Class Notes for Fall, 2000

Week of Aug 23 Sep 6 Oct 2 Nov 6 Phy 205 page
Aug 28 Sep 11 Oct 9 Nov 13
Sep 18 Oct 16 Nov 20 Notes for Fall, 1999
Sep 25 Oct 23 Nov 27
Oct 30

Phy 205, Week 5

18 Sep Review for the exam. Do some more of the homework and related problems.

20 Sep Go over the definitions of velocity and acceleration as they are in three dimensions.

v = dr / dt      and      a = dv / dt.

Better:      v( t ) = dr( t ) / dt      and      a( t ) = dv( t ) / dt.

These derivatives are defined as a limit in just the same way as other derivatives are. df/dx is the limit as Dx approaches zero of

[ f( x+Dx ) - f( x ) ] / Dx

Similarly the definitions of dx/dt in terms of x( t + Dt ) and x( t ).

The only change here is that r(t) is a vector, the displacement vector of an object from some fixed origin.

Draw a lot of pictures to emphasize what this looks like. Also write it out in terms of the basis (unit) vectors, i and j. The equation for the displacement from the origin is

r( t ) = x( t )i + y( t )j

Differentiate this to get

v( t ) = dr( t ) / dt = i dx( t ) / dt + j dy( t ) / dt = i vx( t ) + j vy( t ).

A similar equations obtains for the acceleration:

a( t ) = dv( t ) / dt = i dvx( t ) / dt + j dvy( t ) / dt = i ax( t ) + j ay( t ).

Start to work out the special case in which the acceleration is known to be simple: gravity. If the only force acting is gravity, and air resistance can be neglected, then the acceleration of a dropped or thrown object near the Earth's surface is the vector g = 9.8m/s2(down).

22 Sep Do a couple of problems involving trajectories. The only force on the object thrown is assumed to come from gravity: Fgrav = mg. The acceleration that an object experiences comes from Newton's equation

a = Ftotal / m

In this case, Ftotal / m = mg / m = g.

From the definition of acceleration, this is dv( t ) / dt = g. For the example where I throw something horizontally, set up the usual coordinate system and basis vectors. I then broke this vector equation into two component equations:

dvx( t ) / dt = 0,     and     dvy( t ) / dt = - g.

Use the usual method of taking anti-derivatives to get information about the velocity and position, with the accompanying arbitrary constants. When I took the origin to be the initial position of the ball, and then I threw the ball horizontally, this led to some initial equations

x( 0 ) = 0,     y( 0 ) = 0,     vx( 0 ) = v0,     vy( 0 ) = 0.

These are enough to determine the four unknown constants that come from the preceeding anti-derivatives. The next step is to find when and where it hits the floor. To do that I need the equation for the floor. With the coordinate system chosen, the equation for the floor is y = -h. From this I can get when it lands, and then where.

In this case the results predict that the time to drop is independent of the horizontal velocity. I compared this to an experiment. (It worked.)

Repeat the calculation, but with different initial conditions, so that the ball I throw has both initial components of velocity. After getting the results parametrized by time, I eliminated this variable to obtain a single equation between x and y. It describes a parabola, and I was able to compare this prediction to what happens when I compare the trajectory to the graph of a parabola. It's hard to get the initial conditions right, but it more-or-less agreed.

Back Next