NATIONAL UNIVERSITY HO CHI MINH CITY
UNIVERSITY OF TECHNOLOGY
GENERAL PHYSICS 1
Dr. Pham Thi Hai Mien
Department of Applied Physics
Faculty of Applied Science
COURSE SYLLABUS
Course hours: 65 hours
- Theory: 36 hours
- Exercise: 12 hours
- Project: 24 hours
Assessment: 4 graded components
- Midterm test: 30% (Writing and Multiple choice, 60 min)
- Assignment: 10%
- Project: 10% (Matlab)
- Final exam: 50% (Writing and Multiple choice, 90 min)
Textbooks:
- R.A. Serway and J.W. Jewett, Physics for Scientists and Engineers,
6th Ed., Thomson 2004.
- Halliday, Resnick, Walker, Fundamental physics , Edu. Pub., 2000.
- Nguyen Thi Be Bay et al., General Physics A1, HCMUT Textbook,
2009.
CONTENTS
1. Kinematics
2. Dynamics of particles
midterm test
3. Mechanics of particle systems and rigid bodies
4. Thermodynamics
5. Electric fields
6. Magnetic fields
final exam
CHAPTER 1. KINEMATICS
1. FUNDAMENTAL QUANTITIES
1.1. Position vector
1.2. Velocity
1.3. Acceleration
2. MOTION WITH A CONSTANT ACCELERATION
2.1. Rectilinear motion
2.2. Projectile motion
2.3. Circular motion
2.4. Relative motion
1.1. POSITION VECTOR
• Position vector is a vector that
goes from the origin of the
coordinate system Oxyz to the
particle M and determines the
object’s
position
in
the
coordinate system:
r = OM = xi + yj + zk (m)
z
z
i
k
r
M (x,y,z)
O
j
y
y
x
x
• When a object moves, its
position changes with time
=> vector position is a function of time
path
r(t) = x(t)i + y(t) j + z(t)k
5
DISPLACEMENT AND DISTANCE
• During a time interval ∆t, the particle moves from M1 to M2
The particle’s displacement (the change in position) is:
∆ r = r2 − r1
•
∆ r points from initial position to final
position.
M1
M2
• If a particle has travelled a closed path:
∆r = 0
but distance ∆S ≠ 0
1.2. VELOCITY
a) Average velocity
• Average velocity during a time
interval:
∆r
(m/s)
vavg =
∆t
• Average speed during a time
interval:
∆S
(m/s)
vavg =
∆t
∆ r= 0 → vavg= 0
• When a particle makes a closed path :
∆S ≠ 0 → vavg ≠ 0
7
b) Instantaneous velocity (or simply velocity)
• Velocity of a particle at any instant is the rate at which its position
is changing with time.
Velocity is the derivative of position vector with respect to time.
∆ r dr
=
=
=
v lim
vavg lim
∆t →0
∆t →0 ∆t
dt
dx dy dz
=
i+
j + k = vx i + v y j + vz k
dt
dt
dt
• Direction of velocity: velocity is always
tangent to the path and points in the
direction of motion.
• Magnitude of velocity (or speed):
v=
v 2x + v 2y + v 2z
SAMPLE PROBLEM 1
A particle moves along the x axis according to the equation x = 2.00 +
3.00t − 1.00t2 , where x is in meters and t is in seconds. At t = 3.00 s,
find:
a) the position of the particle.
b) the velocity of the particle.
SOLUTION
a) x(t =
3) =+
2 3x3 − 1x32 =
2 m
dx
b) v = =−
3 2t → v(t =
3) =−
3 2x3 =
−3 m / s
dt
SAMPLE PROBLEM 2
A person walks first at a constant speed of 5.00 m/s along a straight
line from point A to point B and then back along the line from B to
A at a constant speed of 3.00 m/s.
a) What is her average velocity over the entire trip?
b) What is her average speed over the entire trip?
