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Part One We have taken a look at some basic quantities: time and location. We have taken a look at how to deal with 3-D quantities – vectors. We have taken a look at the basics of motion: position, velocity and acceleration.

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**In Part Two, we will look at:**

how to work with 2-D motion (combine motion and vectors); how we control or predict motion (what causes motion); how a third fundamental property: mass, is involved; what are the basic forces and how do they work.

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**Two Dimensional Motion**

How do we work with two dimensions when we consider motion?

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**Two Dimensional Motion**

How do we work with two dimensions when we consider motion? We work with vectors by working with the components!

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**Two Dimensional Motion**

Does it matter which form we use: rectangular or polar?

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**Two Dimensional Motion**

Does it matter which form we use: rectangular or polar? Since vectors add nicely in rectangular, we need to work with rectangular components!

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**Two Dimensional Motion**

But if we work in components, how are the components “connected” ?

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**Two Dimensional Motion**

But if we work in components, how are the components “connected” ? By time! Time is the same for the x and for the y. For each time, there is only one x and one y. If we wanted, we could eliminate the time and express y in terms of x: convert x(t) into t(x), and then convert y(t) into y(x). This is sometimes used, but we do lose information about time if we do this.

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**Special Case: Trajectory Motion**

We have equations for objects falling under the influence of gravity if we neglect air resistance and we don’t go too far from the earth’s surface: y = yo + vyot + (1/2)ayt2 vy = vyo + ayt

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**Special Case: Trajectory Motion**

If we again consider no air resistance, there should be zero acceleration in the x direction. This leads to the following equations for x: x = xo + vxot + (1/2)*0*t2 vx = vxo + 0*t which become simply: vx = vxo and x = xo + vxt .

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**Special Case: Trajectory Motion**

In effect, then, we have three equations: x = xo + vxt y = yo + vyot + (1/2)ayt2 vy = vyo + ayt . In these three equations we have nine quantities: xo, x, yo, y, vx, vyo, vy, ay and t. This means that we need to know 6 of these in order to solve for the other 3.

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**Example of Trajectory Motion**

Problem: If you are at the top of a building 12 meters above the ground, and you throw a ball out with a speed of 33 m/s at an angle of 40o above the horizontal, how far away from the building will the ball land (assuming the ground is level at the base of the building)?

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**Example of Trajectory Motion**

The first thing we recognize is that this is a trajectory problem, so we have the three equations: x = xo + vxt y = yo + vyot + (1/2)ayt2 vy = vyo + ayt . This also means that there is no acceleration in the x, so vxo = vx.

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**Example of Trajectory Motion**

The second thing we recognize is that this takes place on the earth, so that ay = -9.8 m/s2. Next let’s draw the diagram and put our information from the problem on that diagram: (go to the next slide for the diagram).

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**Example of Trajectory Motion**

xo = ay = -9.8 m/s2 yo = vxo = t = vyo = x = y = vx = vxo vy =

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**Example of Trajectory Motion**

Let’s look at the information given in the problem and see how that specifies some of the remaining unknown quantities: Problem: If you are at the top of a building 12 meters above the ground, ... We can identify this value as the initial y value, so yo = +12 meters. We can also choose to measure from the building, so this means xo = 0 meters.

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**Example of Trajectory Motion**

xo = 0 m ay = -9.8 m/s2 yo = 12 m vxo = t = vyo = x = y = vx = vxo vy =

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**Example of Trajectory Motion**

Let’s continue to look at the information: … and you throw a ball out with a speed of 33 m/s at an angle of 40o above the horizontal … This information refers to the initial velocity, but it is in polar form, not rectangular! So we need to use the polar to rectangular transformation equations:

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**Example of Trajectory Motion**

vyo = vo sin (qo) = 33 m/s sin (40o) = m/s, and vxo = vo cos (qo) = 33 m/s cos (40o) = m/s .

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**Example of Trajectory Motion**

xo = 0 m ay = -9.8 m/s2 yo = 12 m vxo = m/s t = vyo = m/s x = y = vx = vxo vy =

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**Example of Trajectory Motion**

Now we look at the last part of the problem statement: … how far away from the building will the ball land … How far away from the building is asking the question: x = ? But we do know it lands on the ground, so this indicates that: y = 0 meters .

