Taylor 3.5-4.2 (today) and 4.2-4.3 (Wednesday).
Consider, again, two point masses, ma = mb = m, connected by a massless rod of length d. This time they are free to move in two dimensions on a frictionless surface. At t=0 both masses are at rest when mass a is struck and given veclocity v0 perpendicular to the rod connecting the masses. Describe the motion for later times.
We can describe the motion as the sum of the motion of the center of mass and rotation about the center of mass. With no forces applied after t=0, the center of mass moves with constant velocity determined using conservation of momentum
giving v = ½v0.
Similarly, no torques are applied after t=0, so the masses rotate about the center of mass with constant angular momentum. The angular momentum about the center of mass is determined from the angular momentum at t=0
At later times the masses rotate about their center of mass with this angular momentum. Using the above prescription, we can write the angular momentum as Iω where I is the moment of inertia about the center of mass
So we have
where we see that for later times the masses rotate with constant angular velocity ω = v0/d
An uniform hoop of mass m and radius r starts from rest on a plane inclined at an angle θ with respect to the horizontal. The hoop rolls down the plane without slipping. What is the hoop's velocity and position as a function of time? Compare to a block sliding without friction.
Solve this problem in terms of motion of the CM and rotation about the CM. The CM of the hoop is its center (notice that this point is not on the hoop!), and its moment of inertia about the CM is
Draw a diagram of the hoop on the inclined plane showing gravity acting from the CM, the normal force of the plane acting on the hoop at its point of contact, and a frictional force that keeps the hoop from slipping (acts at the point of contact and points up the plane). Choose coordinates so that x points down the plane and y points perpendicular to the plane. In the y direction the normal force balances the component of weight perpendicular to the plane, Fn = mg cosθ.
In the x direction we have the other component of gravity and friction, so the equation of motion reads
Notice that the only information about friction comes from the fact that the hoop rolls without slipping. This means that we need more equations to eliminate friction from the above expression, specifically, we need an equation about the rotation. Use the rule that the time derivative of the angular momentum equals the applied torque. Since the weight and normal force both pass through the CM, the only force that produces a torque is friction. Additionally, we can make use of the fact that L=Iω
One revolution of the hoop changes φ by 2π and moves the CM by 2πr, therefore
Therefore we can replace ω in the second equation, and use that to eliminate f from the first to get
The velocity (of the CM) as a function of time is
and the position (of the CM) as a function of time is
where I've used vx=0 and x=0 at t=0.
Compare this to the case of a block of the same mass as the hoop, sliding down a frictionless plane. The velocity and position of the block are given by
The block slides twice as fast as the hoop rolls! Neither object loses energy due to friction, so after falling the same height, the block has twice the speed of the hoop! We'll explore more about this in the next chapter.
Let's summarize. The motion of the center of mass is governed by the external forces acting on the system. Internal forces contribute to internal motion, but don't affect the motion of the CM. The time rate of change of the total angular momentum of a system equals the net external torque. The total angular momentum is composed of two components, the angular momentum due to rotations about the center of mass, and the angular momentum of the orbit of the center of mass.
The overall idea when analyzing the motion of a system is to take a top down approach. Start with the motion of the CM of the system. Then look for rotations about the center of mass (a collective motion). Finally, consider the individual motions of particles relative to the CM.
Energy is another very useful constant, but even more abstract than momentum. Today we take conservation of energy as a fundamental precept, but it wasn't formulated until about 1880 by Helmholtz. To utilize conservation of energy, we must take many forms of energy into account: kinetic energy, potential energy, and thermal energy are of particular importance in mechanics. You will run into many other types in your study of physics: electromagnetic energy, chemical energy, nuclear energy, and still others.