PHY 183 - Program Physics for Scientists and Engineers

Chapter 1 KINEMATICS 7/5Chapter 2 DYNAMICS 7/5 Chapter 3 WORK AND ENERGY 6/4 Chapter 4 ROTATIONAL MOTION 6/4 The first test 40% (2)Chapter 5 PERIODIC MOTION 5/3Chapter 6 WAVE MOTION 5/3 Chapter 7 FLUIDS AND THERMAL PHYSICS

5/3 Chapter 8 GAS LAWS AND KINETIC THEORY

5/3Chapter 9 LIQUID PHASE 6/4 The final examination 60%

1. Net force and Newton's first law2. Newton's second law3. Newton's third law4. Frictional forces5. Gravitation6. Circular motion7. Centripetal force8. Static equilibrium and reference frames

Part 4

Frictional Forces

Forces: Normal Force

Book at rest on table:What are forces on book?

• Weight is downward• System is “in equilibrium” (acceleration = 0 net force = 0)• Therefore, weight balanced by another force

W

FN

• FN = “normal force” = force exerted by surface on object• FN is always perpendicular to surface and outward• For this example FN = W

Learning Check

Book at rest on table titled

an angle of 300 to horizontal

How much is normal force ??

N NY N

0

W F 0 to 0Y : F F

3W.cos30 mg.

2

W

FN

300

Y

Forces: Kinetic Friction

• Kinetic Friction (a sliding Friction):A force, fk, between two surfaces that opposes relative motion. Magnitude: fk = kFN

* k = coefficient of kinetic frictiona property of the two surfaces

FN

W

Ffk

direction of motion

Learning Check

Book (50g) move with a= 2m/s2 on table titled an angle 300 to horizontal

How much is coefficient of kinetic friction ??

N K

0y N Y

0 0x K N

0

N

W F F a

3to 0Y(a 0) : F W mg.cos30 0,05.9,8. (N)

2

to 0Y(a a) : W.sin 30 F mg.sin 30 F a

mg.sin 30 a

F

W

FN

300

Y

Fk

X

Forces: Static Friction

W

FN

Ffs

• Static Friction:A force, fs, between two surfaces that prevents relative motion.

• fs ≤ fsmax= sFN force just before breakaway

s = coefficient of static friction a property of the two surfaces

Exercise

10N

FN

Ffs

• F increases from 30N to 50N (force just before breakaway)

• coefficient of static friction is ??• Fs changes from ……… to………. (N) • What is unit of coefficient of static

friction ??

Forces: Tension

• Tension: force exerted by a rope (or string)

• Magnitude: same everywhere in rope Not changed by pulleys

• Direction: same as direction of rope.

T

example: box hangs from a rope attached to ceiling

T

5N

y

Fy = may

T - W = may

T = W + may

In this case ay = 0

So T = W=5N

How much is rope tension ?

Part 5

Gravitation

Gravity is a very tiny force

• Force between two objects each 1 Kg at a distance of 1 meter is F = G M1 M2 /R2

• G = 6.67 x 10 -11 (Nm2/kg2)1 N is about the weight of one apple

(100g) on the earth• The reason the effects of gravity are so

large is that the masses of the earth, sun, stars, …. are so large -- and gravity extends so far in space

Gravity and Weight

Force on mass:

mg gm mR

GMF

2e

eg

Me

Remass on surfaceof Earth

m

g

m 10 x 6.38 R

m/s 9.81 g and kg 10 x 5.98 M using

g

GM R

6e

224e

e2e

Fg W = mg

Calculate the earth’s radius from the gravity of Earth

(note:M=5,98 .1024 kg; g=9,81 m/s2) How much weight of a toy of 30g mass ??

)N(294,0 0,03.9,8 gmFg

See movie

Part 6

Circular motion

What is CM?(Circular Motion)

• Motion in a circle with: Constant Radius R– Trajectory is circular form

R

vv

x

y

(x,y)

How can we describe CM?

• In general, one coordinate system is as good as any other:– Cartesian:

• (x,y) [position]• (vx ,vy) [velocity]

– Polar:• (R,) [position]• (vR ,) [velocity]

In CM:– R is constant (hence vR = 0).– angular (t) is function of time measured by RAD.– Polar coordinates are a natural way to describe CM!Polar coordinates are a natural way to describe CM!

RR

vv

x

y

(x,y)

Polar and Cartesian Coordinates

x = R cos y = R sin

2 3/2 2

-1

1

0

sincos

RR

x

y

(x,y)

X2 + y2 = R2

tgy / x

Polar Coordinates...

