Taylor 3.5-4.1 (today) and 4.2-4.3 (Monday).
Relations for linear and angular momentum from chapter 3.
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
Round objects (hoops, disks, wheels, cylinders, and spheres) can roll and/or slide across a surface. Pure sliding means that the object doesn't rotate at all. Pure rolling means that the point of contact between the round object and the surface doesn't slide -- close to the surface, that point moves perpendicularly down to the surface then perpendicularly up from the surface, the horizontal motion vanishes as it touches the surface. It is also possible to have a combination of sliding and rolling.
When an object slides across a surface it experiences sliding friction characterized as the product of the normal force and the coefficient of sliding friction f = μN, and directed opposite the sliding.
When an object doesn't slide, it can still experience a frictional force, one we characterize as static friction. Static friction is different from sliding friction in that the size of the force can vary, and since there is no motion, the direction must be determined from the dynamics of the situation, that is, since it isn't sliding, the net force parallel to the surface must be zero. For example, the frictional force is zero for an object sitting on a horizontal surface with no external, horizontal force (the sum of the horizontal forces is zero). If this object now experiences a small horizontal force, small enough that it doesn't slide, then the frictional force must be equal in magnitude and opposite in direction to the applied force. If the horizontal force is increased, eventually the object will begin to slide. The force of static friction has a maximum value, and when the applied force exceeds this maximum value. In analogy with sliding friction, we write the maximum value of static friction as f ≤ μN, where the coefficient μ for static friction can differ from that for sliding friction. Usually the coefficient of static friction is a little larger than the coefficient of sliding friction.
The fact that static friction can take on a range of values means that it is one more unknown in many problems. But we also gain an equation, the one relating the change in angular velocity to the external torque. More complex problems can be considered where a transition is made from rolling with sliding (slipping) to pure rolling. Let's look at an example to get some feeling for this.
A 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
or, after rearranging
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 strategy for analyzing the motion of a system is to take a top down approach. Start with the motion of the center of mass 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 center of mass.