Chapter 3 Class Notes



Chapter 4 Class Notes

This week, we tackle one old topic and two new topics that fall under the heading Motion in Two Dimensions. The old topic is kinematics but now in two dimensions. The two new topics are circular motion and relative motion. The common theme here is two dimensions, which will require you to deal explicitly with a two dimensional coordinate system.

Let’s start with kinematics in two dimensions. This topic is best introduced by considering a specific example; the projectile.

Projectile Motion

Simple projectile motion, that is, projectile motion which is unaffected by air resistance, can be broken down into two independent components; the horizontal, or x component and the vertical, or y component. The word independent is crucial, as it means that whatever motion is going on in the x-direction is completely unaffected by what’s going on in the y-direction. Let’s examine the two motions separately.

Y – motion

The motion in the y or vertical direction is just the same as throwing a ball up in the air and catching it, which is a problem we examined carefully in Chapter 2. The ball is subjected to a constant acceleration that caused by Earth’s gravity.

X - motion

Imagine that we project a ball horizontally in space, away from any gravitating objects. According to Newton’s first law, the ball will continue in a straight line, since it is unaffected by any forces.

Combined X – Y Motion

Now, let’s combine these two motions. The ball will rise and fall with constant acceleration, but it will also move horizontally with constant velocity, since, in the absence of air resistance, there is no acceleration in the horizontal direction. (Gravity acts downwards – remember!). So what we end up with is a curved trajectory that is actually parabolic in shape;

[pic]

Lets analyze the motion using the 3 equations of kinematics

In the y – direction, we have

y – yo = voy t – gt2

2

vy2 – voy2 = -2 g (y – yo)

vy = voy - gt

The notation requires some clarification. ay = -g, because the only acceleration is gravity, acting downwards. Implicit in this choice is a sign convention where up is + and down is - .

y is the vertical displacement after time, t.

yo is the vertical displacement at t = 0, usually zero, but not always. If the projectile is launched from the top of a cliff, then yo could be the height of the cliff, but it all depends on where you choose the origin of your co-ordinate system.

vy and voy are the y components of the velocities after time, t, and at t=0, respectively. You see, the projectile may be launched horizontally, or vertically, or at some angle, so the velocity must be resolved into it’s x and y components.

In the x-direction we have the 3 same equations as above, but with x subscripts instead of y subscripts. Two of the equations reduce to the same equation because there is no acceleration, ax = 0;

Thus,

vx = vox

a trivial result which is just telling you that the x component of the velocity does not change, which you already knew because ax = 0.

So we are left with just one useful equation, which is

x – xo = vox t

Now, let’s use these equations to solve a problem.

Example Problem

A canon fires a canon-ball at an angle of 30 degrees to the horizontal with an initial velocity of 33 m/s. a). How far does the canon ball travel horizontally? and b) how long is it in the air?

Solution

As with all Physics problems, we must start with a simple diagram, note we have both a coordinate system and a sign convention.

+

-

[pic]

The first part of the question asks for the horizontal range which means we need to be looking at our x equation of motion.

x – xo = vox t

in this equation xo = 0, since the origin of the coordinate system is placed on the location where the ball leaves the canon. The variable x is the horizontal range we are looking for, if the time, t, is the total time that the canon ball is in the air. Finally, vox is the x component of the initial velocity; vo cos θ. Note, we are not substituting any numbers at this point.

Unfortunately, we do not know the total time, t, that the canon-ball is in the air, but we can find it by examining the y motion.

The one thing we know for sure about the y motion is that the ball momentarily stops at it’s maximum altitude, which occurs exactly halfway. So, if we can find the time when vy = 0, we can just double it to get the total time of flight!

Of the 3 equations of motion in the y direction, only one equation is going to help us and it is the one linking velocity and time;

vy = voy - gt

Substitute vy = 0, voy = vo sin θ, and we get half the time of flight, t1/2

So, re-arranging for t we get,

t1/2 = vo sin θ

g

which we need to double to get the total time of flight, t,

thus,

t = 2 vo sin θ

g

Now, lets substitute this equation for the total time into the equation for horizontal range. What we get is,

x = 2 vo2 sin θ cos θ

g

this is interesting in that it contains the trigonometric identity, sin 2θ, so, simplifying

x = vo2 sin 2θ

g

Interestingly, this equation has two roots, since there are two different angles that have the same value of sin 2θ. This is most easily visualized by recalling what a sine curve looks like. However, the consequence of this result physically is that there are, in principle, two angles that you can fire the canon-ball at and have it land in exactly the same place!

Let’s substitute some numbers;

x = (33)2 sin (2 . 30) where the dot means “times”

9.8

x = 96.23 m

and the total time of flight

t = 2 . 33 sin ( 30 )

9.8

t = 3.36 s

Question: What other angle would yield the same horizontal range? I’ll leave this as an exercise for the student!

How would this problem change if the canon were instead placed on the top of a cliff of height h?

[pic]

Again, we start with a diagram, and then we would have to think carefully about where we want to put the origin of the coordinate system, on top of the cliff? or at the base of the cliff? It’s an important decision since in the latter yo = h, the height of the cliff (as shown in the diagram), whereas in the former, yo = 0. The choice also impacts the final height of the canon-ball. If the origin is chosen up on top of the cliff, then the final vertical displacement of the canon-ball, y = -h. If, on the other hand the origin is chosen to be at the base of the cliff (as shown in the diagram), the final vertical displacement of the canon-ball is y = 0.

Also, the trajectory is not symmetrical about the maximum altitude any more, since the ball has further to travel vertically on the way down than on the way up. The consequence of this is that the equation for the total time of flight

t = 2 vo sin θ

g

no longer works, since this equation was derived assuming that the trajectory is symmetrical. So, you see, these type of problems get complicated rather quickly.

Let’s do an example to illustrate how to solve one of these more complicated problems.

Problem

A canon fired from the top of a 50m cliff launches a canon-ball at an angle of 45o with respect to the horizontal with an initial velocity of 30 m/s. a) How long is the ball in the air? and b) What is the final velocity with which the canon-ball hits the ground?

Solution

I’m going to adopt the diagram and coordinate system shown on the previous page.

The values for the kinematic variables therefore become

yo = h, yf = 0

xo = 0, xf = ?

vox = vo cos θ, voy = vo sin θ

vf = ?

ax = 0, ay = -g

t = ?

To find the time of flight, we need to choose one of the 2 equations of kinematics that has time in it plus other variables that we know. The only reasonable choice is

y – yo = voy t – gt2

2

upon substituting the numbers, yields a quadratic equation in t,

0 – 50 = 30 sin45 t – 9.8t2

2

or

0 = – 9.8t2 + 30 sin45 t + 50

2

with solution,

t = -21.21 +/- √ (-21.21)2 + 4. 4.9. 50

-9.8

taking the positive root yields,

t = 6.02 s

The second part of the solution has to do with finding the final velocity. The only way to do this is to find the x and y components of the final velocity and add them together using vector addition. The y velocity is found using

vy = voy - gt

which, upon substituting the numbers, yields

vy = 30 sin 45 – 9.8 . 6.02

vy = -37.78 m/s

The minus sign indicates that the canon-ball is travelling downwards.

The x component of the velocity is unchanged, so the vector sum is

21.21m/s

θ

37.78m/s

or

√ (21.21)2 + (37.78)2 = 43.32 m/s

and,

the angle θ = Tan-1 37.78 = 60.68o

21.21

Circular Motion

The next topic in the textbook is circular motion, but we do not have enough time to get into circular motion this week, besides, we encounter circular motion again in the next chapter. So, I am going to postpone talking about circular motion until next week.

Relative Velocity

I would like to spend a few minutes talking about relative velocity because it is careful consideration of this topic that led Einstein to his Special Theory of Relativity. Lets start with a simple train example. Imagine someone is walking along the corridors of a moving train. To someone standing still on the platform, the person on the train appears to be moving by with the velocity of the train plus the velocity with which the person on the train is walking. The velocities just add.

Consider the vectors, the velocity of the person relative to the train is

p t

and the velocity of the train relative to the fixed ground is

t g

So that when you add them together, you get the velocity of the person relative to the ground

p g

Notice how you can label the tips and the tails of the arrows to help you connect them together in the right order. Now, the dilemma that Einstein spotted was that if the first vector is actually a light beam, then, relative to an observer standing still on the platform, the light beam appears to be travelling faster than the speed of light by an amount equal to the speed of the train. The problem with this scenario is that physicists have determined the rather curious result that the speed of light is actually a constant. Another way of saying this is that the speed of light is independent of the velocity of the source emitting the light. So, to get around the problem of the apparently faster than light velocity, encountered for a light beam on a moving train, Einstein invented the Special Theory of Relativity.

I do not have the time to get into the Special Theory of Relativity here other than to say that the relativistic effects do not become noticeable until the train moves with a velocity that is greater than 10% of the speed of light. For velocities less than that we can simply add the velocities using vector algebra as shown above.

Lets try another example where a plane is flying north through the air with some airspeed, and the air is moving east with respect to the ground, as would be encountered on a windy day. The vector diagram for this situation would look like this

w g

p

Note that the vector representing the planes velocity is labelled “pw” since the plane is moving through the air, ie. relative to the wind. The airspeed indicator in the cockpit of the plane records the planes airspeed not the ground speed. Whereas the wind speed is measured relative to the ground using a device called an anemometer. The planes ground speed is the resultant of the two vectors, in the direction indicated by matching up the labels on the tips and tails of the arrows correctly.

One final note, in the homework problems, the angles of the vector triangle will not necessarily be right angles which means that you will have to use the sine and cosine rules to solve for the resultant vector.

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