Transcript Document

PHY 183 - Program
Physics for Scientists and Engineers
Chapter 1 KINEMATICS
7/5
Chapter 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
5/3
Chapter 6 WAVE MOTION
5/3
Chapter 7 FLUIDS AND THERMAL PHYSICS
5/3
Chapter 8 GAS LAWS AND KINETIC THEORY
5/3
Chapter 9 LIQUID PHASE
6/4
The final examination 60%
1. Net force and Newton's first law
2. Newton's second law
3. Newton's third law
4. Frictional forces
5. Gravitation
6. Circular motion
7. Centripetal force
8. Static equilibrium and
reference frames
Part 4
Frictional Forces
Forces: Normal Force
FN
Book at rest on table:
What are forces on book?
W
• Weight is downward
• System is “in equilibrium” (acceleration = 0  net force = 0)
• Therefore, weight balanced by another force
• FN = “normal force” = force exerted by surface on object
• FN is always perpendicular to surface and outward
• For this example FN = W
Learning Check
Y
FN
Book at rest on table titled
an angle of 300 to horizontal
W
How much is normal force ??
300
W  FN  0  to 0Y : FNY  FN
3
 W.cos 30  mg.
2
0
Forces: Kinetic Friction
FN
direction of motion
F
fk
W
• Kinetic Friction (a sliding Friction):
A force, fk, between two surfaces that opposes relative
motion. Magnitude: fk = kFN
*
k = coefficient of kinetic friction
a property of the two surfaces
Learning Check
Y
Book (50g) move with a= 2m/s2 on
table titled an angle 300 to horizontal
FN
How much is coefficient of
kinetic friction ??
Fk
W
300
X
W  FN  FK  a 
3
to 0Y(a y  0) : FN  WY  mg.cos 30  0, 05.9,8.
(N)
2
to 0Y(a x  a) : W.sin 300  FK  mg.sin 30 0  FN  a
0
mg.sin 300  a
 
FN
Forces: Static Friction
FN
fs
F
W
• 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
FN
fs
F
10N
• 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
T
• Tension: force exerted by a rope (or string)
• Magnitude: same everywhere in rope
Not changed by pulleys
• Direction: same as direction of rope.
example: box hangs from a
rope attached to ceiling
y
T
How much is
rope tension ?
5N
SFy = may
T - W = may
T = W + may
In this case ay = 0
So T = W=5N
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
m
mass on surface
of Earth
Re
Me
Force on mass:
 GM e 
Fg   2  m  gm  mg
 Re 
g
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 ??
 GM e 
R e2  

g


using M e  5.98 x 10 24 kg and g  9.81 m/s 2
 R e  6.38 x 10 6 m
Fg  gm  0,03.9,8  0,294( N)
See movie
Part 6
Circular motion
What is CM?
(Circular Motion)
• Motion in a circle with:
Constant Radius R
– Trajectory is circular form
y
v
(x,y)
R
x
How can we describe CM?
• In general, one coordinate system is as good as
y
any other:
– Cartesian:
• (x,y)
• (vx ,vy)
v
[position]
[velocity]
– Polar:
• (R,)
• (vR ,)
(x,y)

R
[position]
[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!
x
Polar and Cartesian Coordinates
y
R
x = R cos 
y = R sin 
(x,y)
X2 + y2 = R2

x
tg y / x
1
cos
sin
/2


0
-1
3/2
2
Polar Coordinates...
• In Cartesian coordinates, we say velocity:
• dx/dt = v and x = vt + x0
• In polar coordinates, angular velocity:
y
d/dt = .
 = t + 0
Has units of radians/second.
Displacement s = vt.
but s = R = Rt, so:
v = R
v
R
t
s
x
MCQ test
•
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/s
What is velocity V ???
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
v
Realize that:
R
– period (T) = seconds / revolution
– So T = 1 / f = 2/
 = 2 / T = 2f

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 ??
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
f
L
Polar Coordinates...
•
•
•
•
In Cartesian coordinates, we say acceleration:
dv/dt = a and v = at + v0
In polar coordinates, angular acceleration :
y
d/dt =b and  = bt + 0
v
Has units of radians/second.
R
For vector of angular velocity
s
 d
b
dt
Has direction of axis looking point A
turning to watch loop
t
x
MCQ test
• 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)
(b)
(c)
650 rad/s2
20 rad/s2
105 rad/s2
What is UCM?
(Uniform Circular Motion)
• Motion in a circle with:
y
v
– Constant Radius R
– Trajectory is circular form
Constant Speed v = |v|
(x,y)
R
x
Acceleration in UCM:
• Even though the speed is constant, velocity
is not constant since the direction is
changing: must be some acceleration!
– Consider average acceleration in time t
aav = v / t
v
v2
R
t
v1
v2
v1
Acceleration in UCM:
• This is called Centripetal
Acceleration.
• Now let’s calculate the magnitude:
Similar triangles:
v
v2
v2
R
R
v1
v1
v R

v
R
But R = vt for small t
v vt

So:
v
R
v v 2

t
R
v2
a
R
Useful Equivalent
2
v
We know that a 
R
and
Substituting for v we find that:
a=
2
R
v = R

R 
a
2
R
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:
1 rot
1 day
rad
6 -1
x
x 2
 2 .66 x10 s
27 .3 day 86400 s
rot
• So  = 2.66 x 10-6 s-1.
Now calculate the acceleration.
^
2
2
– a =  R = 0.00272 m/s = 0.000278
g
– direction of a points at the center of
the Earth (-r ).
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
• First calculate the angular
frequency :
1 rot 1 min
rad
-1

x
x 2  0.00115 s
91 min 60 s
rot
a  2 R 0 
ES
RO
300 km
RE
Home work
Thesis
1. 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
T
Y
T
O
FN
M

O

m
W
w
x’
h
O
Calculating
accelerations
m1 > m 2
x
T
T
m1
W1
m2
W2