Class Notes, Sep 10

Class Notes, Fall 2001

Week of	Aug 22	Sep 5	Oct 1	Nov 5	Phy 205 page
	Aug 27	Sep 10	Oct 8	Nov 12
		Sep 17	Oct 15	Nov 19
		Sep 24	Oct 22	Nov 26
			Oct 29		Notes for Fall, 2000

Phy 205, Week 4

10 Sep Start from the two key definitions of chapter two, those of velocity and acceleration in one dimension, and quickly repeat the derivations of the velocity and acceleration for the special case x(t) = A + Bt + Ct². The results were

v_x( t ) = B + 2 C t and a_x = 2C

Emphasize the peculiar nature of gravity, that the force on a mass is proportional to the mass itself. This implies that (getting ahead of myself a bit) if the only force on a mass is gravity then for

a_x = F_x / m,

the "m" will cancel out of the right-hand side. If gravity is all that's acting then all object will have the same acceleration. This is easy to demonstrate by dropping a couple of things and seeing what happens. The value of this proportionality factor between force and mass is called the gravitational field, and near the Earth's surface it has a value close to 9.8 m/s² = g.

If I throw something straight up, what will its motion be? How high will it go? If I make the approximation that gravity is the only significant force acting on it, then I know its acceleration to be g (down). In setting up this problem, first pick a coordinate system. Here I chose y to be measured positive upward. There are several possible choices, what you have to do is pick one. Now I know that

a_y = dv_y / dt = - g.

From the previous calculation, the one that started with x(t) = A + Bt + Ct², I know that is yields a constant value of the acceleration, 2C. Work backwards and ask what v_y(t) will have this value for a_y? Result:

v_y( t ) = - g t + D,

where this D is an arbitrary constant.

Continue and ask what y(t) will give this value of v_y(t)? The answer is

y( t ) = - g t² / 2 + D t + D'

where D' is another arbitrary constant. The next problem is to figure out these two unknown constants that have shown up. For two unknowns, you need two equations. These equations can appear from a variety of conditions, but I'll start with the simplest, where I know the position and the velocity at one time.

Assume that the position at time t=0 is at the origin of the y-coordinate. That is, y(0)=0. Also suppose that I've thrown something upward at this time with an initial velocity of v_o, which may be something like 5m/s.

y( 0 ) = - g 0² / 2 + D 0 + D' = 0 and v_y( 0 ) = - g 0 + D = v_o.

This provides D and D' and in summary the behavior of the motion is provided by the equations

y( t ) = - g t² / 2 + v_o t. and v_y( t ) = - g t + v_o.

Next I looked at what this said for the special case to find how high the thrown object will rise.

v_y( t ) = 0 implies t = v_o / g and then y( v_o/g ) =v_o² / 2 g.

Finally, analyze this result and plug in some numbers for the special case.

12 Sep I recapitulated the work starting from the two basic equations,

v_y = dy / dt and a_y = dv_y / dt,

applying it to the same example, y(t)=A+Bt+Ct². you get the results,

v_y = dy / dt = B + 2 C t and a_y = dv_y / dt = 2 C.

Apply this to a question: Just how fast is 60 miles per hour? For this rather vague question I'll ask If you push a car off the edge of a tall building, how tall does it have to be so that the car will hit the ground at 60 mph?

Remember, there are only two equations in chapter 2, and I wrote them just above: the definitions of velocity and acceleration (in one dimension).

(1) Draw a sketch. It can be crude; mine certainly was.

(2) Pick a coordinate system to describe the process. The one that I was instructed by the class to use had y measured positive up from ground level.

(3) The acceleration is known, a_y = - g. [This is basically where the physics ends. The rest is mathematics done to solve the problem.]

(4) Here the acceleration is a constant, so compare it to the case just worked out, the one that started with a quadratic x(t)=A+Bt+Ct². It led to a constant acceleration, 2C. Work backward and ask what velocity will give that and it's B+2Ct, so in the present case, where the acceleration is - g, then that is 2C. I conclude that the velocity is

v_y = - g t + B,

where B is some unknown constant.

Repeat this to get the position function of time. As before, the position is a quadratic,

y( t ) = - g t² / 2 + B t + A.

This process of working backwards, from the derivative to the function is called taking the "anti-derivative." It's just a process of recognizing familiar combinations.