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**Example of Trajectory Motion**

xo = 0 m ay = -9.8 m/s2 yo = 12 m vxo = m/s t = vyo = m/s x = ? y = 0 m vx = vxo vy =

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**Example of Trajectory Motion**

Notice that we have three equations, and we have three quantities still unknown: x, t and vy. One of these, x, is explicitly asked for in the problem. The other two, t and vy are not explicitly asked for, but neither are they specified. Now we can use our three equations to solve for our three unknowns!

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**Doing the algebra x = xo + vxt y = yo + vyot + (1/2)ayt2**

vy = vyo + ayt Substituting in the knowns, we have: x = (0 m) + (25.28 m/s * t) 0 m = (12 m) + (21.21 m/s*t) + (1/2 *-9.8 m/s2*t2) vy = (21.21 m/s) + (-9.8 m/s2 * t)

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**Doing the algebra x = (0 m) + (25.28 m/s * t)**

0 m = (12 m) + (21.21 m/s*t) + (1/2 *-9.8 m/s2*t2) vy = (21.21 m/s) + (-9.8 m/s2 * t) Note that the second equation has only one unknown: t, but it has a square on it in one term and is found in another term also. Thus we use the quadratic equation to solve for it: in terms of the quadratic formula: a = -4.9 m/s2, b = m/s, and c = 12 m.

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**Doing the algebra 0 m = (12 m) + (21.21 m/s*t) + (1/2 *-9.8 m/s2*t2)**

If we have ax2 + bx + c = 0, then x = [-b +/- {b2 - 4ac}1/2] / [2a] so with a = -4.9 m/s2, b = m/s, and c = 12 m we get: t = [ m/s +/- {(21.21 m/s)2 - (4*-4.9 m/s2*12 m)}1/2] / [2*-4.9 m/s2] = [ m/s +/- {26.17 m/s}] / -9.8 m/s2 = -0.51 s or s.

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Doing the algebra t = s or s. Which answer is the one we are looking for? The negative answer (-0.51 s) would indicate that it hit the ground before we threw it. This is obviously not the answer we’re looking for. In fact, the equations do not apply before we threw the ball because there were other forces acting on the ball besides gravity! Thus the correct time for the ball to hit is: t = 4.83 s.

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Doing the algebra Having used the second equation to solve for t, we now use the first equation: x = (0 m) + (25.28 m/s * t) to solve for x: x = (0 m) + (25.28 m/s * 4.83 s) = meters This does appear to be a reasonable distance for the ball to go (neglecting air resistance)! This answers the question posed. If we want, we could use the third equation to solve for vy.

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Trajectories The previous example was showing how we could predict where the ball was going to go based on how we threw it. We can also determine how to throw a ball so that it hits where we want it to. This is the point of the computer homework program: Trajectories (Vol 1, #6).

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Trajectories (Vol 1, #6) In the trajectories program, the setup is like the previous example, except in what is given and in what is asked for. Instead of giving the initial speed and angle, you are given the final x position. Thus we have x as known, but we have vo and qo (which is equivalent to vxo and vyo) as unknown - NOT A FAIR TRADE! We now have four unknowns instead of three!

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Trajectories (Vol 1, #6) What does it mean to have four unknowns (vo, qo, t and vy) with three equations? It means that there is more than one right way of hitting the target (solving the problem). Does that correspond to what we expect to happen? Yes - we can hit the target by various combinations of vo and qo. In this case, you have one free choice!

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**Trajectories x = xo + vo*cos(qo)*t y = yo + vo*sin(qo)*t – ½*g*t2**

vy = vo*sin(qo) – g*t Blue quantities are known; red are unknown. Can choose one of the unknowns: vo, qo, t, vy. The first two equations have three unknowns, the third equation has all four unknowns. - I like to choose the angle as my choice.

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**Another Special Case of 2-D Motion: Circular Motion**

What does it mean to go in a circle? The radius must be constant! This indicates that polar coordinates might be easier to use than rectangular! In order to work with vectors and motion, we need to work in rectangular coordinates to derive the results, but once the results are there, it does turn out that polar form is nicer!

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**Uniform Circular Motion**

Even though the radius is constant, the angle changes. For right now, we will consider the special case where the angle changes at a constant rate. This type of motion is called uniform circular motion. Note that since the radius doesn’t change, there should be no radial component of the velocity! (r = constant, vr = zero).

