physics 1: mechanicswebdirectory.hcmiu.edu.vn/portals/25/userdata/dnhtam/physics 1/phy1... ·...
TRANSCRIPT
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Physics 1: Mechanics
Đào Ngọc Hạnh Tâm Office: A1.503,
Email: [email protected] HCMIU, Vietnam National University
Acknowledgment: Most of these slides are supported by Prof. Phan Bao Ngoc
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● 02 credits (30 teaching hours)
Course Requirements ● Attendance + Discussion/activity + Assignment: 30%
● Mid-term exam: 30%
● Final: 40%
Notice: Finish homework & Read text ahead of time
No cellphone & laptop in class
Absence rate >20% => not
allowed to take the final
exam.
● Textbook: Principles of Physics, 9th edition, Halliday/Resnick/Walker (2011),John Willey & Sons, Inc.
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Part A Dynamics of Mass Point Chapter 1 Bases of Kinematics
Chapter 2 Force and Motion (Newton’s Laws)
Assignment
Part B Laws of Conservation Chapter 3 Work and Mechanical Energy
Midterm exam
Chapter 4 Linear Momentum and Collisions
Part C Dynamics and Statics of Rigid Body Chapter 5 Rotation of a Rigid Body About a Fixed Axis
Assignment
Chapter 6 Equilibrium and Elasticity
Chapter 7 Gravitation
Final exam
Contents of Physics 1
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Part A Dynamics of Mass Point
Chapter 1: Bases of Kinematics
1. 1. Motion in One Dimension
1.1.1. Position, Velocity, and Acceleration
1.1.2. One-Dimensional Motion with Constant Acceleration
1.1.3. Freely Falling Objects
1. 2. Motion in Two Dimensions
1.2.1. Position, Velocity, and Acceleration Vectors
1.2.2. Two-Dimensional Motion with Constant Acceleration. Projectile Motion
1.2.3. Circular Motion. Tangential and Radial Acceleration
1.2.4. Relative Velocity and Relative Acceleration
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Part A Dynamics of Mass Point
Chapter 1: Bases of Kinematics
1. 1. Motion in One Dimension
1.1.1. Position, Velocity, and Acceleration
1.1.2. One-Dimensional Motion with Constant Acceleration
1.1.3. Freely Falling Objects
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Measurements
• Use laws of Physics to describe our understanding of nature
• Test laws by experiments • Need Units to measure physical quantities • Three SI “Base Quantities”:
– Length – meter – [m]
– Mass – kilogram – [kg]
– Time – second – [s]
Systems:
– SI: Système International [m kg s]
– CGS: [cm gram second]
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1.1. Motion in one dimension Kinematics • Kinematics: describes motion
• Dynamics: concerns causes of motion
To describe motion, we need to measure:
– Displacement: x = xt – x0 (measured in m or cm)
– Time interval: t = t – t0 (measured in s)
amF
dynamic kinematic
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A. Position: determined in a reference frame.
Motion of an mouse
Space vs. time graph
t=0 s: x=-5 m t=3 s: x=0 m x=0-(-5)=5 m
1.1.1. Position, Velocity and Acceleration
Two features of displacement: - its direction (a vector) - its magnitude
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B. Velocity: describing how fast an object moves
Unit: m/s or cm/s
B.1. Average velocity:
Theavg of the mouse:
2m/s3s
6mavgv
B.2. Average speed:
Δt
distancetotalsavg
Note: average speed does not include direction
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•If a motorcycle travels 20 m in 2 s, then its average velocity is:
•If an car travels 45 km in 3 h, then its average velocity is:
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Sample Problem (average velocity vs average speed):
A car travels on a straight road for 40 km at 40 km/h. It then continues in the opposite direction for another 20 km at 40 km/h. (a) What is the average velocity of the car during this 60 km trip? (b) What is the average speed?
