py1dr1 physical properties and applications unit (metres per second) m/s or ms-1 velocity velocity...
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Recommended Text:
"Introduction to Biological Physics for the Health
and Life Sciences.“
Authors: Kirsten Franklin et al
Published by Wiley, 2010.
Junior Freshman
Course: PY1H01
PHYSICS FOR HEALTH SCIENCES
(Dental Science)
Physics –fundamental science
Knowledge of it is required in many fields:
chemistry, medicine, biology, dentistry etc
For example, the length of a football
pitch is 100m.
To say it is 100 has no meaning.
100 inches, 100m or 100 apples?
Dimensions and Units
Measuring a physical property
Any physical quantity has two parts,
a number and a unit.
Example, length and time
physical properties are called dimensions.
They denote the physical nature of a quantity
Dimensions Units
Time seconds, hours, etc
length metres, feet, miles ,etc
(SI) also called the Metric System,
Units
Mass: kilogram
Length: metre
Time: second
Fundamental Physical
Quantities. (Dimensions) (SI units)
Almost all measureable quantities can be
expressed as a combination of
dimensions: mass, length and time.
In 1960, international committee agreed on
a standard system of units called
Systéme International (SI)
One kg is defined as the mass of a certain
piece of metal kept at IBWM in France.
Some other fundamental units:
ampere (current), kelvin (temperature) etc
Measurement Standards
Definition of SI units:
When we measure a quantity we compare it
with some reference.
One second defined as exactly
9,192,631,700 times the period of
oscillation of radiation emitted by
a cesium atom. Atomic clock
Time
One metre defined as the distance light
travels in a vacuum in
Length
Mass
1
299,792,458s
Comprehensive, easy (shorter) system for
very large or very small numbers using the
powers of 10.
3.470.00001= 3.4710-5
Scientific Notation
Distance from the earth to the sun is
150,000,000 kilometres = 1.50x108 km.
To write 51,000,000 in scientific notation we note
51,000,000 = 5.110,000,000 = 5.1107
or
127,000 = 1.27100,000 = 1.27105
or
0.0000347 =
SI notation, prefixes & Abbreviations
Multiplying
factor
SI prefix
and
abbreviation
Scientific
notation
1 000 000 000 000 tera (T) 1012
1 000 000 000 giga (G) 109
1 000 000 mega (M) 106
1 000 kilo (k) 103
0.001 milli (m) 10-3
0.000 001 micro (µ) 10-6
0.000 000 001 nano (n) 10-9
0.000 000 000 001 pico (p) 10-12
Example:
1 kilometer = 1km = 1000m = 103 m
1 nanometer =1nm = 0.000 000 001m =10-9m
1milligram = 1 mg = 0.001gram = 10-3 gram
1microsecond = 1ms = 0.000 001s = 10-6 s
Equations, Unit Consistency,
Conversions
Numbers and units must equate
Example: speed = distance / time.
In equation form s = d / t
If a car travels 100 m in 20s, its speed is 5m/s
That is 100 m / 20 s = (100/20)(m/s) = 5 m/s
Units divide and multiply just like numbers
Obviously, numbers on each side of the
equation must equate. 2 = 5 is wrong
Units must also equate. 5 apples ≠ 5 oranges.
In physical equations
Each physical quantity has 2 parts,
a number and a unit.
11 100036 / 36 36 10 / 10
1 3600
km mkm h m s ms
hr s
How to change units?
Example
1km = 1000m
1h = 60min = 6060s = 3600s
Exercise: Convert 1m/s to km/h?
hrkmhrkmh
msm /6.3/
1000
36001
1
136001/1
Units and change of units
Convert km/h m/s
Convert m/s km/h
Mechanics
Objectives
to link, time, displacement, distance,
velocity, speed and acceleration
An athlete can run at a steady
speed of 36km/h and can stop in
2.5s. What is the average
acceleration of the athlete while
stopping?
Study of Motion without regard to its cause is
called kinematics.
The relationship of motion to the forces which
cause it is called dynamics.
In this section we will consider motion in a
straight line.
Motion is concerned with the displacement of
an object from one position in space and time
to another.
Displacement (Dx) of an object is defined as
its change in position and is given by
2 1x x xD where x1 is its initial position and x2 is its
final position. (Greek letter delta (D) is used to
denote a change in any physical quantity)
Displacement
Dx can be positive or negative
positive, in the positive x direction
negative, in the negative x direction
1km
3km
4km
A person walks 4km east and then 3 km west.
What is her displacement ?
Displacement
Displacement has both magnitude and
Direction: it is a vector quantity
Answer: Displacement is 1km east.
Even though the distance travelled is 7km,
displacement is 1km east.
Distance and displacement are different.
Distance has magnitude only but no given
direction and is called a scaler.
Quantities that can be described by a single
number (with unit) are called scalars, while
quantities also needing directional information
are called vectors.
