SPH4UI Lecture 1 Notes - The Burns Home Page

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SPH4UI Lecture 1 Notes - The Burns Home Page

SPH4UI: Lecture 1

Course Info & Advice

• Course has several components:

Lecture: (me talking, demos and you asking questions)

• Discussion sections (tutorials, problem solving, quizzes)

Homework Web based

• Labs: (group exploration of physical phenomena)

• What happens if you miss a lab or class test

• Read notes from online

• What if you are excused?? (What you need to do.)

• That topic of your exam will be evaluated as your test

“Kinematics”

The first few weeks of the course should be review, hence the pace is

fast. It is important for you to keep up!

Then, watch out….

Fundamental Units

• How we measure things!

• All things in classical mechanics can be expressed in terms of the

fundamental units:

• Length: L

• Mass : M

• Time : T

• For example:

• Speed has units of L / T (e.g. miles per hour).

• Force has units of ML / T 2 etc... (as you will learn).

Page 1


Units...

• SI (Système International) Units:

• mks: L = meters (m), M = kilograms (kg), T = seconds (s)

• cgs: L = centimeters (cm), M = grams (gm), T = seconds (s)

• British Units:

• Inches, feet, miles, pounds, slugs...

• We will use mostly SI units, but you may run across some

problems using British units. You should know where to look to

convert back & forth.

Ever heard of Google

Converting between different

systems of units

• Useful Conversion factors:

• 1 inch = 2.54 cm

• 1 m = 3.28 ft

• 1 mile = 5280 ft

• 1 mile = 1.61 km

• Example: convert miles per hour to meters per second:

mi mi ft 1 m 1 hr m

1 1 5280 0.447

hr hr mi 3.28 ft 3600 s s

Page 2


Dimensional Analysis

• This is a very important tool to check your work

• It’s also very easy!

• Example:

Doing a problem you get the answer distance

d = vt 2 (velocity x time 2 )

Units on left side = L

Units on right side = L / T x T 2 = L x T

• Left units and right units don’t match, so answer must be

wrong!!

Dimensional Analysis

The period P of a swinging pendulum depends only on

the length of the pendulum d and the acceleration of

gravity g.

• Which of the following formulas for P could be

correct ?

(a) P = 2 (dg) 2

P 2 d

d

(b) (c) P 2

g

g

Given: d has units of length (L) and g has units of (L / T 2 ).

Solution

Solution

• Realize that the left hand side P has units of

time (T )

• Try the first equation

• Realize that the left hand side P has units of

time (T )

• Try the second equation

(a)

2 4

L L

L T

2

4

T T

Not Right!

(b)

L

L

2

T

2

T

T

Not Right!

(a) P 2

dg

2 (b) P 2

d (c)

g

P 2

d

g

(a) P 2

dg

2 (b) P 2

d (c)

g

P 2

d

g

Page 3


Solution

• Realize that the left hand side P has units of

time (T )

• Try the first equation

(c)

L

L

2

T


T

2

T

Dude, this is it

(a) P 2

dg

2 (b) P 2

d (c)

g

d

P 2

g

Vectors

• A vector is a quantity that involves both

magnitude and direction.

• 55 km/h [N35E]

• A downward force of 3 Newtons

• A scalar is a quantity that does not involve

direction.

• 55 km/h

• 18 cm long

Vector Notation

• Vectors are often identified with arrows

in graphics and labeled as follows:


Displacement

Displacement is an object’s change in position.

Distance is the total length of space traversed

by an object.

We label a vector with a variable.

This variable is identified as a vector either by an arrow

above itself :

A

Or

By the variable being BOLD: A

1m

3m

Displacement:

Distance:

6.7m

5m

6m 2 3m

2

6.7m

5m 3m 1m 9m

Start

Finish = 500 m

Displacement = 0 m

Distance = 500

m

Page 4


E

R

D

Vector Addition

A

B

C

D

R

E

B

A

C

B A

D

R

E

A + B + C + D + E = Distance

R = Resultant = Displacement

C

R A B

Rectangular Components

2 2

-x

A opp

sin


R hyp

B adj

cos


R hyp

A opp

tan


B adj

Quadrant II

Rsin


Quadrant III

y

A

-y

Quadrant I

R

B

Rcos


Quadrant IV

x

Vectors...

The components (in a particular coordinate system) of r,

the position vector, are its (x,y,z) coordinates in that

coordinate system

• r = (r x ,r y ,r z ) = (x,y,z)

• Consider this in 2-D (since it’s easier to draw):

• r x = x = r cos where r = |r |

• r y = y = r sin

y (x,y)

r arctan( y / x )


x

Vectors...