SOLUTION
a) vavg = 0
2SAB
2v1v 2
∆S SAB + SBA
b) v=
=
=
=
= 3.75 m / s
avg
∆t t AB + t BA SAB + SAB v1 + v 2
v1
v2
1.3. ACCELERATION
∆v v 2 − v1
• Average acceleration over a time interval ∆t: a =
=
avg
∆t
t 2 − t1
• Instantaneous acceleration (or simply acceleration) of a particle
at any instant is the rate at which its velocity is changing at that
instant:
2
∆v dv d r
2)
(m/s
=
a lim a avg
= lim = =
2
∆t →0
∆t →0
∆t
dt
dt
Acceleration is the first derivative of velocity or the second
derivative of position with respect to time.
• If the signs of the velocity and acceleration of a particle are the
same, the speed of the particle increases.
• If the signs of the velocity and acceleration of a particle are
opposite, the speed of the particle decreases.
11
TANGENTIAL ACCELERATION & CENTRIPETAL ACCELERATION
2
dv v
τ+ n
• Acceleration has 2 components: a= a t + a n=
dt
R
• Tangential acceleration a t is tangent to the path and reflects the
change in speed.
• Centripetal (or normal) acceleration a n is toward the center of
the path and indicates the change in velocity’s direction. a
t
τ - tangential unit vector
n - normal unit vector
R - curvature radius
an
R
a
v
at
an
v
a
R
an
at
v
Accelerated motion
Decelerated motion
dv
> 0, at ↑↑ v
dt
dv
< 0, at ↑↓ v
dt
2. MOTION WITH A CONSTANT ACCELERATION
v v0 + at
• Velocity at any instant: =
1 2
• Position at any instant: r =r0 + v 0 t + at
2
• Equation of path (trajectory): f (x, y, z) = C
v0 - initial velocity
r0 - initial position
2.1. RECTILINEAR MOTION
• A particle moving along a straight line has no centripetal acceleration:
an = 0
a = at
a) Uniform rectilinear motion
• Total acceleration: a = 0
• Velocity is constant: v = const
x x 0 + vt , where x0 - initial position.
• Position at any instant: =
• Position and velocity can be either positive or negative depending on
the chosen coordinate system.
b) Uniformly accelerated / decelerated rectilinear motion
• Total acceleration is constant: a = const
Uniformly accelerated motion : a.v > 0
Uniformly decelerated motion : a.v < 0
v
• Velocity at any instant : =
v0 + at
1 2
• Position at any instant : x = x 0 + v 0 t + at
2
• Special relation: v 2 − v 02= 2a(x − x 0 )
16
SAMPLE PROBLEM 3
A baseball is hit so that it travels straight upward after being struck by
the bat. A fan observes that it takes 3.00 s for the ball to reach its
maximum height. Find (a) the ball’s initial velocity and (b) the height
it reaches.
SOLUTION
y
a) v = v0 − gt = 0 → v0 = gt = 9.8 × 3 = 29.4 m / s
1 2
b) Using y = y0 + v0t + at
2
1 2
1
→ H = y = 0 + v0t − gt = 29.4 × 3 − 9.8 × 32 = 44.1m
2
2
g
H
v0
Or using v 2 − v02= 2a ( y − y0 )
2
v
→ 0 − v02 = 2(− g )( y − 0) → y = 0 = 44.1m
2g
O
2.2. PROJECTILE MOTION
• An object is projected with some
initial velocity v0 making with the
horizontal axis an angle α.
• Assume that air has no effect on
the projectile.
• Projectile’s acceleration is always
the free fall acceleration g ,
which is directed vertically
downward.
L
• Projectile motion can be divided into 2 separate motions:
Horizontal motion with zero acceleration.
Vertical motion with constant downward acceleration.
• Projectile’s path is parabolic.
• Projectile’s position:
=
x v0 cos α.t
1 2
=
y v0 sin α.t − 2 gt
The initial velocity is above the horizontal
• Projectile’s equation of the path:
g
2
y = tan α.x − 2
x
2v0 cos 2 α
L
• Projectile’s velocity:
=
=
α const
v x v0 cos
v0 sin α
=
v y 0 at=
tH
v y v0 sin α − gt →=
g
The y-component of velocity v y is directed upward initially and
its magnitude steadily decreases to zero, which marks the maximum
height H. Then it reverses direction and its magnitude becomes
larger with time.
