Motion in Three Dimensions

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Chapter 6. Newton's Laws in Three Dimensions

6.1 Forces Have No Perpendicular Effects

Suppose you could shoot a rifle and arrange for a second bullet to be dropped from the same height at the exact moment when the first left the barrel. Which would hit the ground first? Nearly everyone expects that the dropped bullet will reach the dirt first, and Aristotle would have agreed. Aristotle would have described it like this. The shot bullet receives some forced motion from the gun. It travels forward for a split second, slowing down rapidly because there is no longer any force to make it continue in motion. Once it is done with its forced motion, it changes to natural motion, i.e. falling straight down. While the shot bullet is slowing down, the dropped bullet gets on with the business of falling, so according to Aristotle it will hit the ground first.

a / A bullet is shot from a gun, and another bullet is simultaneously dropped from the same height. 1. Aristotelian physics says that the horizontal motion of the shot bullet delays the onset of falling, so the dropped bullet hits the ground first. 2. Newtonian physics says the two bullets have the same vertical motion, regardless of their different horizontal motions.

Luckily, nature isn't as complicated as Aristotle thought! To convince yourself that Aristotle's ideas were wrong and needlessly complex, stand up now and try this experiment. Take your keys out of your pocket, and begin walking briskly forward. Without speeding up or slowing down, release your keys and let them fall while you continue walking at the same pace.

You have found that your keys hit the ground right next to your feet. Their horizontal motion never slowed down at all, and the whole time they were dropping, they were right next to you. The horizontal motion and the vertical motion happen at the same time, and they are independent of each other. Your experiment proves that the horizontal motion is unaffected by the vertical motion, but it's also true that the vertical motion is not changed in any way by the horizontal motion. The keys take exactly the same amount of time to get to the ground as they would have if you simply dropped them, and the same is true of the bullets: both bullets hit the ground simultaneously.

These have been our first examples of motion in more than one dimension, and they illustrate the most important new idea that is required to understand the three-dimensional generalization of Newtonian physics:

Forces have no perpendicular effects.

When a force acts on an object, it has no effect on the part of the object's motion that is perpendicular to the force.

In the examples above, the vertical force of gravity had no effect on the horizontal motions of the objects. These were examples of projectile motion, which interested people like Galileo because of its military applications. The principle is more general than that, however. For instance, if a rolling ball is initially heading straight for a wall, but a steady wind begins blowing from the side, the ball does not take any longer to get to the wall. In the case of projectile motion, the force involved is gravity, so we can say more specifically that the vertical acceleration is 9.8 m/s², regardless of the horizontal motion.

self-check: In the example of the ball being blown sideways, why doesn't the ball take longer to get there, since it has to travel a greater distance? (answer in the back of the PDF version of the book)

Relationship to relative motion

These concepts are directly related to the idea that motion is relative. Galileo's opponents argued that the earth could not possibly be rotating as he claimed, because then if you jumped straight up in the air you wouldn't be able to come down in the same place. Their argument was based on their incorrect Aristotelian assumption that once the force of gravity began to act on you and bring you back down, your horizontal motion would stop. In the correct Newtonian theory, the earth's downward gravitational force is acting before, during, and after your jump, but has no effect on your motion in the perpendicular (horizontal) direction.

If Aristotle had been correct, then we would have a handy way to determine absolute motion and absolute rest: jump straight up in the air, and if you land back where you started, the surface from which you jumped must have been in a state of rest. In reality, this test gives the same result as long as the surface under you is an inertial frame. If you try this in a jet plane, you land back on the same spot on the deck from which you started, regardless of whether the plane is flying at 500 miles per hour or parked on the runway. The method would in fact only be good for detecting whether the plane was accelerating.

Discussion Questions

◊ The following is an incorrect explanation of a fact about target shooting:

“Shooting a high-powered rifle with a high muzzle velocity is different from shooting a less powerful gun. With a less powerful gun, you have to aim quite a bit above your target, but with a more powerful one you don't have to aim so high because the bullet doesn't drop as fast.”

What is the correct explanation?

dq-target-shooting

◊ You have thrown a rock, and it is flying through the air in an arc. If the earth's gravitational force on it is always straight down, why doesn't it just go straight down once it leaves your hand?

◊ Consider the example of the bullet that is dropped at the same moment another bullet is fired from a gun. What would the motion of the two bullets look like to a jet pilot flying alongside in the same direction as the shot bullet and at the same horizontal speed?

b / This object experiences a force that pulls it down toward the bottom of the page. In each equal time interval, it moves three units to the right. At the same time, its vertical motion is making a simple pattern of +1, 0, -1, -2, -3, -4, ... units. Its motion can be described by an x coordinate that has zero acceleration and a y coordinate with constant acceleration. The arrows labeled x and y serve to explain that we are defining increas- ing x to the right and increasing y as upward.

6.2 Coordinates and Components

c / The shadow on the wall shows the ball's y motion, the shadow on the floor its x motion.

d / Example 1.

e / A parabola can be defined as the shape made by cutting a cone parallel to its side. A parabola is also the graph of an equation of the form y∝ x².

f / Each water droplet follows a parabola. The faster drops' parabolas are bigger.

'Cause we're all
Bold as love,
Just ask the axis. -- Jimi Hendrix

How do we convert these ideas into mathematics? Figure b shows a good way of connecting the intuitive ideas to the numbers. In one dimension, we impose a number line with an x coordinate on a certain stretch of space. In two dimensions, we imagine a grid of squares which we label with x and y values, as shown in figure b.

But of course motion doesn't really occur in a series of discrete hops like in chess or checkers. The figure on the left shows a way of conceptualizing the smooth variation of the x and y coordinates. The ball's shadow on the wall moves along a line, and we describe its position with a single coordinate, y, its height above the floor. The wall shadow has a constant acceleration of -9.8 m/s². A shadow on the floor, made by a second light source, also moves along a line, and we describe its motion with an x coordinate, measured from the wall.

The velocity of the floor shadow is referred to as the x component of the velocity, written v_x. Similarly we can notate the acceleration of the floor shadow as a_x. Since v_x is constant, a_x is zero.

Similarly, the velocity of the wall shadow is called v_y, its acceleration a_y. This example has a_y=-9.8 m/s².

Because the earth's gravitational force on the ball is acting along the y axis, we say that the force has a negative y component, F_y, but F_x=F_z=0.

The general idea is that we imagine two observers, each of whom perceives the entire universe as if it was flattened down to a single line. The y-observer, for instance, perceives y, v_y, and a_y, and will infer that there is a force, F_y, acting downward on the ball. That is, a y component means the aspect of a physical phenomenon, such as velocity, acceleration, or force, that is observable to someone who can only see motion along the y axis.

All of this can easily be generalized to three dimensions. In the example above, there could be a z-observer who only sees motion toward or away from the back wall of the room.

Example 1: A car going over a cliff

◊ The police find a car at a distance w=20 m from the base of a cliff of height h=100 m. How fast was the car going when it went over the edge? Solve the problem symbolically first, then plug in the numbers.

◊ Let's choose y pointing up and x pointing away from the cliff. The car's vertical motion was independent of its horizontal motion, so we know it had a constant vertical acceleration of a=-g=-9.8 m/s². The time it spent in the air is therefore related to the vertical distance it fell by the constant-acceleration equation

$Delta y = frac{1}{2}a_y Delta t^2 qquad , text{or} -h = frac{1}{2}(-g)Delta t^2 qquad .$

Solving for Δ t gives

$Delta t = sqrt{frac{2h}{g}} qquad .$

Since the vertical force had no effect on the car's horizontal motion, it had a_x=0, i.e., constant horizontal velocity. We can apply the constant-velocity equation

$v_x = frac{Delta x}{Delta t} qquad , text{i.e.,} v_x = frac{w}{Delta t} qquad .$

We now substitute for Δ t to find

$v_x = w / sqrt{frac{2h}{g}} qquad ,$

which simplifies to

$v_x = w sqrt{frac{g}{2h}} qquad .$

Plugging in numbers, we find that the car's speed when it went over the edge was 4 m/s, or about 10 mi/hr.

Projectiles move along parabolas.

What type of mathematical curve does a projectile follow through space? To find out, we must relate x to y, eliminating t. The reasoning is very similar to that used in the example above. Arbitrarily choosing x=y=t=0 to be at the top of the arc, we conveniently have x=Δ x, y=Δ y, and t=Δ t, so

$y = frac{1}{2}a_y t^2 qquad (a_y<0)$

$x = v_x t$

We solve the second equation for t=x/v_x and eliminate t in the first equation:

$y = frac{1}{2}a_yleft(frac{x}{v_x}right)^2 .$

Since everything in this equation is a constant except for x and y, we conclude that y is proportional to the square of x. As you may or may not recall from a math class, y∝ x² describes a parabola.

◊ Solved problem: A cannon — problem 5

Discussion Question

◊ At the beginning of this section I represented the motion of a projectile on graph paper, breaking its motion into equal time intervals. Suppose instead that there is no force on the object at all. It obeys Newton's first law and continues without changing its state of motion. What would the corresponding graph-paper diagram look like? If the time interval represented by each arrow was 1 second, how would you relate the graph-paper diagram to the velocity components v_x and $v_y?$

◊ Make up several different coordinate systems oriented in different ways, and describe the a_x and a_y of a falling object in each one.

6.3 Newton's Laws in Three Dimensions

g / Example 2.

It is now fairly straightforward to extend Newton's laws to three dimensions:

Newton's first law

If all three components of the total force on an object are zero, then it will continue in the same state of motion.

Newton's second law

The components of an object's acceleration are predicted by the equations

$a_x = F_{x,total}/m qquad ,$

$a_y = F_{y,total}/m qquad , qquad text{and}$

$a_z = F_{z,total}/m qquad .$

Newton's third law

If two objects A and B interact via forces, then the components of their forces on each other are equal and opposite:

$F_{text{A on B},x} = -F_{text{B on A},x} qquad ,$

$F_{text{A on B},y} = -F_{text{B on A},y} qquad , qquad text{and}$

$F_{text{A on B},z} = -F_{text{B on A},z} qquad .$

Example 2: Forces in perpendicular directions on the same object

◊ An object is initially at rest. Two constant forces begin acting on it, and continue acting on it for a while. As suggested by the two arrows, the forces are perpendicular, and the rightward force is stronger. What happens?

◊ Aristotle believed, and many students still do, that only one force can “give orders” to an object at one time. They therefore think that the object will begin speeding up and moving in the direction of the stronger force. In fact the object will move along a diagonal. In the example shown in the figure, the object will respond to the large rightward force with a large acceleration component to the right, and the small upward force will give it a small acceleration component upward. The stronger force does not overwhelm the weaker force, or have any effect on the upward motion at all. The force components simply add together:

$F_{x,total} = F_{1,x} + cancelto{0}{F_{2,x}}$

$F_{y,total} = cancelto{0}{F_{1,y}} + F_{2,y}$

Discussion Question

◊ The figure shows two trajectories, made by splicing together lines and circular arcs, which are unphysical for an object that is only being acted on by gravity. Prove that they are impossible based on Newton's laws.

Summary

Vocabulary

component — the part of a velocity, acceleration, or force that would be perceptible to an observer who could only see the universe projected along a certain one-dimensional axis

parabola — the mathematical curve whose graph has y proportional to x²

Notation

x, y, z — an object's positions along the x, y, and z axes

v_x, v_y, v_z — the x, y, and z components of an object's velocity; the rates of change of the object's x, y, and z coordinates

a_x, a_y, a_z — the x, y, and z components of an object's acceleration; the rates of change of v_x, v_y, and v_z

Summary

A force does not produce any effect on the motion of an object in a perpendicular direction. The most important application of this principle is that the horizontal motion of a projectile has zero acceleration, while the vertical motion has an acceleration equal to g. That is, an object's horizontal and vertical motions are independent. The arc of a projectile is a parabola.

Motion in three dimensions is measured using three coordinates, x, y, and z. Each of these coordinates has its own corresponding velocity and acceleration. We say that the velocity and acceleration both have x, y, and z components

Newton's second law is readily extended to three dimensions by rewriting it as three equations predicting the three components of the acceleration,

$a_x=F_{x,total}/m qquad ,$

$a_y=F_{y,total}/m qquad ,$

$a_z=F_{z,total}/m qquad ,$

and likewise for the first and third laws.

Homework Problems

1. (a) A ball is thrown straight up with velocity v. Find an equation for the height to which it rises.
(b) Generalize your equation for a ball thrown at an angle θ above horizontal, in which case its initial velocity components are v_x=v cos θ and v_y=v sin θ.

2. At the Salinas Lettuce Festival Parade, Miss Lettuce of 1996 drops her bouquet while riding on a float moving toward the right. Compare the shape of its trajectory as seen by her to the shape seen by one of her admirers standing on the sidewalk.

3. Two daredevils, Wendy and Bill, go over Niagara Falls. Wendy sits in an inner tube, and lets the 30 km/hr velocity of the river throw her out horizontally over the falls. Bill paddles a kayak, adding an extra 10 km/hr to his velocity. They go over the edge of the falls at the same moment, side by side. Ignore air friction. Explain your reasoning.
(a) Who hits the bottom first?
(b) What is the horizontal component of Wendy's velocity on impact?
(c) What is the horizontal component of Bill's velocity on impact?
(d) Who is going faster on impact?

4. A baseball pitcher throws a pitch clocked at v_x=73.3 mi/h. He throws horizontally. By what amount, d, does the ball drop by the time it reaches home plate, L=60.0 ft away?
(a) First find a symbolic answer in terms of L, v_x, and g.
(b) Plug in and find a numerical answer. Express your answer in units of ft. [Note: 1 ft=12 in, 1 mi=5280 ft, and 1 in=2.54 cm](answer check available at lightandmatter.com)

h / Problem 4.

5. A cannon standing on a flat field fires a cannonball with a muzzle velocity v, at an angle θ above horizontal. The cannonball thus initially has velocity components v_x=v cos θ and v_y=v sin θ.
(a) Show that the cannon's range (horizontal distance to where the cannonball falls) is given by the equation R=(2v²/g)sinθcosθ .
(b) Interpret your equation in the cases of θ =0 and θ =90°.(solution in the pdf version of the book)

6. Assuming the result of problem 5 for the range of a projectile, R=(2v²/g)sinθcosθ, show that the maximum range is for θ =45°. ∫

7. Two cars go over the same bump in the road, Maria's Maserati at 25 miles per hour and Park's Porsche at 37. How many times greater is the vertical acceleration of the Porsche? Hint: Remember that acceleration depends both on how much the velocity changes and on how much time it takes to change.(answer check available at lightandmatter.com)