(5) Use some more of the given information to determine the unknown constants, A and B. The problem now is one of translating colloquial English into colloquial mathematics. What other information is known? You push the car off the roof. You may as well take that to be time t=0, then the fact that the car isn't thrown down, just dropped, means

v_y( 0 ) = 0, but v_y( 0 ) = - g 0 + B = B,

so B = 0.

The initial position of the car is on the roof. Where is that? Unknown, so give it a name, call it h. This translates to

y( 0 ) = h, but y( 0 ) = - g 0² / 2 + B 0 + A = A,

so A = h.

(6) Put it all together:

y( t ) = - g t² / 2 + h and v_y( t ) = - g t.

All the motion of the car until it hits the ground is contained in these two equations, so if there is any information to be determined, this is where to look.

(7) It hits the ground at some time (unknown), so give that a name. I'll call it t_f. "It hits the ground" is an English sentence that has to be translated into mathematics. It say that at time t_f the position of the car and the position of the ground are the same. This isn't colloquial English, but it is closer to colloquial mathematics:

y( t_f ) = 0.

The other information concerns the velocity. When it hits the ground it is supposed to be moving at 60 mph; call this speed v_f. This translates to

v_y( t_f ) = - v_f.

These two equations are then

y( t_f ) = - g t_f² / 2 + h = 0 and v_y( t_f ) = - g t_f = - v_f.

These are two equations in the two unknowns h and t_f. Eliminate the latter and solve for the former:

t_f = v_f / g, so h = g t_f² / 2 = g ( v_f/g )² / 2 = v_f² / 2 g.

(8) At the end you can plug in the numbers. For the particular case that v_f = 60 mph, you have to convert some units. The result is about 121 feet. A 10 or 12 story building.

Another example is from one of the old exams that I passed out. The drag racer starts from rest and ends a speed v_f after a distance L. What is the acceleration?

This problem is done the same way. Choose the obvious coordinate system with x measured positive along the track and the origin at the start. The acceleration is unknown, but the basic equations are the same:

v_x = dx / dt and a_x = dv_x / dt,

where a_x is an unknown constant. As before I use that fact that I've done the derivative of a special polynomial for which the acceleration came out to be a constant, the quadratic, so I can write down the general form of the position and the velocity for this case:

v_x( t ) = a_x t + B and x( t ) = a_x t² / 2 + B t + A.

A and B are two arbitrary constants to be determined later. Now use the other information stated in the problem (or a result of my choice of coordinates). Everything starts at time t=0:

x( 0 ) = 0 and v_x( 0 ) = 0.

These two equations determine the two unknowns A and B, both zero.

Next what happens at the end of the track? I don't know the time at which this happens, so I give it a name; call it t_f.

x( t_f ) = L and v_x( t_f ) = v_f.

These are two equations in the two unknowns, v_f and t_f. Solve them.

14 Sep Do a couple of homework problems, emphasizing that several of the apparently different problems that I've done are really the same problem. They all start the same way and the manipulations are the same. All that seems to change is the data. Sometimes one quantity is given and another is to be found; sometimes the reverse.

The difference between chapters two and three is in the use of vectors. The world is three dimensional, so I can't simply use the single coordinate "x." It is replaced by the coordinate vector r(t), the displacement vector from some chosen origin. You can describe objects in two or three dimensions using coordinate systems, but this is a more unified notation and it makes some of the manipulations easier.

In one dimension the definition of velocity was v_x=dx/dt. Now it becomes the vector equation.

v = dr / dt.

The definition of dx/dt is as a specified limiting process, and the definition here is the same

d ->
r

dt
=
lim
Δ t -> 0

->
r

( t + Δ t ) - ->
r

( t )
Δ t

[I'm switching back and forth between the arrow notation and the boldface notation for vectors because I can type the boldface version easily but the other has to be processed.]

I examined the geometric interpretation of this velocity, drawing a lot of pictures. The key result was that the velocity vector is tangent to the path representing the motion of the object being described. It's just about what you would expect any reasonable definition of velocity to look like.

The acceleration is defined similarly. It is the time-derivative of the velocity vector,

a = dv / dt.

Again, I looked at the geometric interpretation and the result is not this time so intuitive. A car turning a corner to the right has an acceleration vector pointing more or less right.

Back Next