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 5

17 Sep Definitions of velocity and acceleration in three dimensions. Write them in terms of components, where I pick an x-y coordinate system and the corresponding basis vectors i and j. Then the displacement vector from the origin is

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

When you differentiate these, the unit vectors are constants and it is the sum of two terms, so the velocity is just

v = dr / dt = i dx / dt + j dy / dt = vx i + vy j

Similarly for the acceleration.

Apply these equations to the particular case of gravity. Pick a coordinate system and corresponding basis with x horizontal and y vertical as usual and the i and j in accordance. For gravity the force is proportional to the mass, so if gravity is the only force acting then all objects would have the same acceleration. Take this case.

a = g,     or in this basis,     i dvx/dt + j dvy/dt = - g j

For two vectors to be equal, their respective components must agree. This provides two scalar equations from the one vector equation,

dvx/dt = 0     and     dvy/dt - g.

You can now handle these equations separately. Take anti-derivatives of each, just as with the one-dimensional problems of the preceeding chapter. This gives

vx(t) = C,     x(t) = C t + D     and
vy(t) = - g t + C',     y(t) = - g t2/2 + C' t + D'

For the demonstration apparatus that I used, I threw a ball horizontally. Pick the coordinate system with the origin where the ball starts and call the initial speed v0. The initial conditions translate into the equations

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

These determine C, D, C', and D'. The resulting equations that describe the motion are then

x( t ) = vy(0) t,     y( t ) = - g t2/2.

When does it hit the floor? When the coordinates of the ball match the coordinates of the floor? The floor is level, so its equation is y = -h. Here h is about a meter and a half. The equation for the time of impact is then

- g t2/2 = -h,     with solution     t = ( 2 h / g )1/ 2.

Notice that this doesn't depend on v0. Check this experimentally with the apparatus that let me simultaneously drop a ball and project one horizontally. To the ear at least, they hit the floor at the same time.

19 Sep Another example: Throw a tennis ball at a given initial speed and angle, and find out how it will move. The basic equations are the same, only the other data is different.

v = dr / dt,     and     a = dv / dt

When I break them into components, I get a equations for the time-derivatives of vx and vy and x and y. As usual, find anti-derivatives with all their attendant constants; then use information given in the problem to evaluate the constants. In this case, I found the values to be

x( 0 ) = 0       y( 0 ) = 0       vx( 0 ) = v0 cos θ       vy( 0 ) = v0 sin θ .

This gives the equations for the position in terms of the parameter t:

x = ( v0 cos θ ) t       y = - g t2/ 2 + ( v0 sin θ ) t

One way to analyze these equations is to see what its shape is. You can do this by eliminating the variable "t" between the two equations. The result is

y = - ( g/2 ) ( x/ v0 cos θ )2 + x tan θ.

As a check, what is this if you turn off gravity?

Next observe that this is the equation of a parabola and I tested this by comparing the motion of a thrown tennis ball to the picture of a parabola.

21 Sep Go over a key homework problem, the derivation of acceleration of an object in circular motion. The component method. Then show the relationship of this way to look at it to the more geometric method. The latter involves less algebra, but is much trickier to see.

Newton's laws of motion. I'm starting with the second law, and will come back to the other two very soon. A simple example of an object sliding down an incline will be the first thing to examine.

The only things acting on the cart are (1) the track, (2) the air, and (3) the gravitational pull by the Earth. The basic equation (Newton's 2nd law) is that after you've added all the forces together to get the total force, then

a = Ftotal / m

I'll neglect the force by the air, as it's pretty small. The bearings on the cart are very good, so friction is small enough to neglect also. I have to write three vectors out, two forces and an acceleration, and I'll use components to do it. Use the same basis that I used on the exam, horizontal and vertical. Then

a = i a cos θ - j a sin θ,       Fgrav = - m g j,       Ftrack = i Ftr sin θ + j Ftr cos θ

Put these into Newton's equation and you have a relationship between the applied force and the acceleration of the cart. To manipulate it, equate the like components:

a cos θ = Ftr sin θ / m

- a sin θ = - g + Ftr cos θ / m

These are two equations for the two unknowns, a and Ftr.

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