(b)
savg
=total distance
Dt=
40 + 20
1.5= 40 (km/h)
(a)
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B.3. Instantaneous Velocity and Speed
The average velocity at a given instant (t 0), which approaches
a limiting value, is the velocity:
dt
dx(t)
Δt
Δx(t)limv(t)
0Δt
The slope (tanθ) of the tangent line gives v(t)
Speed is the magnitude of velocity,
ex: v=±40 km/h, so s=40 km/h.
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Sample Problem : The position of an object described by:
x = 4-12t+3t2 (x: meters; t: seconds)
(1) What is its velocity at t =1 s?
(2) Is it moving in the positive or negative direction of x just then?
(3) What is its speed just then?
(4) Is the speed increasing or decreasing just then?
(5) Is there ever an instant when the velocity is zero? If so, give the time t; if not answer no.
(6) Is there a time after t= 3 s when the object is moving in the negative direction of x? if so, give t; if not, answer no.
v=dx/dt=-12+6t=-6 (m/s)
negative
S=6 (m/s)
0<t<2: decreasing; 2<t: increasing
t=2 s
no
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C. Acceleration:
C1. Average acceleration:
The rate of change of velocity:
Unit: m/s2 (SI) or cm/s2 (CGS)
C2. Instantaneous acceleration:
At any instant:
The derivative of the velocity (or the second one of the position) with respect to time.
12
12
avg
tt
vv
Δt
Δva
2
2
0Δt dt
xd
dt
dx
dt
d
dt
dv(t)
Δt
Δv(t)lima(t)
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1.1.2. Constant acceleration:
If t0 = 0: (1)
If t0=0: (2)
)]dtta(t[vxvdtxx 0
t
t00
t
t0
00
dt
dxv
constadt
dva
t
t0
0 adtvv )ta(tvv 00
2
)ta(t)t(tvxx
2
0000
200
2
1tvxx at
atvv 0
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Specialized equations:
From Equations (1) & (2):
v 2 - v 0
2 = 2a(x - x 0 )
tvv )(2
1xx 00
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Example: An electron has a=3.2 m/s2 At t (s): v=9.6 m/s
Find: v at t1=t-2.5 (s) and t2=t+2.5 (s)?
Key equation: v = v0+at (v0 is the velocity at 0 s)
• At time t: v = v0+at
• At t1: v1 = v0 + at1
v1=v + a(t1-t) = 9.6+3.2x(-2.5) = 1.6 (m/s)
• At t2: v2 = v0 + at2
v2= v +a (t2-t) = 9.6+3.2(2.5) = 17.6 (m/s)
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1.1.3. Freely falling objects: • “Free-fall” is the state of an object
moving solely under the influence of gravity.
• The acceleration of gravity near the Earth’s surface is a constant, g=9.8 m/s2 toward the center of the Earth.
Free-fall on the Moon
Free-fall in vacuum
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Example (must do):
A ball is initially thrown upward along a y axis, with a velocity of 20.0 m/s at the edge of a 50-meters high building. (1) How long does the ball reach its maximum height? (2) What is the ball’s maximum height? (3) How long does the ball take to return to its release point? And its velocity at that point? (4) What are the velocity and position of the ball at t=5 s? (5) How long does the ball take to hit the ground? and what is its velocity when it strikes the ground?
y
Using two equations: atvv 0
200 at
2
1tvyy
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(1) How long does the ball reach its maximum height?
(2) What is the ball’s maximum height?
y
gtvatvv 00
(s) 04.28.9
20vt 0
g
200 at
2
1tvyy
2max 4)(-9.8)(2.0
2
12.04200y
(m) 4.20y max
At its maximum height, v = 0:
We choose the positive direction is upward
v0 = 20.0 m/s, y0 = 0, a = -9.8 m/s2
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We can use: At the ball’s maximum height:
(3) How long does the ball take to return to its release point? And its velocity at that point?