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**Uniform Circular Motion**

What about the angle, q ? In circular motion the angle does change! In uniform circular motion the angle changes at a constant rate! We call the change in angle with respect to time the angular speed, w. w = Dq / Dt Note that the units of angular speed, w are: radians/second.

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**Uniform Circular Motion**

If we wish to use cycles instead of radians, we define a new symbol, f: f = w * (1 cycle / 2p radians), or more simply: f = w / 2p . Note that the units of f are cycles/second. This unit also has its own name called a Hertz. Note also that we almost never use the unit of degrees/second when indicating angular speed.

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**Uniform Circular Motion**

Also note that the inverse of f, measured in cycles/sec, is what we call the period, T: T = 1/f , where T is measured in seconds/cycle; that is, it is the time for the object to complete one revolution. For the earth spinning on its axis, T is one day; for the earth orbiting around the sun, T is one year.

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**Uniform Circular Motion**

It might seem, then, that since the radius is constant and the radial speed is zero; and that since the angular speed is constant for uniform circular motion, it might seem that there is no acceleration in uniform circular motion! But that is not looking at the motion from the rectangular point of view:

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**Uniform Circular Motion**

From the rectangular point of view, the x motion goes forward, slows down and stops, and then reverses until it slows down and stops and the the whole process repeats. The same is true for the y motion. From this point of view it is easy to see that there is acceleration in both the x and y components.

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**Uniform Circular Motion**

How do we understand the acceleration from the polar point of view? Acceleration is the change in velocity. The velocity is a vector, and in polar form has a magnitude and a direction. Although the magnitude of the velocity does not change in uniform circular motion, the direction changes! There is an acceleration when an object turns.

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**Uniform Circular Motion**

We need to consider two more aspects: the formula for this turning acceleration, and the formula for the speed in polar form. We already have the angular speed (in radians/sec), but this is not the same as velocity (in meters/second).

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**Uniform Circular Motion**

If we go back to the definition of angles measured in radians (q = s/r), we can see that the circular speed (or tangential speed): vq = Ds / Dt = D(rq) / Dt = r(Dq / Dt) = rw. By using the calculus in rectangular form, we come up with the following accelerations in polar form: aq = 0 , and ar = -rw2 , where the minus indicates that the acceleration is toward the center.

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**Uniform Circular Motion**

Putting all of this together, we have: r = constant; q changes with time vr = zero; vq = rw where w = Dq / Dt ar = -rw2 ; aq = zero with f = w/2p and T = 1/f . NOTE: the direction of the acceleration is in the direction of the turning - which is toward the center of the circle.

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**Uniform Circular Motion**

Note that for solving circular motion problems, we have four equations: vq = rw ar = -rw2 f = w/2p and T = 1/f . In these four equations, there are six quantities: r, vq, ar, w, f and T. Thus we need to identify two of these to solve for the remaining four.

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**Example of Uniform Circular Motion**

How fast are we moving now since we are “riding” the earth as it orbits the sun? Facts you should know: period for earth’s orbit is one year, distance to the sun (the radius of the circle) is 93 million miles. [Actually the earth goes in an ellipse, but it is very close to being a circle. For a first try, this will give very good results.]

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Example (cont.) First step in reading the problem is to recognize this as uniform circular motion. Thus we have the four equations: vq = rw ar = -rw2 f = w/2p and T = 1/f . Next from the problem we are given: r = 93 million miles = 1.49 x 1011 m; T = 365 days = 3.15 x 107 sec.

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**Example, cont. vq = rw ar = rw2 f = w/2p and T = 1/f .**

r = 93 million miles = 1.49 x 1011 m; T = 365 days = 3.15 x 107 sec. Since we have four equations (above) and four unknowns (vq, w, ar and f), we should be able to solve for the unknowns. Note that we drop the minus sign for ar since it merely indicates that the acceleration is towards the center.