The magnitude (length) of r is found using the

Pythagorean theorem:

r

x

y

r r x y

2 2

The length of a vector clearly does not depend on its direction.

Page 5


• Vector A = (0,2,1)

• Vector B = (3,0,2)

• Vector C = (1,-4,2)

Vector Example

What is the resultant vector, D, from adding A+B+C?

(a) (3,5,-1) (b) (4,-2,5) (c) (5,-2,4)

Resultant of Two Forces

• force: action of one body on another;

characterized by its point of application,

magnitude, line of action, and sense.

• Experimental evidence shows that the

combined effect of two forces may be

represented by a single resultant force.

The resultant is equivalent to the

diagonal of a parallelogram which

contains the two forces in adjacent legs.

• Force is a vector quantity.

P

Q

P

-P

Vectors

• Vector: parameters possessing magnitude and direction

which add according to the parallelogram law. Examples:

displacements, velocities, accelerations.

• Scalar: parameters possessing magnitude but not

direction. Examples: mass, volume, temperature

• Vector classifications:

- Fixed or bound vectors have well defined points of

application that cannot be changed without affecting

an analysis.

- Free vectors may be freely moved in space without

changing their effect on an analysis.

- Sliding vectors may be applied anywhere along their

line of action without affecting an analysis.

• Equal vectors have the same magnitude and direction.

• Negative vector of a given vector has the same magnitude

and the opposite direction.

P

P

Q

Q

Q -Q

P-Q

Q

Addition of Vectors

P

P

• Trapezoid rule for vector addition

• Triangle rule for vector addition

• Law of cosines,

2 2 2

R P Q 2PQcos

B

R P Q

• Law of sines,

sin A sin B sinC


Q R A

• Vector addition is commutative,

P Q Q P

• Vector subtraction

P Q P Q



Page 6


P

Q

Addition of Vectors

S

• Addition of three or more vectors through

repeated application of the triangle rule

Resultant of Several Concurrent Forces

• Concurrent forces: set of forces

which all pass through the same

point.

P

P

Q

S

2P

-1.5P

The polygon rule for the addition of three or

more vectors.

• Vector addition is associative,

P Q S P Q S P Q S

• Multiplication of a vector by a scalar

increases its length by that factor (if scalar

is negative, the direction will also change.)

A set of concurrent forces applied

to a particle may be replaced by

a single resultant force which is

the vector sum of the applied

forces.

• Vector force components: two or

more force vectors which,

together, have the same effect as

a single force vector.

Sample Problem

Sample Problem Solution

Two forces act on a bolt

at A. Determine their

resultant.

• Graphical solution - construct a

parallelogram with sides in the

same direction as P and Q and

lengths in proportion.

Graphically evaluate the

resultant which is equivalent in

direction and proportional in

magnitude to the diagonal.

• Trigonometric solution - use the

triangle rule for vector addition

in conjunction with the law of

cosines and law of sines to find

the resultant.

Q


P

R 98 N 35

R

• Graphical solution - construct a

parallelogram with sides in the same

direction as P and Q and lengths in

proportion. Graphically evaluate the

resultant which is equivalent in

direction and proportional in

magnitude to the diagonal.

Page 7


Sample Problem Solution

• Trigonometric solution

From the Law of Cosines,

2 2 2

R P Q 2PQcos

B

2 2

40N 60N 240N60Ncos155

R 97.73N

From the Law of Sines,

sin A sin B


Q R

Q

sin A

sin B R

60N

sin155

97.73N

A 15.04

20 A

35.04

Motion in 1 dimension

• In 1-D, we usually write position as x(t).

• Since it’s in 1-D, all we need to indicate direction is + or .

Displacement in a time t = t 2 - t 1 is x = x(t 2 ) - x(t 1 ) = x 2 - x 1

x

x

x

x

2

1

t 1 t 2

t

some particle’s trajectory

in 1-D

t

1-D kinematics

1-D kinematics...

• Velocity v is the “rate of change of position”

• Average velocity v av in the time t = t 2 - t 1 is:

x

x

x

x

2

1

x( t2) x( t1)

x

vav



t t t

t 1 t 2

t

2 1

trajectory

V av = slope of line connecting x 1 and x 2 .

t

• Consider limit t 1 t 2

• Instantaneous velocity v is defined as:

dx()

t

vt ()

dt

so v(t 2 ) = slope of line tangent to path at t 2 .

x

x

x

x

2

1

t 1 t 2

t

t

Page 8


1-D kinematics...

