Medical Physics, Lecture-1 [Compatibility Mode]

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Physics for Medical Sciences

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Dr. Mohamed FadhaliAssist. Professor of Photonics And Laser Physics

For Dentistry

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Physics: The most basic of all sciences!

• Physics = The study of the behavior of and the structure of matter and energy and of the interaction between matter and energy.

• This course will include the following topics and focusing on their medical applications:– Motion (MECHANICS) & Vectors– Force & Energy– Fluids– Heat – Sound & Ultrasound applications)– Optics & light and Laser Therapy applications– Medical Imaging and Instrumentations

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Physics Principles are used in many practical applications, including construction.

Communication between Architects & Engineers is essential if disaster is to be avoided.

Units, Standards, SI System

• All measured physical quantities have units.• Units are very important in physics!!• In this course we will use (almost) exclusively the SI

system of units. SI = “Systéme International”(French)More commonly called the “MKS system” (meter-kilogram-second) or more simply, “the metric system”

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SI or MKS System• Defined in terms of standards for length, mass, and time.

• Length unit: Meter (m) (kilometer = km = 1000 m)

– Standard meter. Newest definition in terms of speed of light ≡Length of path traveled by light in vacuum in (1/299,792,458) of a second!

• Time unit: Second (s)– Standard second. Newest definition ≡ time required for

9,192,631,770 oscillations of radiation emitted by cesium atoms!

• Mass unit: Kilogram (kg)– Standard kilogram ≡ Mass of a specific platinum-iridium alloy

cylinder kept at Intl Bureau of Weights & Measures in France

January 3, 2012 Physics 114A - Lecture 1 8/21

The SI Time Unit: second (s)The SI Time Unit: second (s)

The second was originally defined as (1/60)(1/60)(1/24) of a mean solar day. Currently, 1 second is defined as 9,192,631,770 oscillations of the radio waves absorbed by a vapor of cesium-133 atoms. This is a definition that can be used and checked in any laboratory to great precision.

13th Century Water Clock Cesium Fountain Clock

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The SI Length Unit: meter (m)The SI Length Unit: meter (m)The meter was originally defined as

1/10,000,000 of the distance from the Earth’s equator to its North pole on the line of longitude that passes through Paris. For some time, it was defined as the distance between two scratches on a particular platinum-iridium bar located in Paris.

Currently, 1 meter is defined as the distance traveled by light in 1/299,792,458 of a second


The kilogram was originally defined as the mass of 1 liter of water at 4oC.

Currently, 1 kilogram is the mass of the international standard kilogram, a polished platinum-iridium cylinder stored in Sèveres, France. (It is currently the only SI unit defined by a manufactured object.)

Question: In a “telephone” conversation, could you accurately describe to a member of a alien civilization how big a kilogram was?

Answer: More or less. Avagadro’s number of carbon-12 atoms (6.02214199… x 1023) has a mass of exactly

12.00000000000… kg.

The SI Mass Unit: kilogram (kg)

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Prefixes


Dimensions and Units

Any valid physical equation must be dimensionally consistent – each side must have the same dimensions.

From the Table:

Distance = velocity × time

Velocity = acceleration × time

Energy = mass × (velocity)2

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January 3, 2012 13/21

Example: The period P (T) of a swinging pendulum depends only on the length of the pendulum d (L) and the acceleration of gravity g (L/T2). Which of the following formulas for P could be correct ?

P dg

= 2πP dg

= 2π(a) (b) (c)P = 2π (dg)2

L LT

LT

T⋅

= ≠2

2 4

4

LL

T

T T2

2= ≠

( )P dg= 2 2π(a) (b) (c)P dg

= 2π

Try equation (a). Try equation (b). Try equation (c).

TT

TLL 2

2

==

P dg

= 2π

Displacement & Distance • Distance traveled by an object

≠ displacement of the object!