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**Example, cont. vq = rw ar = rw2 f = w/2p and T = 1/f .**

r = 93 million miles = 1.49 x 1011 m; T = 365 days = 3.15 x 107 sec. Using T = 1/f , we can find f: f = 1/T = 1/ 3.15 x 107 sec = x 10-8 Hz. Using f = w/2p, we can find w: w = 2pf = 2*3.14* 3.17 x 10-8 Hz = 1.99 x 10-7 rad/sec

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**Example, cont. vq = rw ar = rw2 f = w/2p and T = 1/f .**

r = 1.49 x 1011 m; T = 3.15 x 107 sec. f = 3.17 x 10-8 Hz. w = 1.99 x 10-7rad/sec Using vq = rw, we can find vq: vq = r*w = 1.49 x 1011 m * 1.99 x 10-7rad/sec = 29,670 m/s = 66,340 miles/hour !

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**Example, cont. vq = rw ar = rw2 f = w/2p and T = 1/f .**

r = 1.49 x 1011 m; T = 3.15 x 107 sec. f = 3.17 x 10-8 Hz. w = 1.99 x 10-7rad/sec vq = 29,670 m/s . Completing the problem, we can use ar = rw2 to find ar: ar = rw2 = 1.49 x 1011 m * (1.99 x 10-7rad/sec)2 = m/s2

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Example, cont. A couple of notes: we have a very high speed, but a very low acceleration in the past example. Note that this is the case because w was very small, and the acceleration involved w2 . In many cases, we can have a very high acceleration if w is very high. This is the principle behind the centrifuge.

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Relative Velocity How do we work with a situation in which an object moves on something that is itself moving? Examples: Flying an airplane in air that is moving (flying in a wind), running a boat on the river, walking inside an airplane that is moving.

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Relative Velocity Since velocity is a vector, we get the total velocity by simply adding the velocity of the object in or on the medium to the velocity of the medium. But we must add these two velocities as vectors! That means, we must work in rectangular coordinates.

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**Example: Boat on a river**

Problem: A boat is to cross a river such that it gets to a point due East of where it starts. The boat goes 5 m/s and the river flows South at 2 m/s. We’ll assume here that the river flows at a uniform speed - not faster in the middle and slower at the edges. Also, if the river is 100 meters wide, how long will it take to cross the river?

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**Example: Boat on a river**

First we draw vR = 2 m/s a diagram: since the river will push the boat downstream we should head vb = 5 m/s somewhat upstream q = ? w = 100 m

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**Example: Boat on a river**

The problem, then, is what is the angle, q? The basic principle we use is vector addition. vb + vR = vTotal But what do we want the two vectors to add up to (what do we want vTotal to be)? What does the phrase “due East” mean here? It means that vTotal should have a zero North-South component!

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**Example: Boat on a river**

Converting to rectangular components from polar, and then adding the components, gives: vb cos(q) + vR (0) = vTotal-x vb sin(q) + vR (-1) = vTotal-y = 0 From the y component equation, we have an equation for q: (5 m/s) sin(q) - (2 m/s) = 0 m/s.

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**Example: Boat on a river**

(5 m/s) sin(q) - (2 m/s) = 0 m/s. Solving for q gives: q = Arcsin(2 m/s / 5 m/s) = 23.6o .

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**Example: Boat on a river**

The time to cross the river depends on the width of the river - but this is purely in the x direction. This means that we need the x component of the total speed: vb cos(q) + vR (0) = vTotal-x = (5 m/s) cos(23.6o ) + (0 m/s) = m/s. Since vx = Dx/Dt, where Dx = w, Dt = t = w/vx = 100 m / 4.58 m/s = seconds.

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**Example #2: Boat on a river**

Suppose that we change the problem slightly to become: what angle should we head the boat to cross the river in the least time? Will the shortest distance (see previous problem) give us the shortest time? (If so, then the answer to the previous problem is the answer to this problem as well.) How is time related to the angle? How is time related to the velocity?

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Relative Velocity Again the key is to consider the two-dimensional nature of the motion. What we need to do to cross the river is to go in the x direction. Going in the y direction does not help us cross the river. Looked at from this point of view, what we want is the biggest x-speed so that the time to cross the x distance is as small as possible. This we see can only be if we head straight across!

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Relative Velocity To see if this checks out, consider the time for going the shortest distance (previous problem, going straight across): seconds. If we head straight across ( = 0o), then vx = vboat cos() = 5 m/s cos (0o) = 5 m/s. t = w/vx = 100 m / (5 m/s) = seconds. In heading straight across we will go a further distance (due to our y motion - downstream) but we will go faster, and the faster “wins”. In going straight across, we used some of our boat velocity to “fight” the river.

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