• Acceleration a is the “rate of change of velocity”

• Average acceleration a av in the time t = t 2 - t 1 is:

v( t2) v( t1)

v

aav


t t t

2 1

• And instantaneous acceleration a is defined as:

dv t

at ()

dt

2

( ) d x( t)

dt

2

dx()

t

using vt ()

dt

Calculus way of saying t

gets very very small

Recap

• If the position x is known as a function of time, then we can

find both velocity v and acceleration a as a function of

time!

x

dx

v

a

x( t )

dt

dv

dt

2

d x


2

dt

Calculus (don’t worry you will

understand this in next year.)

x

v

a

t

t

t

More 1-D kinematics

Recap

• We saw that v = dx / dt

• In “calculus” language we would write dx = v dt, which we

can integrate to obtain:

x( t ) x( t ) v( t)

dt

2 1

• Graphically, this is adding up lots of small rectangles:

v(t)

+ +...+

= displacement

t

t2


t1

• So for constant acceleration we

find:

1 2

x x0 v0t

at

2

v v0

v

at

a const

a

x

t

t

t

Page 9


Motion in One Dimension

Question

Solution

• When throwing a ball straight up, which of the following is

true about its velocity v and its acceleration a at the

highest point in its path?

(a) Both v = 0 and a = 0.

(b) v 0, but a = 0.

(c) v = 0, but a 0.

y

• Going up the ball has positive velocity, while coming down

it has negative velocity. At the top the velocity is

momentarily zero.

x

• Since the velocity is

continually changing there must

be some acceleration.

‣In fact the acceleration is caused

by gravity (g = 9.81 m/s 2 ).

‣(more on gravity in a few lectures)

The answer is (c) v = 0, but a 0.

v

a

t

t

t

Recap:

• For constant acceleration:

This is just

1 2

x x0 v0t

at

v v0

2

at

a

for constant

• From which we know:

acceleration!

2 2

v v0

2a(x x0

)

1

vav

(v0

v)

2

Recap:

• For constant acceleration:

1

x x0 v0t at

2

2

Page 10


• Read Carefully!

Problem Solving Tips:

• Before you start work on a problem, read the problem

statement thoroughly. Make sure you understand what

information is given, what is asked for, and the meaning of

all the terms used in stating the problem.

• Using what you are given, set up the algebra for the problem and

solve for your answer algebraically

• Invent symbols for quantities you know as needed

• Don’t plug in numbers until the end

• Watch your units !

• Always check the units of your answer, and carry the units

along with your formula during the calculation.

• Understand the limits !

• Many equations we use are special cases of more general

laws. Understanding how they are derived will help you

recognize their limitations (for example, constant

acceleration).

1-D Free-Fall

• This is a nice example of constant acceleration (gravity):

• In this case, acceleration is caused by the force of gravity:

• Usually pick y-axis “upward” y

• Acceleration of gravity is “down”:

a

y

g

v

y

= v0y

- gt

1

y y0 v0y

t g t

2

y

a y = g

2

v

a

t

t

t

Gravity facts:

• g does not depend on the nature of the material!

• Galileo (1564-1642) figured this out without fancy clocks &

rulers!

• On the surface of the earth, gravity acts to give a constant

acceleration

• demo - feather & penny in vacuum

• Nominally, g = 9.81 m/s 2

• At the equator g = 9.78 m/s 2

• At the North pole g = 9.83 m/s 2

• More on gravity in a few lectures!

Penny

& feather

Gravity facts:

• Actually, gravity is a “fundamental force”.

• Other fundamental forces: electric force, strong and weak

forces

• It’s a force between two objects, like me and the earth. or earth

and moon, or sun and Neptune, etc

• Gravitational Force is proportional to product of masses:

• F(1 acting on 2) proportional to M 1 times M 2

• F(2 acting on 1) proportional to M 1 times M 2 too!

• Proportional to 1/r 2

• r is the separation of the 2 masses

• For gravity on surface of earth, r = radius of earth

• Example of Gauss’s Law (more on this later)

At the surface of earth gravitational force attracts “m” toward the

center of the earth, is approximately constant and equal to mg.

The number g=9.81 m/s 2 contains the effect of M earth and r earth .

Page 11


Question:

Solution:

The pilot of a hovering helicopter

drops a lead brick from a height

of 1000 m. How long does it

take to reach the ground and

how fast is it moving when it gets

there? (neglect air resistance)

1000 m

• First choose coordinate system.

• Origin and y-direction.

• Next write down position equation:

1

y y0 v0yt gt

2

2

1000 m

• Realize that v 0y = 0.

