|
AN INTRODUCTION TO NEWTONIAN
MECHANICS by Edward Kluk Dickinson State University, Dickinson ND |
A few words about vectors
A vertical projectile motion
y(t) = vot -
(g/2)t2
and velocity
v(t) = vo - gt .
According to the last formula the motion should start upward (t
= 0) with velocity 0.45 m/s . This velocity due to
gravity acceleration should gradually decrease reaching zero at the time
tmh = vo / g = 45 s because on the
surface of this strange planet g = 0.01 m /
s2 . At this particular instant our object
will reach its maximal height ymh . We can express
this height with help of vo and
g inserting t = tmh = vo
/ g into formula for y(t) . A simple
algebra leads to the following result ymh =
vo2 / (2g) = 10.1 m . Right after the object
reaches its maximal height the velocity v(t) becomes increasingly
negative which means that it falls faster and faster down. Finally at
a particular time instant tf it will reach
its starting point with y = 0. Inserting the total time
of flight tf into our expression for
y(t) we obtain 0 =
votf -
(g/2)tf2. Solving it for
tf we obtain tf = 2vo
/ g which is twice of tmh
. This means that the flight up takes the same amount of time
as the fall down. The final velocity vf
at the instance when the object hits the ground equals to
v(tf) = - vo .
Now find "experimentally" ymh . To
make it easier set the tracer on and start the motion. Notice that initially
y = 0 for the bottom of the object. Consequently
you must find highest position of the bottom of the object and compare it
with the theoretical prediction for ymh . Try to
remember what height in the vision field was reached by the top of
the object. It will help you to measure more precisely
tmh . Difficulties with precise measurements of
tmh arise because close to its maximum height
the object moves so slowly that it is not ease to catch the turning point.
For next two runs switch the tracer off, measure tmh
and tf , and compare them with theoretical
predictions.
A slant projectile motion
Vectors are mathematical objects with two attributes,
magnitude and direction. Take a speed, it has a magnitude telling
how fast something with this speed is moving. But there is no information
about direction of this motion there. Consequently a speed is not a vector.
Talking about horizontal projectile motion we were using vectors without
naming them. As a matter of fact, instead using there terms "horizontal component
of speed" and "vertical component of speed" we should be using horizontal
and vertical components of velocity. Velocity is a vector showing both speed
(which is its magnitude) and direction of motion at a given instant. It is
intuitively clear that the velocity must be at every point along (tangential
to) the path of the motion. If you are a vector wizard it is clear for you
that any horizontal projectile motion is a vectorial superposition (or a
vectorial sum) of two motions, horizontal uniform motion and vertical free
fall. If you are not, but you still understand how horizontal projectile
motion is composed of these two motions, it is good enough.
Let us introduce the same two dimensional Cartesian frame of
reference as before, with an origin at zero height and zero displacement
which is in the lower left corner of the vision field, x
axis directed horizontally to the right and y axis
directed vertically up. Set the height of the object in the applet to zero,
its initial speed vo to 0.45 m/s
and the projection angle to +90o. If there were no gravity
acceleration the object should be moving up with the constant speed
equal to vo and its y coordinate
should change proportionally with time according to the formula
y(t) = vot . In the presence of gravity
acceleration g, but with the initial speed equal to zero and
no support the object should be falling down with y(t) = -
(g/2)t2. Thus, with the gravity acceleration
g and the initial velocity vo we may
expect to see a simple combination of these two motions with
If the initial velocity of the object creates an angle ß with
the horizontal, then beside of a vertical component the object motion also
has a horizontal component. We may try to assume that each of these
components depends only on a proper component of the initial velocity. This
means, the vertical component of the motion depends only on vertical component
of the initial velocity whereas horizontal component of the motion depends
only on horizontal component of the initial velocity. Being a vector wizard
you know what does it mean. If, however, you are not familiar with vectors
you have a problem with understanding what exactly are these vertical and
horizontal components of velocity. You certainly understand that they are
directed respectively vertically and horizontally, but what about their
magnitudes? In times of Galileo and Newton people who investigated this kind
of motion had the same problem. A formal theory of vectors was created later.
A possible solution of our problem is shown in the first figure. An initial
velocity vin which creates an
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angle ß with the horizontal direction is resolved there onto two components vinx and viny with help of perpendicular projections on x and y axes. If such model of resolution will work only experiments can answer.