• Displacement = change in position of object.• Displacement is a vector (magnitude & direction). Distance is a scalar

(magnitude).• Figure: distance = 100 m, displacement = 40 m East

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Displacement

x1 = 10 m, x2 = 30 mDisplacement ≡ ∆x = x2 - x1 = 20 m

The arrow represents the displacement (in meters).

t1↓

t2↓

← times

x1 = 30 m, x2 = 10 mDisplacement ≡ ∆x = x2 - x1

= - 20 mDisplacement is a VECTOR

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• Distance vs. Displacement

Vectors and Scalars • Many quantities in physics, like

displacement, have a magnitude and a direction. Such quantities are called VECTORS.– Other quantities which are vectors: velocity,

acceleration, force, momentum, ...• Many quantities in physics, like distance,

have a magnitude only. Such quantities are called SCALARS.– Other quantities which are scalars: speed,

temperature, mass, volume, ...

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Average Velocity

Average Speed ≡ (Distance traveled)/(Time taken)

Average Velocity ≡ (Displacement)/(Time taken)

• Velocity: Both magnitude & direction describing how fast an object is moving. A VECTOR. (Similar to displacement).

• Speed: Magnitude only describing how fast an object is moving. A SCALAR. (Similar to distance).

• Units: distance/time = m/s

ûIn general:

∆x = x2 - x1 = displacement∆t = t2 - t1 = elapsed timeAverage Velocity:

= (x2 - x1)/(t2 - t1)

← timest2↓

t1↓

Bar denotes average

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smsm

txvaverage

130.3

40

+=

+=

∆∆

=rr

Velocity can be determined from a position-time graph

Average velocity equals the slope of the line joining the initial and final positions

Instantaneous Velocity≡ velocity at any instant of time

≡ average velocity over an infinitesimally short time

ûMathematically, instantaneous velocity:≡ ratio considered as a whole for smaller & smaller ∆t.

As you should know, mathematicians call this a derivative.

⇒ Instantaneous velocityv ≡ time derivative of displacement x

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Acceleration ûVelocity can change with time. An object with

velocity that is changing with time is said to be accelerating.ûDefinition: Average acceleration = ratio of change in

velocity to elapsed time.a ≡ = (v2 - v1)/(t2 - t1)

¡ Acceleration is a vector.ûInstantaneous acceleration

ûUnits: velocity/time = distance/(time)2 = m/s2

• Average Acceleration

• Instantaneous Acceleration

• Units:

if

ifave tt

vvtva

−−

=∆∆

=

tva

∆∆

= lim∆tà0

m/ss

ms2

à =

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Example : Average Acceleration

A car accelerates along a straight road from rest to 90 km/h in 5.0 s. Find the magnitude of its average acceleration. Note: 90 km/h = 25 m/s

a = = (25 m/s – 0 m/s)/5 s = 5 m/s2

Example : Car Slowing DownA car moves to the right ona straight highway (positive x-axis). The driver puts on the brakes. If the initial velocity (when the driver hits the brakes) is v1 = 15.0 m/s.It takes 5.0 s to slow down to v2 = 5.0 m/s. Calculate the car’s average acceleration.

a = = (v2 – v1)/(t2 – t1) = (5 m/s – 15 m/s)/(5s – 0s)a = - 2.0 m/s2

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Dr. M. Fadhali

Motion with Constant AccelerationThe average velocity of an object during a time interval t is

The acceleration, assumed constant, is

In addition, as the velocity is increasing at a constant rate, we know that

Combining these last three equations, we find:

Dr. M. Fadhali

We can also combine these equations so as to eliminate t:

We now have all the equations we need to solve constant-accelerationproblems.

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Since a is constant, we can integrate this using the above rule to find:

• Similarly, since we can integrate again to get:

v = a dt = a dt = at + v0∫∫v =

dxdt

x = v dt = (at + v0 )dt =12∫∫ at 2 + v0t + x0

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Eliminating t:

x = x0 + v0t +12

at 2l Solving for t:

x = x0 + v0v − v0

a

+12

a v − v0

a

2

v2 − v02 = 2a(x − x0 )

• Simplifying:

Dr. Mohamed Al- Fadhali

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Dr. M. Fadhali

Ex. 1 At highway speeds, a particular automobile is capable of an acceleration of about 1.6 m/s2. At this rate, how long does it take to accelerate from 80 km/h to 110 km/h?