1

y y0

gt

2

2

y

y = 0

Solution:

1D Free Fall

• Solve for time t when y = 0 given

that y 0 = 1000 m.

1 2

y y0

- gt 2

t

2y

21000m

14.3s

g 9.81m s

0


2

y 0 = 1000 m

• Alice and Bob are standing at the top of a cliff of

height H. Both throw a ball with initial speed v 0 , Alice

straight down and Bob straight up. The speed of the

balls when they hit the ground are v A and v B

respectively. Which of the following is true:

• Recall:

• Solve for v y :

vy

v -v 2 a( y - y )

2 2

y 0y

0

2gy

0

140 m/

s

y = 0

y

(a) v A < v B (b) v A = v B (c) v A > v B

v 0

v A

Alice

v 0

v B

Bob

H

Page 12


1D Free fall

• Since the motion up and back down is symmetric, intuition

should tell you that v = v 0

• We can prove that your intuition is correct:

Bob

v 0

Equation: v 2 v 2

0

g H H

v = v 0 the speed at the bottom should

2( ) 0

H

This looks just like Bill threw the

ball down with speed v 0 , so

be the same as Alice’s ball.

y = 0

Does motion in one direction affect

motion in an orthogonal direction?

2 ball drop

• For example, does motion in the y-direction affect motion in

the x-direction?

• It depends….

• For simple forces, like gravitational and electric forces, NO

• For more complicated forces/situations, like magnetism, YES

• In any case, vectors are the mathematical objects that we

need to use to describe the motion

• Vectors have

• Magnitude

• Units (like meters, Newtons, Volts/meter,

meter/sec 2 …)

• Direction

Vectors:

• In 1 dimension, we could specify direction with a + or - sign.

For example, in the previous problem a y = -g etc.

• In 2 or 3 dimensions, we need more than a sign to specify the

direction of something:

• To illustrate this, consider the position vector r in 2 dimensions.

Example: Where is Waterloo?

‣ Choose origin at Toronto

‣ Choose coordinates of

distance (km), and

direction (N,S,E,W)

‣ In this case r is a vector that

points 120 km north.

Waterloo

Toronto

A vector is a quantity with a magnitude and a direction

r

2-D Kinematics

• Most 3-D problems can be reduced to 2-D problems when

acceleration is constant:

• Choose y axis to be along direction of acceleration

• Choose x axis to be along the “other” direction of

motion

• Example: Throwing a baseball (neglecting air resistance)

• Acceleration is constant (gravity)

• Choose y axis up: a y = -g

• Choose x axis along the ground in the direction of the

throw

Page 13


“x” and “y” components of motion are independent.

• A man on a train tosses a ball straight up in the air.

‣View this from two reference frames:

Reference frame

on the moving train.

Problem:

• David Eckstein clobbers a fastball toward center-field. The

ball is hit 1 m (y o ) above the plate, and its initial velocity is

36.5 m/s (v ) at an angle of 30 o () above horizontal. The

center-field wall is 113 m (D) from the plate and is 3 m (h)

high.

• What time does the ball reach the fence?

• Does David get a home run?

Reference frame

on the ground.

y 0

v


h

D

• Choose y axis up.

Problem...

Problem...

• Use geometry to figure out v 0x and v 0y :

g

y

v

v x = v 0x v y = v 0y - gt

y 0


x = v x t y = y 0 + v 0y t - 1 / 2 gt 2

• Choose x axis along the ground in the direction of the hit.

• Choose the origin (0,0) to be at the plate.

• Say that the ball is hit at t = 0, x(0) = x 0 = 0. y(0) = y 0 = 1m

• Equations of motion are:

Find v 0x = |v| cos .

and v 0y = |v| sin .

v 0y

v 0x

remember, we were told that = 30 deg

x

Page 14


Problem...

The time to reach the wall is: t = D / v x (easy!)

• We have an equation that tell us y(t) = y 0 + v 0y t + a t 2 / 2

• So, we’re done....now we just plug in the numbers: a = -g

• Find:

• v x = 36.5 cos(30) m/s = 31.6 m/s

• v y = 36.5 sin(30) m/s = 18.25 m/s

• t = (113 m) / (31.6 m/s) = 3.58 s

• y(t) = (1.0 m) + (18.25 m/s)(3.58 s)

- (0.5)(9.8 m/s 2 )(3.58 s) 2

= (1.0 + 65.3 - 62.8) m = 3.5 m

Motion in 2D

• Two footballs are thrown from the same point on a flat field.

Both are thrown at an angle of 30 o above the horizontal. Ball 2

has twice the initial speed of ball 1. If ball 1 is caught a

distance D 1 from the thrower, how far away from the thrower

D 2 will the receiver of ball 2 be when he catches it?

