Forces; Projectile Motion

Class Notes Physics I, 10/23/98

Forces; Projectile Motion

If we drop a single coffee filter of mass m, it quickly reaches its terminal velocity, where the air resistance it encounters is equal and opposite to its weight mg.

If we drop two such coffee filters, one inside the other, then their terminal velocity will be reached when the air resistance is equal and opposite to the total weight 2*m*g of the two filters.
Thus the air resistance on the single filter is half that on the doubled filter.

We observe that when we drop the doubled filter from twice the height of the single filter, releasing the objects at the same instant, they strike the floor at very nearly the same time.

Since the heavier filter covers double the distance in the same time, its average velocity will be double that of the first.
If the time required for the objects to reach terminal velocity is small, then these average velocities will be near the terminal velocities of the two objects.
We can therefore say that, if terminal velocity is reached quickly, the terminal velocity of the doubled filter appears to be very nearly double the terminal velocity of the single filter.

We therefore conclude that doubled weight, and therefore doubled air resistance, seems to be associated with doubled velocity.

If doubled velocity implies doubled air resistance, then it is plausible that air resistance is proportional to velocity.

We could then write fAir = k v, where fAir stands for air resistance and k is a constant proportionality.
This one observation does not prove that air resistance is proportional to velocity, even within the general range of speeds observed in the demonstration. Thus the question mark in the figure below.

http://youtu.be/tdoqbQHloH0

http://youtu.be/xFeBFG22sEE

Recall that an object resting or sliding down an inclined plane is generally subject to at least three forces:

the force of gravity,

a normal force exerted by the incline,

and a frictional force related to the normal force.

The gravitational force, also called the weight of the object, is usually broken into components parallel to and perpendicular to the incline, as indicated in the figure below.

We can mimic this situation by attaching strings supporting a known weight to force balances.

If the balances are moved into a position where the strings are perpendicular, with one string oriented so as to mimic the direction of an incline, then the forces indicated by the balances will exactly (with an experimental error) support the weight.
The string force parallel to the incline will be equal and opposite to the weight component parallel to the incline, and the string force perpendicular to the incline will be equal and opposite to the weight component perpendicular to the incline.

The figure below shows forces observed for certain angle, represented in blue.

The red vectors represent the weight components inferred from these results.

We note that the Pythagorean Theorem is very nearly consistent with these results, with `sqrt(1.5 ^ 2 + 1.2 ^ 2) = 1.9, very nearly corresponding to the actual 2 Newton force.

http://youtu.be/xMAzSp5GwVQ

We now consider a projectile problem in which the projectile has an initial velocity which is not horizontal.

Assume an initial velocity of five meters/second and angle 30 degrees above horizontal.

Assume that the projectile leaves the ramp in the figure below at a point 10 meters above a level plane.

We wish to analyze the motion of the projectile.

This means we wish to conclude everything we can about its motion.

We begin with our knowledge that the x motion and the y motion are independent, and that both are subject to uniform accelerations.

We therefore seek to find initial and final velocities, accelerations, and displacements for each direction as well as the common time interval `dt.

We easily find the initial x and y velocities, which are the x and y components of the initial velocity.

These velocities are found in the figure below to be v0x = .433 m/s and v0y = 2.5 m/s.

The horizontal motion is characterized by the fact that the velocity remains constant, implying zero horizontal acceleration ax = 0 and equal initial and final velocities vfx = v0x.

The only thing we don't know about the horizontal motion is the time interval `dt and the displacement `ds.
If we knew either of these quantities we could find the other, but we don't so we can't.

We seem to be stuck in our attempt to analyze the horizontal motion so we turn to the vertical motion.

The initial velocity is v0y = 2.5 m/s, as noted earlier. Since this positive velocity is upward, the positive direction is implicitly chosen to be up.
The vertical acceleration is then negative, equal to -9.8 m/s/s.
We also see that the vertical displacement is negative, equal to -10 m.

This last statement about the vertical displacement caused some confusion in class.

It seemed to most students that the vertical displacement should be measured from the maximum altitude, and not from the point which the object leaves the ramp.

It would in fact be possible to analyze this problem by breaking the motion into two segments, one beginning when the object leaves the ramp and ending when the object reaches its maximum altitude (where vy = 0), and the other beginning at this instant an ending when the object crashes to the ground.

This is, however, unnecessary:

The equations of motion work just find if we only know the initial velocity, acceleration and displacement.

It doesn't matter that the displacement is in the opposite direction to the initial velocity.

However, this situation does require careful interpretation, as we will see shortly.

Knowing the initial velocity, acceleration and displacement, we can determine the time interval `dt using the equation `ds = v0 `dt + .5 a `dt^2, as shown below for the y quantities.

We first note that the equation is quadratic and put it into standard form A `dt^2 + B `dt + C = 0 by subtracting the displacement from both sides.
The solutions of this equation will give us values for `dt.
Our solution will be `dt = [ -B +- `sqrt( B^2 - 4 A C)] / (2 A).

For this equation we have A = 1/2 ay, B = v0y and C = -`dsy.

http://youtu.be/PV-XD_9h6rM

Substituting our values for A, B and C into the quadratic formula we obtain the expression for `dt shown and simplified below.

We obtain `dt = .26 s +- 1.43 s.

The .26 s comes from dividing the negative of the initial velocity by the acceleration.

This corresponds to dividing the change in velocity, from the first instant until velocity reaches 0, by the acceleration.

This gives the time required for the projectile to reach its maximum altitude, were the vertical velocity is of course zero.

The 1.43 seconds is the time required for the projectile to then reached the ground. The reasons for this are somewhat more complicated, but are well worth thinking about.

The resulting values for `dt are -1.17 sec and +1.69 sec.

Since `dt is the time required to reach the ground after being projected from the ramp, the negative value is clearly not what we are after.

We therefore conclude that it takes 1.69 seconds for the projectile to reach the ground.

Though it is not shown in the figure below, it is now very easy to calculate the horizontal range.

We merely multiply the 1.69 second time interval by the constant horizontal velocity 4.33 m/s and obtain horizontal displacement `dsx.

http://youtu.be/y3nehh54K_U

The -1.17 second value of `dt is somewhat interesting in its own right.

This is the clock time at which a projectile which is 10 meters above the ground at clock time t = 0, and moving at 5 m/second at an angle 30 degrees above horizontal was at ground level, assuming that it was moving freely before t = 0.
That is, the motion of the real projectile is no different than it would have been if the platform holding the ramp had not been there at all, and the projectile had been fired from just the right point on the ground 1.17 seconds before the real projectile left the ramp.

http://youtu.be/jeciKUWzWNM