• In Cartesian coordinates, we say velocity:

• dx/dt = v and x = vt + x0

• In polar coordinates, angular velocity:

d/dt = . = t +

Has units of radians/second.

Displacement s = vt.

but s = R = Rt, so:

RR

vv

x

y

st

v = R

• A fighter pilot flying in a circular turn (with 500m in radius) follow the equation: = 5t+10t2 (rad). • It’s position at time t=2s is: a) 1432,3 rad b) 1423,3 degree c) 50 degree

• The angular velocity at time t=3s is:(a) 650 m(b) 65 rad/s(c) 105 rad/sWhat is velocity V ???

MCQ test

Period and Frequency

• Recall that 1 revolution = 2 radians– frequency (f) = revolutions / second (a)

– angular velocity () = radians / second (b) – By combining (a) and (b) = 2 f

Realize that:– period (T) = seconds / revolution– So T = 1 / f = 2/ = 2 / T = 2f

RR

vv

s

Example

• The Wings of motor spins with frequency f = 3000 rpm. The length of each wings is

L = 80cm ?

what is angular velocity ?

Total path of point A travel after 10 s ??

What is period ??

f

L

A

= 2 f = 6000 (rad/m)S=Lt = 2 fL t =1000 .0,8=800 (m)T=1/f=1/50(rps)=0.02 s

Polar Coordinates...

• In Cartesian coordinates, we say acceleration:• dv/dt = a and v = at + v0 • In polar coordinates, angular acceleration :• d/dt = and = t +

Has units of radians/second.For vector of angular velocity

Has direction of axis looking point Aturning to watch loop

RR

vv

x

y

st

dt

d

• A fighter pilot flying in a circular turn (with 500m in radius) follow the equation: =5t+10t2 (rad). The angular acceleration at time t=3s is:

(a) 650 rad/s2

(b) 20 rad/s2

(c) 105 rad/s2

MCQ test

What is UCM?(Uniform Circular Motion)

• Motion in a circle with: – Constant Radius R– Trajectory is circular form

Constant Speed v = |vv|

R

vv

x

y

(x,y)

Acceleration in UCM:

• Even though the speedspeed is constant, velocityvelocity is notnot constant since the direction is changing: must be some acceleration!– Consider average acceleration in time t

aav = vv / tvv2

tvv1

vv1vv2

vv

RR

Acceleration in UCM:

• This is called This is called Centripetal Acceleration.

• Now let’s calculate the magnitude:

vv2

vv1

vv1vv2

vv

RRRR

vv

RR

Similar triangles:

But R = vt for small t

So:vt

vR

2 v

vv tR

avR

2

Useful Equivalent

R

Ra

2

We know that and v = R

Substituting for v we find that:

avR

2

a = 2R

Centripetal Acceleration

• The Space Shuttle is in Low Earth Orbit (LEO) about 300 km above the surface. The period of the orbit is about 91 min. What is the acceleration of an astronaut in the Shuttle in the reference frame of the Earth? (The radius of the Earth is 6.4 x 106 m.)

(a) 0 m/s2

(b) 8.9 m/s2

(c) 9.8 m/s2

Example: Newton & the Moon

• What is the acceleration of the Moon due to its motion around the Earth?

• What we know (Newton knew this also):– T = 27.3 days = 2.36 x 106 s (period ~ 1 month)– R = 3.84 x 108 m (distance to moon)

– RE = 6.35 x 106 m (radius of earth)

R RE

Moon...

• Calculate angular velocity:

• So = 2.66 x 10-6 s-1.Now calculate the acceleration.

– a = 2R = 0.00272 m/s2 = 0.000278 g– direction of aa points at the center of

the Earth (-r r ).

1

27 3

1

864002 2 66 10 6

..

rot

dayx

day

sx

rad

rotx s-1

^

• First calculate the angular

frequency :

Exercise: Centripetal Acceleration of ES ??

Given: RO = RE + 300 km = 6.4 x 106 m + 0.3 x 106 m = 6.7 x 106 m Period = 91 min

RO

300 km

RE

-1

20

1 rot 1 min radx x 2 0.00115 s

91 min 60 s rot

a R

ES

Home work

Thesis1. Hair tension and it’s applications

2. Frictions and their applications

3. Frictional reduction

4. The moon movements

5. Water moving by moon gravitation

6. Solar system movements

Calculating accelerations

x’

O

x

O

hmM

TFN

W

Y

w

T

Calculating accelerations

m1 > m2

m1

m2

O

x

T

W1

T

W2