• Assume the receiver and QB are the same height

(a) D 2 = 2D 1 (b) D 2 = 4D 1 (c) D 2 = 8D 1

• Since the wall is 3 m high, Eckstein gets the homer!!

Thinking deeper: Can you figure out what angle gives the longest fly ball?

To keep things simple, assume y 0 = 0, and go from there…

Solution

Solution

The distance a ball will go is simply

x = (horizontal speed) x (time in air) = v 0x t

• To figure out “time in air”, consider the

1

equation for the height of the ball: y y0 v0y

t g t

2

• When the ball is caught, y = y 0

1

t v0

y

g t 0

2

two

solutions

1 2

v0

y

t g t 0

2

t

v

0

2 y

t 0

g

(time of catch)

(time of throw)

2

• So the time spent in the air is proportional to v 0y t:


v

0

2 y

g

• Since the angles are the same, both v 0y and v 0x for ball 2

are twice those of ball 1.

v 0,2

v 0,1 ball 2

ball 1

v 0y ,2

v 0y ,1

v 0x ,1 v 0x ,2

• Ball 2 is in the air twice as long as ball 1, but it also has twice

the horizontal speed, so it will go 4 times as far!!

Page 15


Projectile Motion

Motion in 2D Again

As you can see, it can become difficult to solve problems

that involve motion in both the x and y axis. Lucky for you,

people from all over the world have had the same

difficulties. Therefore a complete set of equations have

been created that will help solve these problems. These are

known as ballistic formulas. They assume launch height and

landing height are the same.

Range

v

2 i

sin 2


2v sin

dx


i


Distance: Travel Time: t

g

g

Maximum:

v sin 2

i


h

2g

Height

Time to top:

t


v i

sin


g

• Two footballs are thrown from the same point on a flat field.

Both are thrown at an angle of 30 o above the horizontal. Ball 2

has twice the initial speed of ball 1. If ball 1 is caught a

distance D 1 from the thrower, how far away from the thrower

D 2 will the receiver of ball 2 be when he catches it?

• Assume the receiver and QB are the same height

(a) D 2 = 2D 1 (b) D 2 = 4D 1 (c) D 2 = 8D 1

2

2vo

sin 2


dx


g

2

vi

sin 2


4

g

v

2 i

sin 2


dx


g

Example

A golfer hits a golf ball so that it leaves the club with an

initial speed v 0 =37.0 m/s at an initial angle of 0 =53.1 o .

a) Determine the position of the ball when t=2.00s.

b) Determine when the ball reaches the highest point

of its flight and find its height, h, at this point.

c) Determine its horizontal range, R.

a) v0 x

v0 cos 0

37.0 m / scos 53.1

22.2 m / s

v0 y

v0 sin 0

37.0 m / ssin 53.1

29.6 m/

s

The x-distance: x v0 xt

22.2 m/ s2.00s

44.4m

1 1

The y-distance: y v 2 2

2

0 yt gt 29.6 m / s 2.00s 9.80 m / s 2.00s 39.

6m

2

2

Example

A golfer hits a golf ball so that it leaves the club with an

initial speed v 0 =37.0 m/s at an initial angle of 0 =53.1 o .

a) Determine the position of the ball when t=2.00s.

b) Determine when the ball reaches the highest point

of its flight and find its height, h, at this point.

c) Determine its horizontal range, R.

v sin 2

i


h

v

2 i

sin 2


2g

dx


g

2

37.0 m/ ssin 53.1

2


37.0 m/ s sin 253.1

2

29.80 m/

s


2

9.80 m/

s

44.7m

134m

v i

sin


t


g

37.0 m/ ssin 53.1


3.02s

2

9.80 m/

s


Page 16


Shooting the Monkey

(tranquilizer gun)

Shooting the Monkey...

• Where does the zookeeper

aim if he wants to hit the monkey?

( He knows the monkey will

let go as soon as he shoots ! )

• If there were no gravity, simply aim

at the monkey

r = r 0

r =v 0 t

Shooting the Monkey...

• With gravity, still aim at the monkey!

r = r 0 - 1 / 2 g t 2

Recap:

Shooting the monkey...

x = v 0 t

y = - 1 / 2 g

t 2

r = v 0 t - 1 / 2 g t 2

Dart hits the

monkey!

• This may be easier to think about.

It’s exactly the same idea!!

They both have the same V y (t) in this case

x = x 0

y = - 1 / 2 g

t 2

Page 17


Feeding the Monkey

Kinematics Flash Review

Page 18

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