitrary extended rigid mass.

~

Consider the force F in figure 72, where I drew a regular shape (a disk) only to make it easy to see

M

~

Figure 72: A force F is applied in an arbitrary direction at an arbitrary point on an arbitrary rigid object, decomposed in a center of mass coordinate frame. A disk is portrayed only because it makes it easy to see where the center of mass is.

and draw an useful center of mass frame – it could just as easy be a force applied to the blob-shaped mass above in figure 71.

~

I have decomposed the force F into two components:

Suppose that this force acts for a short time dt, beginning (for convenience) with the mass at rest. We expect that the work done will consist of two parts:

where ds = r dθ. This is then:

We know that:

and if we integrate these independently, we get:

or the work decomposes into two parts! The work done by the component of the force through the center of mass accelerates the center of mass and changes the kinetic energy of the center of mass of the system as if it is a particle! The work done by the component of the force perpendicular to the line connecting the center of mass to the point where the force is applied to the rigid object increases the rotational kinetic energy, the kinetic energy in the center of mass frame.

Hopefully this is all making a certain amount of rather amazing, terrifying, sense to you. One reason that torque and rotational physics is so important is that we can cleanly decompose the physics of rotating rigid objects consistently, everywhere into the physics of the motion of the center of mass and rotation about the center of mass. Note well that we have also written the WKE

theorem in rotational terms, and are now justified in using all of the results of the work and energy chapter/week in (fully or partially) rotational problems!

Before we start, though, let’s think a teeny bit about the rolling constraint and work, as we will be solving many rolling problems.

5.7.2: The Rolling Constraint and Work

A car is speeding down the highway at 50 meters per second (quite fast!) being chased by the police. Its tires hum as they roll down the highway without sliding. Fast as it is going, there are four points on this car that are not moving at all relative to the ground! Where are they?

The four places where the tires are in contact with the pavement, of course. Those points aren’t sliding on the pavement, they are rolling, and “rolling” means that they are coming down at rest onto the pavement and then lifting up again as the tire rolls on.

If the car is travelling at a constant speed (and we neglect or arrange for their to be no drag/friction) we expect that the road will exert no force along the direction of motion of the car – the force exerted by static friction will be zero. Indeed, that’s why wheels were invented – an object that is rolling at constant speed on frictionless bearings requires no force to keep it going – wheels are a way to avoid kinetic/sliding friction altogether!

More reasonably, the force exerted by static friction will not be zero, though, when the car speeds up, slows down, climbs or descends a hill, goes around a banked turn, overcomes drag forces to maintain a constant speed.

What happens to the energy in all of these cases, when the only force exerted by the ground is static friction at the points where the tire touches the ground? What is the work done by the force of static friction acting on the tires?

Zero! The force of static friction does no work on the system.

If you think about this for a moment, this result is almost certain to make your head ache. On the one hand, it is obvious:

because dx = 0 in the frame of the ground – the place where the tires touch the ground does not move, so the force of static friction acts through zero distance and does no work.

Um, but if static friction does no work, how does the car speed up (you might ask)? What else

could be doing work on the car? Oooo, head-starting-to-huuuurrrrrt...

Maybe, I dunno, the motor?

In fact, the car’s engine exerts a torque on the wheels that is opposed by the pivot force at the road – the point of contact of a rolling object is a natural pivot to use in a problem, because forces exerted there, in addition to doing no work, exert no torque about that particular pivot. By fixing that pivot point, the car’s engine creates a net torque that accelerates the wheels and, since they are fixed at the pivot, propels the car forward. Note well that the actual source of energy, however, is the engine, not the ground. This is key.

In general, in work/energy problems below, we will treat the force of static friction in rolling problems (disks, wheels, tires, pulleys) where there is rolling without slipping as doing no work and hence acting like a normal force or other force of constraint – not exactly a ”conservative force” but one that we can ignore when considering the Generalized Non-Conservative Work – Mechanical Energy theorem or just the plain old WKE theorem solving problems.

Later, when we consider pivots in collisions, we will see that pivot forces often cause momentum not to be conserved – another way of saying that they can cause energy to enter or leave a system

– but that they are generally not the source of the energy. Like a skilled martial artist, they do not provide energy themselves, but they are very e ective at diverting energy from one form to another. In fact, very much like a skilled martial artist, come to think about it.

It isn’t really a metaphor...

Example 5.7.1: Work and Energy in Atwood’s Machine

Figure 73: Atwood’s Machine, but this time with a massive pulley of mass M and radius R (and moment of inertia I = βM R2), this time solving a “standard” conservation of mechanical energy problem.

We would like to find the speed v of m1 and m2 (and the angular speed ω of mass M ) when mass m2 > m1 falls a height H, beginning from rest, when the massless unstretchable string connecting the masses rolls without slipping on the massive pulley. We could do this problem using a from the solution to the example above, finding the time t it takes to reach H, and backsubstituting to find v, but by now we know quite well that it is a lot easier to use energy conservation (since no non-conservative forces act if the string does not slip) which is already time-independent.

Figure ?? shows the geometry of the problem. Note well that mass m1 will go up a distance H at the same time m2 goes down a distance H.

Again our solution strategy is almost identical to that of the conservation of mechanical energy problems of two weeks ago. We simply evaluate the initial and final total mechanical energy including the kinetic energy of the pulley and using the rolling constraint and solve for v. We can choose the zero of potential energy for the two mass separately, and choose to start m2 a height H above its final position, and we start mass m1 at zero potential. The final potential energy of m2 will thus be zero and the final potential energy of m1 will be m1gH. Also, we will need to substitute the rolling constraint into the expression for the rotational kinetic energy of the pulley in the little patch of algebra below:

(529)

Thus: