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Week 1: Newton’s Laws |
Now the big question: Does the car hit the bike? If it does, it does so at some real time, call it th. “Hitting” means that there is no distance between them – they are at the same place at the same time, in particular at this time th. Turning this sentence into an equation, the condition for a collision is algebraically:
xb(th) = v0bth + D = − |
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2 acth2 + v0cth = xc(th) |
(72) |
Rearranged, this is a quadratic equation:
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acth2 − (v0c − v0b) th |
+ D = 0 |
(73) |
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and therefore has two roots. If we write down the quadratic formula: |
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th = |
0c − |
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± p |
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ac |
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(74) |
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(v0c |
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we can see that there will only be a real (as opposed to imaginary) time th that solves the collision condition if the argument of the square root is non-negative. That is:
(v0c − v0b)2 ≥ 2acD |
(75) |
If this is true, there will be a collision. If it is false, the car will never reach the bike.
There is actually a second way to arrive at this result. One can find the time ts that the car is travelling at the same speed as the bike. That’s really pretty easy:
v0b = vc(ts) = −acts + v0c |
(76) |
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or |
(v0c − v0b) |
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ts = |
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(77) |
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ac |
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Now we locate the car relative to the bike. If the collision hasn’t happened by ts it never will, as afterwards the car will be slower than the bike and the bike will pull away. If the position of the car is behind (or barely equal to) the position of the bike at ts, all is well, no collision occurs. That is:
xc(ts) = − |
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2 acts2 + v0cts ≤ v0bts + D |
(78) |
if no collision occurs. It’s left as an exercise to show that this leads to the same condition that the quadratic gives you.
Next, let’s see what happens when we have only one object but motion in two dimensions.
1.8: Motion in Two Dimensions
The idea of motion in two or more dimensions is very simple. Force is a vector, and so is acceleration. Newton’s Second Law is a recipe for taking the total force and converting it into a di erential equation of motion:
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~r |
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~a = |
d |
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F tot |
(79) |
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dt2 |
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In the most general case, this can be quite di cult to solve. For example, consider the forces that act upon you throughout the day – every step you take, riding in a car, gravity, friction, even the wind exert forces subtle or profound on your mass and accelerate you first this way, then that as you move around. The total force acting on you varies wildly with time and place, so even though your trajectory is a solution to just such an equation of motion, computing it algebraically is out of
Week 1: Newton’s Laws |
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the question. Computing it with a computer would be straightforward if the forces were all known, but of course they vary according to your volition and the circumstances of the moment and are hardly knowable ahead of time.
However, much of what happens in the world around you can actually be at least approximated by relatively simple (if somewhat idealized) models and explicitly solved. These simple models generally arise when the forces acting are due to the “well-known” forces of nature or force rules listed above and hence point in specific directions (so that their vector description can be analyzed) and are either constant in time or vary in some known way so that the calculus of the solution is tractable45.
We will now consider only these latter sorts of forces: forces that act in a well-defined direction with a computable value (initially, with a computable constant value, or a value that varies in some simple way with position or time). If we write the equation of motion out in components:
ax |
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d2x |
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Ftot,x |
(80) |
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dt2 |
m |
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ay |
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d2y |
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Ftot,y |
(81) |
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dt2 |
m |
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az |
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d2z |
= |
Ftot,z |
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(82) |
dt2 |
m |
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we will often reduce the complexity of the problem from a “three dimensional problem” to three “one dimensional problems” of the sort we just learned to solve in the section above.
Of course, there’s a trick to it. The trick is this:
Select a coordinate system in which one of the coordinate axes is aligned with the total force.
We won’t always be able to do this, but when it can it will get us o to a very good start, and trying it will help us understand what to do when we hit problems where this alone won’t quite work or help us solve the problem.
The reason this step (when possible) simplifies the problem is simple enough to understand: In this particular coordinate frame (with the total force pointing in a single direction along one of the coordinate axes), the total force in the other directions adds up to zero! That means that all acceleration occurs only along the selected coordinate direction. Solving the equations of motion in the other directions is then trivial – it is motion with a constant velocity (which may be zero, as in the case of dropping a ball vertically down from the top of a tower in the problems above). Solving the equation of motion in the direction of the total force itself is then “the problem”, and you will need lots of practice and a few good examples to show you how to go about it.
To make life even simpler, we will now further restrict ourselves to the class of problems where the acceleration and velocity in one of the three dimensions is zero. In that case the value of that coordinate is constant, and may as well be taken to be zero. The motion (if any) then occurs in the remaining two dimensional plane that contains the origin. In the problems below, we will find it useful to use one of two possible two-dimensional coordinate systems to solve for the motion: Cartesian coordinates (which we’ve already begun to use, at least in a trivial way) and Plane Polar coordinates, which we will review in context below.
As you will see, solving problems in two or three dimensions with a constant force direction simply introduces a few extra steps into the solution process:
45“Tractable” here means that it can either be solved algebraically, true for many of the force laws or rules, or at least solved numerically. In this course you may or may not be required or expected to explore numerical solutions to the di erential equations with e.g. matlab, octave, or mathematica.
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Week 1: Newton’s Laws |
•Decomposing the known forces into a coordinate system where one of the coordinate axes lines up with the (expected) total force...
•Solving the individual one-dimensional motion problems (where one or two of the resulting solutions will usually be “trivial”, e.g. constant motion)...
•Finally, reconstructing the overall (vector) solution from the individual solutions for the independent vector coordinate directions...
and answering any questions as usual.
1.8.1: Free Flight Trajectories – Projectile Motion
Perhaps the simplest example of this process adds just one small change to our first example. Instead of dropping a particle straight down let us imagine throwing the ball o of a tower, or firing a cannon, or driving a golf ball o of a tee or shooting a basketball. All of these are examples of projectile motion
– motion under the primary action of gravity where the initial velocity in some horizontal direction is not zero.
Note well that we will necessarily idealize our treatment by (initially) neglecting some of the many things that might a ect the trajectory of all of these objects in the real world – drag forces which both slow down e.g. a golf ball and exert “lift” on it that can cause it to hook or slice, the fact that the earth is not really an inertial reference frame and is rotating out underneath the free flight trajectory of a cannonball, creating an apparent deflection of actual projectiles fired by e.g. naval cannons. That is, only gravity near the earth’s surface will act on our ideal particles for now.
The easiest way to teach you how to handle problems of this sort is just to do a few examples
– there are really only three distinct cases one can treat – two rather special ones and the general solution. Let’s start with the simplest of the special ones.
Example 1.8.1: Trajectory of a Cannonball
y
m
v0 mg
θ
R |
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Figure 9: An idealized cannon, neglecting the drag force of the air. Let x be the horizontal direction
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and y be the vertical direction, as shown. Note well that F g = −mgyˆ points along one of the coordinate directions while Fx = (Fz =)0 in this coordinate frame.
A cannon fires a cannonball of mass m at an initial speed v0 at an angle θ with respect to the ground as shown in figure 9. Find:
a)The time the cannonball is in the air.
b)The range of the cannonball.
Week 1: Newton’s Laws |
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We’ve already done the first step of a good solution – drawing a good figure, selecting and sketching in a coordinate system with one axis aligned with the total force, and drawing and labelling all of the forces (in this case, only one). We therefore proceed to write Newton’s Second Law for both coordinate directions.
Fx |
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max = 0 |
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(83) |
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may = m |
d2y |
= −mg |
(84) |
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We divide each of these equations by m to obtain two equations of motion, one for x and the other for y:
ax |
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0 |
(85) |
ay |
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−g |
(86) |
We solve them independently. In x:
ax = |
dvx |
= 0 |
(87) |
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The derivative of any constant is zero, so the x-component of the velocity does not change in time. We find the initial (and hence constant) component using trigonometry:
vx(t) = v0x = v0 cos(θ) |
(88) |
We then write this in terms of derivatives and solve it:
vx = |
dx |
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v0 cos(θ) |
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dx |
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v0 cos(θ) dt |
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v0 cos(θ) Z |
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We evaluate C (the constant of integration) from our knowledge that in the coordinate system we selected, x(0) = 0 so that C = 0. Thus:
x(t) = v0 cos(θ)t |
(89) |
The solution in y is more or less identical to the solution that we obtained above dropping a ball, except the constants of integration are di erent:
ay = |
dvy |
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−g |
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dvy |
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= −gt + C′ |
(90) |
For this problem, we know from trigonometry that: |
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= v0 sin(θ) |
(91) |
so that C′ = v0 sin(θ) and: |
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vy (t) = −gt + v0 sin(θ) |
(92) |
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Week 1: Newton’s Laws |
We write vy in terms of the time derivative of y and integrate: |
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gt2 + v0 sin(θ)t + D |
(93) |
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Again we use y(0) = 0 in the coordinate system we selected to set D = 0 and get: |
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gt2 + v0 sin(θ)t |
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y(t) = − |
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(94) |
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Collecting the results from above, our overall solution is thus:
x(t) |
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v0 cos(θ)t |
(95) |
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gt2 + v0 sin(θ)t |
(96) |
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y(t) |
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vx(t) |
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v0x = v0 cos(θ) |
(97) |
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vy (t) |
= −gt + v0 sin(θ) |
(98) |
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We know exactly where the cannonball is at all times, and we know exactly what its velocity is as well. Now let’s see how we can answer the equations.
To find out how long the cannonball is in the air, we need to write an algebraic expression that we can use to identify when it hits the ground. As before (dropping a ball) “hitting the ground” in algebra-speak is y(tg ) = 0, so finding tg such that this is true should do the trick:
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gtg2 + v0 sin(θ)tg |
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y(tg ) = − |
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µ−2 gtg + v0 sin(θ)¶ tg |
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tg,1 |
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2v0 sin(θ) |
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(99)
(100)
are the two roots of this (factorizable) quadratic. The first root obviously describes when the ball was fired, so it is the second one we want. The ball hits the ground after being in the air for a time
tg,2 = |
2v0 sin(θ) |
(101) |
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Now it is easy to find the range of the cannonball, R. R is just the value of x(t) at the time that the cannonball hits!
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2v2 sin(θ) cos(θ) |
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R = x(tg,2) = |
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(102) |
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Using a trig identity one can also write this as: |
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v2 sin(2θ) |
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(103) |
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The only reason to do this is so that one can see that the range of this projectile is symmetric: It is the same for θ = π/4 ± φ for any φ [0, π/4].