So:
y
)(2 02 yya 2
0vv
m ax2 8.9200 y 2
(m) 4.20max y
200 at
2
1tvyy
29.8t2
120t00
(s) or 08.40 t t
(s) 08.4t
At the release point: y = 0
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(4) What are the velocity and position of the ball at t=5 s?
y gtvatvv 00
(m /s) 209.8(4.08)20v
)(2 02 yya 2
0vv
dow nw ardvvvv 20 :0
2
gtvv 0
(m) 5.229.8t2
120ty 2
(m/s) 0.2959.8-20
You can also use:
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(5) How long does the ball take to hit the ground? and what is its velocity when it strikes the ground?
so
y
(m /s) 1.37(5.83)9.8-20gtvv 0
509.8t2
120ty 2
(s) (s); 75.183.5 tt
(s) 83.5t
When the ball strikes the ground, y = -50 m
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1. Displacement (m): x = xt – x0
2. Velocity (m/s): v = x/t ; speed (m/s): 3. Acceleration (m/s2): a = v/t 4. Instantaneous velocity and acceleration: 5. Constant acceleration: 6. Free falling: affected by only gravity (g=9.8 m/s2)
Δt
x)(totals
dt
dx(t)v(t)
atvv 0 2
002
1tvxx at
2
2
dt
xd
dt
dx
dt
d
dt
dv(t)a(t)
Review lesson 1: Motion in one Dimension
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Homework: (1) Read Session 2-10, page 27
(2) Problems 1, 2, 3, 4, 7, 30, 34, 48, 50
(Page 30 - 35)
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Part A Dynamics of Mass Point
Chapter 1 Bases of Kinematics
1. 2. Motion in Two Dimensions
1.2.1. Position, Velocity, and Acceleration Vectors
1.2.2. Two-Dimensional Motion with Constant Acceleration. Projectile Motion
1.2.3. Circular Motion. Tangential and Radial Acceleration
1.2.4. Relative Velocity and Relative Acceleration
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Vectors (Recall) R1. Vectors and scalars: • A vector has magnitude and direction; vectors follow certain rules of combination. •Some physical quantities that are vector quantities are displacement, velocity, and acceleration. •Some physical quantities that does not involve direction are temperature, pressure, energy, mass, time. We call them scalars. R2. Components of vectors: •A component of a vector is the projection of the vector on an axis. •If we know a vector in component notation (ax and ay), we determine it in magnitude-angle notation (a and θ):
ax =acosq a y=asinq
a= ax2 + ay
2tanq =
ay
ax
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R3. Adding vectors: R3.1. Adding vectors geometrically: • Vector subtraction:
R3.2. Adding vectors by components:
θ
s2 = a2 + b2 - 2abcos(180 -q )
sx = ax + bx; sy = ay + by
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R4. Multiplying a vector by a vector: ϕ is the smaller of the two angles between and R4.1. The scalar product (the dot product): R4.2. The vector product (the cross product):
c = absinf
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The direction of is determined by using the right-hand rule: Your fingers (right-hand) sweep into through the smaller angle between them, your outstretched thumb points in the direction of In the right-handed xyz coordinate system:
k = i ´ j
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1.2. Motion in Two Dimensions
1.2.1. The Position, Velocity, and Acceleration Vectors Position:
y
x
A particle is located by a position vector:
jyixr
ix and jy are vector components of r
x and y are scalar components of r
M
O X
Y
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• Displacement:
12 rrr
)jyi(x)jyi(xrΔ 1122
jyixj)y-(yi)x-(xrΔ 1212
•Three dimensions: kzjyixr
kzjyixk)z-(zj)y-(yi)x-(xrΔ 121212
y
x
M
O X1
Y1
M
X2
Y2
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Average Velocity and Instantaneous Velocity:
intervaltime
ntdisplacemevelocityaverage
t
rvavg
• Instantaneous Velocity, t0:
dt
rd
t
rlimv
0Δt
•The direction of the instantaneous velocity of a particle is always tangent to the particle’s path at the particle position.
