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)

• **Home**work 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!

•**The**n, 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. **The**refore a complete set of equations have

been created that will help solve these problems. **The**se are

known as ballistic formulas. **The**y 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!!

**The**y 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