Expanding our theory we may assume that the vertical component of a slant projectile motion is governed only by viny . Consequently the vertical coordinate of the object
y(t) = vinyt - (g/2)t2
and its vertical component of velocity
vy(t) = viny - gt .
Similarly for horizontal coordinate of the object
x(t) = vinxt
and its horizontal component of velocity
vx(t) = vinx .
The last two expressions differ from the former two by lack of the gravity acceleration factor because this acceleration occurs only in the vertical direction. As a matter of fact the vertical direction is defined by the direction of gravity acceleration.
Since we have assumed the vertical and horizontal components of slant projectile motion independent, then all results obtained for vertical projectile motion are valid for the vertical component of slant projectile motion. Only thing we have to change there is to replace vo by viny . Thus tmh = viny / g , ymh = viny2 / (2g) and tf = 2viny / g . Another interesting magnitude related to the horizontal component of the motion is the range of the projection xr = x(tf) = 2vinxviny /g .
Making our "experiments" with projectile motion we control magnitude of initial velocity (or speed) and angle of projection. In the theoretical formulae which were derived above we operate with vinx and viny . If you know basic trigonometry and look at the first figure above the following relations become obvious
vinx = vin cos ß viny = vin sin ß .
If you do not, just use a calculator with cos and sin functions to find both components of initial velocity. It is also possible to solve this problem graphically. Draw a large scale graph like in the first figure above with a required angle ß assigning to a magnitude of required initial velocity a length of 10 cm. Find how many m/s represents 1 cm on your drawing. Measure components of velocity in cm from the drawing and recalculate them in m/s .
Preparing an "experiment" with projection angles 30o, 45o and 60o and a magnitude of initial velocity of 0.30 m/s you should obtain the following values for velocity components and theoretical predictions for tf , xr and ymh
# | angle | vinx [m/s] | viny [m/s] | tf [s] | xr [m] | ymh [m] |
1 | 30o | 0.260 | 0.150 | 30.0 | 7.8 | 1.125 |
2 | 45o | 0.212 | 0.212 | 42.4 | 9.0 | 2.247 |
3 | 60o | 0.150 | 0.260 | 52.0 | 7.8 | 3.380 |
Now make this "experiment" setting the tracer on and compare the theoretical predictions with your "experimental" results. If they are reasonably close (up to 5% of difference) then the theory constructed above, including the method of resolving velocity onto the vertical and horizontal components, works.
Analysing both, theoretical and experimental data for xr you probably already have noticed that xr for 30o and 60o is the same, and for 45o xr is the largest. Is it coincidence or rule? Look at the second figure above and notice the following relations: if OB = OD then BF = DE, angle FOB = angle EOD and area OBFC = area OAED. If we assume that OF represents an initial velocity for projectile starting at the angle FOB from x axis, then OE must represent an initial velocity for another projectile starting with the same speed and at the same angle FOB from y axis. Moreover, the area OBFC for the first projectile can be expressed as vinxviny for this projectile. Similarly, the area OAED for the second projectile can be expressed as vinxviny for this projectile. But these areas are equal. Therefore the products vinxviny for both projectiles must be also equal in spite of fact that (with exception of angle FOB = angle EOD = 45o ) these projectiles have different x components of their initial velocities and different y components of their initial velocities. Consequently xr for both projectiles is the same because xr = 2vinxviny /g . Therefore if two projectiles start with the same speed, the first at angle ß with respect to a horizontal axis and the second at the angle ß with respect to vertical axis, their ranges are the same. Now it should be intuitively clear why for a fixed speed the projection angle of 45o gives the longest range. A formal proof of it is rather simple. From Pythagorean theorem we obtain
vin2 = vinx2 + viny2.
It is also ease to verify the following identity
2vinxviny = vin2 - (vinx - viny)2.
For a fixed speed of a projectile vin2 does not depend on the projectile angle. Therefore the left hand side of the identity reaches maximal value if vinx = viny which occurs if the projection angle is equal to 45o.
Evaluation
If at this point you can solve the following
problems:
the objectives of this lesson are fully achieved. If you have doubts try to read it once more concentrating on them, but do not try to memorize this text. Physics is not about memorizing, it is about understanding.
Last update: Jan 10, 1997 | E - mail to Edward Kluk |
Copyright (c) 1996 Edward Kluk |