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jvivv
jdt
dyi
dt
dx)jyi(x
dt
dv
yx
The scalar components of v
dt
dyv,
dt
dxv yx
Three dimensions:
kvjvivv zyx
dt
dzvz
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Average Acceleration and Instantaneous Acceleration:
intervaltime
yin velocit changeonacceleratiaverage
t
va avg
• Instantaneous Acceleration, t0:
dt
vd
t
vlima
0Δt
a
The scalar components of
jaiaa yx
,,dt
dva
dt
dva
yy
xx
Three dimensions:
dt
dva z
z
;kajaiaa zyx
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1.2.2. Two-Dimensional Motion with Constant Acceleration. Projectile Motion Key point: To determine velocity and position, we need to determine x and y components of velocity and position
Along the x axis:
vx =v0x + axt; x = x0 + v0xt + 1
2axt
2
Along the y axis:
v y = v0y + a yt; y = y0 + v0yt + 1
2ayt
2
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Sample Problem: A particle with velocity (m/s) at t=0 undergoes a constant acceleration of magnitude a = 3.0 m/s2 at an ngle = 1300 from the positive direction of the x axis. What is the particle’s velocity at t=5.0 s, in unit-vector notation and in magnitude-angle notation?
j4.0i2.0v0
a
v
• Key issues: This is a two-dimensional motion, we can apply equations of 2 straight-line motions separately
tavv;tavv y0yyx0xx v0x=-2.0 (m/s) and v0y=4.0 (m/s)
)(m/s2.30)sin(1303.0sinaa
)(m/s-1.93)cos(1303.0cosaa
20
y
20
x
y
a
• At t= 5 s: (m/s) 15.5v;(m/s)-11.7v yx
j15.5i1.71v
•The magnitude and angle of : x (m/s)4.19vvv 2y
2x
0
x
y127θ1.33
v
v)θtan(
v
v
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θ0: launch angle R: horizontal range
Projectile motion
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Projectile Motion: A particle moves in a vertical plane with some initial velocity but its acceleration is always the free-fall acceleration.
• Ox, horizontal motion (no acceleration, ax = 0):
x = x0 +v0xt = x0 +v0cosq0t
• Oy, vertical motion (free fall, ay = -g if the positive y direction is upward):
2000 gt
2
1-tsinθvyy
vy = v0y + ayt = v0sinq0 -gt
•The equation of the path (trajectory):
200
2
0θcosv2
gx-xtanθy
•Horizontal range: 0
20 sin2θ
g
vR
vx = v0x = v0cosq0 = constant
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Example: A projectile is shot from the edge of a cliff 115m above ground level with an initial speed of 65.0 m/s at an angle of 350 with the horizontal (see the figure below). Determine: (a) the maximum height of the projectile above the cliff; (b) the projectile velocity when it strikes the ground (point P); (c) point P from the base of the cliff (distance X).
x
y (a) At its maximum height:
0gt-sinθvv 00y
g00sinθv
t
2000 gt
2
1-tsinθvyy
max
2
00
2
sinθvy H
g
Hmax
(m) 9.70max H
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Hmax x
(b) its velocity:
(m/s) 25.53cosθvv 00x
ya 2vv 20y
2y
(m/s) 28.37sinθvv 000y
)(m/s 8.9 2a(m) 1150 yyy P
(m/s) 36.60vy
j (m/s) 36.60i (m/s)25.53v
(c) Calculate X: g
y0
0y
v-sinvtatsinvv
(s) 96.9t (m) 37.530cosθvvX 00x tt
𝑽 = 𝑽𝒙𝟐 + 𝑽𝒚
𝟐 = 𝟔𝟎. 𝟓𝟗 (𝒎
𝒔)
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Sample Problem (page 70): Figure below shows a pirate ship 560 m from a fort defending the harbor entrance of an island. A defense cannon, located at sea level, fires balls at initial speed v0= 82 m/s. (a) At what angle 0
from the horizontal must a ball be fired to hit the ship? (b) How far should the pirate ship be from the cannon if it is to be beyond the maximum range of the cannon balls?