Vectors and Scalers
Directional information is important, for example;
Scalar Quantities (magnitude but no direction)
e.g. mass, temperature, time etc. Single number and unit completely
specifies each
Vector Quantities (both magnitude and direction)
e.g. displacement, velocity, acceleration.
magnitude, direction and unit required
Orthodontics: teeth must not only be moved
but moved in a particular direction
The velocity is the change in displacement
(Ds) divided by the corresponding change in
time (Dt):
Velocity can be positive or negative:
A B
-80km/h
80km/h
At the end, your average velocity is zero!
Displacement Ds is zero
Velocity is a vector quantity: it has a
magnitude and a direction
s
t
DD
Velocity
SI unit (metres per second)
m/s or ms-1
Velocity
Velocity is a vector quantity
It has magnitude and direction
Example: 30km/hour west.
Velocity and speed are different
Speed is a scaler quantity
Example: 30km/hour.
No direction specified.
Acceleration can be positive or negative:
Acceleration is a vector quantity (magnitude
and direction)
Accelerating from 0m/s to 20m/s in 10s:
Decelerating from 20m/s to 0m/s in 10s:
Acceleration
1 1220 0
210
ms msa ms
s
1 120 20
210
ms msa ms
s
Examples:
Acceleration is the change in velocity divided
by the corresponding change in time
≡ ms-2 SI units
0v va
t t
D D
v0 = initial velocity
v = final velocity
t = time taken
Final velocity (v) is initial velocity (vo) plus
change due to acceleration*:
A runner accelerates at a rate of 8.0 ms-2 in the
first 0.75 s of a race. What is the magnitude of
her velocity at the end of this period?
Example:
0v va
t
ov v at
Linking velocity with acceleration
and time
1 2 10 8 0.75 6v ms ms s ms
Distance is average velocity multiplied by time:
A runner has an acceleration of 8.0 m.sec-2 in
the first 0.75 sec of a race. How far has the
runner traveled in the period?
Example:
2 210 8 sec (0.75sec) 2.25
2s m m
sv
t
0
2
v vs vt t
Linking distance with
velocity, acceleration and time
Average velocity = displacement/time
ov v at
0
2
0
( )
2
1
2
ov v ats t
s v t at
Distance is average
velocity multiplied by
time.
A car accelerates from rest at 16 m.s-2 over a
distance of 400 m; what is the final velocity?
Example:
2 2 2 2
1
0 2 16 400 12800
113.3
v ms m m s
v ms
0
2
v vs t
0v va
t
Linking velocity with
acceleration and distance
Acceleration is change
in velocity divided by
time. 2 2
0 0 0
2 2
v v v v v vas t
t
sv
t
2 2
0 2v v as
Problem solving: Depending on information
given , choose one or more of the 4 equations
0v v at
2 2
0 2v v as
2
0
1
2s v t at
Summary: 4 useful equations
0
2
v vs vt t
Human nerve impulses are propagated at
a rate of 102m/s. Estimate the time it takes for
a nerve impulse, generated when your foot
touches a hot object, to travel to your brain.
dv
t
2 1
1.80.018
10
d mt s
v ms
If your dentist touches a nerve in your tooth the
nerve impulse generated travels to your brain
in 1ms. Estimate the speed of the nerve impulse.
Exercise:
Exercise:
18 ms
dv
t
2 1
3
0.110
1 10
d mv ms
t s
In orthodontic treatment a tooth when subjected
to a certain force moves a distance of 2 mm in
a period of 0.75 years. Estimate the average
speed (in ms-1) of the tooth.
dv
t
6
0.75 0.75 365 24 3600sec
23.65 10
t yrs
t s
Exercise:
3
6
2 10
23.65 10
mv
s
12 184.57 10v ms
Time in seconds
A person moves as shown in the velocity versus
time graph. Find the distance travelled by the
person during each of the segments A, B, C.
dv
t
d vt
A 15 0
20 502
d ms s metres
112.5 10 125d ms s metres B
C
2
final initial
average
v vv
120 520 250
2d ms s metres
0 10 20 30 40 50
0
5
10
15
20
Velo
cit
y (m
s-1)
Time (seconds)
A
B C
Exercise:
0 10 20 30 40 50
0
5
10
15
20
Ve
loc
ity
(ms-1
)
Time (seconds)
B
A
B
C
A car moves as shown in the velocity versus
time graph. Find the acceleration of the car
during each of the segments A, B, C.
1
220 0
210
msa ms
s
12(20 20)
0.020
msa ms
s
B
C 1 1
2(10 20) 100.5
20 20
ms msa ms
s s
A
Exercise:
0v va
t
An athlete can run at a steady speed of
36km/h (!!) and can stop in 2.5s. What is
the average acceleration of the athlete
while stopping?
Exercise:
0v v at
1
12
0 10 2.5
104
2.5
ms a s
msa ms
s
First convert 36km/h to ms-1
11 100036 / 36 36 10 / 10
1 3600
km mkm h m s ms
hr s
a is negative since athlete is decelerating