.

The time can be found from the average acceleration, va

t∆

=∆

( )2 2

1m s30 km h3.6 km h110 km h 80 km h

5.208s 5 s1.6 m s 1.6 m s

vt

a∆ −

∆ = = = = ≈

Ex.2 A world-class sprinter can burst out of the blocks to essentially top speed (of about 11.5 m/s) in the first 15.0 m of the race. What is the average acceleration of this sprinter, and how long does it take her to reach that speed?

The sprinter starts from rest. The average acceleration is found from

( )( )

( )( )

22 22 2 2 20

0 00

11.5 m s 02 4.408 m s 4.41m s

2 2 15.0 mv v

v v a x x ax x

−−= + − → = = = ≈

−

the elapsed time is found by solving

00 2

11.5 m s 0 2.61 s

4.408 m sv v

v v at ta− −

= + → = = =

Ex.3 A car slows down uniformly from a speed of 21.0 m/s to rest in 6.00 s. How far did it travel in that time?

The words “slowing down uniformly” implies that the car has a constant acceleration. The distance of travel is found form

( )00

21.0m s 0m s6.00 sec 63.0 m

2 2v v

x x t+ +

− = = =

2

20

0

0

0

0

vvtvxx

vvvbut

tvxxtvxx

+==−

+=

=−+=

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An electron in a cathode-ray tube accelerates from rest with a constant acceleration of 5.33 x 1012 m/s2 for 0.150 µs, then drifts with a constant velocity for 0.200 µs, then slows to a stop with a negative acceleration of −2.67 x 1013 m/s2. (Note: 1 µs = 10-6 s)How far does the electron travel?

1. Draw the electron (as a dot)in its various positions xi.

2. Calculate the displacement ∆xi and velocity vi for each part of the path:1 12 -6 12 2 -6 2

1 0 2 2(0 m/s)(0.150 10 s)+ (5.33 10 m/s )(0.150 10 s) 0.06 mxx v t a t∆ = + = × × × =12 2 -6 5

15 -6

2 1 2

(5.33 10 m/s )(0.150 10 s) 8.0 10 m/s(8.0 10 m/s)(0.150 10 s) 0.16 m

xv a tx v t= = × × = ×

∆ = = × × =2 2 2 5 2

2 2 00 3 13 2

(0 m/s) (8.0 10 m/s)2 ; 0.012 m2 2( 2.67 10 m/s )

x xx x x

x

v vv v a x xa− − ×

= + ∆ ∆ = = =− ×

1 1 1 (0.06 m) (0.16 m) (0.012 m) 0.232 m =23.2 cmx x x x= ∆ + ∆ + ∆ = + + =

Example: A Traveling Electron

Galileo Free Falling

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Dr. M. Fadhali

Free Falling of Objects

Near the surface of the Earth, all objects experience approximately the same acceleration due to gravity.

This is one of the most common examples of motion with constant acceleration.


Example: Speed of a Lava BombA volcano shoots out blobs of molten lava (lava bombs) from its

summit. A geologist observing the eruption uses a stopwatch to time the flight of a particular lava bomb that is projected

straight upward. If the time for it to rise

and fall back to its launch height is 4.75 s,

what is its initial speed and how high did it go?

(Use g = 9.81 m/s2.)1 12

0 0 02 20 ( )x x v t gt x t v gt= + − ⇒ ∆ = = −1

0 2Either 0 or 0t v gt= − =1 1 2

0 2 2 (9.81 m/s )(4.75 s) 23.3 m/sv gt= = =2 2

0At maximum height, 0 2v v g x= = − ∆2 20

2(23.3 m/s) 27.7 m

2 2(9.81 m/s )vxg

∆ = = =

Medical Physics, Lecture-1 [Compatibility Mode]

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