0
20 sin2θ
g
vR
20
1
v
gRsin2θ
00 63θor 27θ
(a)
(b)
(m)686)452sin(8.9
82sin2θ
g
vR
2
0
20
max
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20
22
10
11
000
cosθv
Rt;
cosθv
Rt
tcosθvxx
R1 R2
It is likely answer 4
However
g
θsin2vt;
g
θsin2vt
t θsinvv
202
101
00y
g when the shells hit the ships, 00y θsinvv
1>2 t2 < t1:
the answer is B
the farther ship gets hit first
Vo
Vo 1 2
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1.2.3. Circular Motion. Tangential and Radial Acceleration
Uniform Circular Motion: A particle moves around a circle or a circular arc at constant speed. The particle is accelerating with a centripetal acceleration:
Where r is the radius of the circle v the speed of the particle
r
va
2
v
r2T
(T: period)
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1.2.3. Circular Motion. Tangential and Radial Acceleration
If the speed is not constant, means velocity vector changes both in magnitude and in direction at every point then the acceleration includes radial and tangential components.
tr aaa
ra
a
ta
Tangential acceleration Radial (centripetal) acceleration
The path of a particle’s motion
ra
a
ta
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The tangential acceleration causes the change in the speed of the particle: parallel to the instantaneous velocity:
T
d va
dt
The radial acceleration arises from the change in direction of the velocity vector: perpendicular to the path
2
R
va
R
(R : radius of curvature of the path at the point)
The magnitude of the acceleration vector : 2 2
T Ra a a
Uniform circular motion (v is constant ) : aT = 0 a = aR
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1.2.4. Relative Velocity and Relative Acceleration
• An is parked, watching a moving car P. Bao is driving at constant speed and also watching P:
BAPBPA xxx
dt
)d(x
dt
)d(x
dt
)d(x BAPBPA
BAPBPA vvv
The velocity vPA of P as measured by A is equal to the velocity vPB of P as measured by B plus the velocity vBA of B as measured by A.
If car P is moving with an acceleration: dt
)d(v
dt
)d(v
dt
)d(v BAPBPA
:constant avBA PBPA aa
Observers on different frames of reference that move at constant velocity relative to each other will measure the same acceleration for a moving object.
A. In one dimension:
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B. In two dimensions:
BAPBPA rrr
BAPBPA vvv
dt
)vd(
dt
)vd(
dt
)vd( BAPBPA
PBPA aa
Note:
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Example: A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North. What is the resultant velocity of the motorboat? If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore? What distance downstream does the boat reach the opposite shore?
R
2
eriver/shor
2
boat/riverboat/shore
eriver/shorboat/riverboat/shore
vvv
vvv
09.364
3 tan(m/s); 5 R
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Time to cross the river:
boat/shoreeriver/shorboat/river v
distanceC
v
distanceB
v
distanceAt
(s) 204
80
v
distanceA
boat/river
t
distance downstream:
(m) 60203tvdistanceB eriver/shor
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Homework: 6, 11, 20, 27, 29, 54, 66, 70, 76 From Page 78, in the Chapter 4, Principle of Physics
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Review: Motion in two Dimensions
Position:
dt
vd
t
vlima
0Δt
Instantaneous velocity and acceleration:
Velocity:
Acceleration:
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Motion with a constant acceleration: Projectile motion:
tcosθvxx 000
• Ox: Horizontal motion (no acceleration):
• Oy: Vertical motion (free fall)
gt-sinθvv 00y
2000 gt
2
1-tsinθvyy
•Horizontal range:
0
20 sin2θ
g
vR
constantcosθvv 00x
g
H2
sinθv2
00max
•Maximum height:
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Where r is the radius of the arc
v the speed of the particle
Uniform Circular Motion:
r
va
2
Relative Velocity and Relative Acceleration:
BAPBPA vvv BAPBPA vvv
T
d va
dt
2
R
va
R
Circular Motion:
r
va
2
tr aaa
2 2
T Ra a a