• Flight time:
2v0 sin α
=
t L 2t=
H
g
• Maximum height:
v02 sin 2 α
=
H y(t
=
H)
2g
• Horizontal range:
v02 sin 2α
=
=
L x(t
L)
g
Horizontal range is maximum for a launch angle of 450
• Projectile’s position:
The initial velocity is below the horizontal
=
x v0 cos α.t
1 2
y
v
sin
.t
gt
=
α
+
0
2
• Projectile’s equation of the path:
g
2
y = tan α.x + 2
x
2v0 cos 2 α
=
v x v0 cos α
• Projectile’s velocity:
=
v y v0 sin α + gt
The x-component of velocity is constant.
The y-component of velocity increases with time.
21
• Projectile’s position:
x = v0 t
1 2
y = 2 gt
The initial velocity is in the horizontal
• Projectile’s equation of the path:
g 2
y= 2 x
2v0
v x = v0
• Projectile’s velocity:
v y = gt
The x-component of velocity is constant.
The y-component of velocity increases with time.
22
SAMPLE PROBLEM 4
A projectile is fired in such a way that its horizontal range is equal to
three times its maximum height. What is the angle of projection?
SOLUTION
v0 sin α )
(
v sin 2α
sin 2 α 2
L=
3H →
3
=
→
=
g
2g
sin 2α 3
2
0
2
sin 2 α
tan α 2
→
=
= → α = 53.10
2 sin α cosα
2
3
2.3. CIRCULAR MOTION
A particle moving along a circle always has some centripetal
acceleration.
a) Uniform circular motion
• Speed is constant: v = const
v2
= const, a t = 0 → a = a n
• Acceleration: a n =
R
2πR
• Period of revolution: T =
v
• Frequency: f=
1 ω
=
T 2π
(Hz)
(s)
b) Non-uniform circular motion
• Velocity changes in magnitude and direction with time:
v ≠ const
• Acceleration changes in magnitude and direction with time :
v2
a=
≠ const
n
R
at ≠ 0
=
a an + at
SAMPLE PROBLEM 5
The figure represents the total acceleration of a particle moving
clockwise in a circle of radius 2.50 m at a certain instant of time. For
that instant, find (a) the radial acceleration of the particle, (b) the speed
of the particle, and (c) its tangential acceleration, (d) does the speed of
the particle increase or decrease with time?
SOLUTION
=
a ) an acos
=
300 15=
cos300 13 m / s 2
=
b) v
=
an R 5.7 m / s 2
300 7.5 m / s 2
=
c) at asin
=
d) The angle between a and v is less than 900
This is an accelerated circular motion.
The speed of the particle increases with time.
2.4. RELATIVE MOTION
• 2 reference frames (Oxyz) and (O’x’y’z’).
• RF O’ moves with some velocity relative
to RF O.
OM
= OO ' + O ' M
(Relative position)
=
r OO ' + r '
dr dr ' d
v M/O' + vO'/O (Relative velocity)
=+
OO ' → v=
M/O
dt dt dt
dv M/O dv M/O' dvO'/O
(Relative
a
=
a
+
a
=
+
→ M/O
M/O'
O'/O
acceleration)
dt
dt
dt
(
)
SAMPLE PROBLEM 6
Snow is falling vertically at a constant speed of 8.0 m/s. At what
angle from the vertical do the snowflakes appear to be falling as
viewed by the driver of a car traveling on a straight, level road with
a speed of 50 km/h?
SOLUTION
vSE = 8 m / s
vCE 50
km / h 13.9 m / s
=
+
v=
v
v
SC
SE
EC
vEC 13.9
α =
tan=
vSE
8
→α =
600
vEC
vCE
α
vSC
vSE
vSE - Velocity of snow relative to the Earth
vCE - Velocity of the car relative to the Earth
vSC - Velocity of snow relative to the car
REFERENCES
1. R.A. Serway and J.W. Jewett, Physics for Scientists and
Engineers, 6th Ed., Thomson 2004.
2. Halliday, Resnick, Walker, Fundamental physics , 9th Ed., Wiley,
2011.
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