Physics 1: Mechanics

Đào Ngọc Hạnh Tâm Office: A1.503,

Email: dnhtam@hcmiu.edu.vn HCMIU, Vietnam National University

Acknowledgment: Most of these slides are supported by Prof. Phan Bao Ngoc

● 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.

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

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

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

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]

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

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

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

•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:

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)

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.

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

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)

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

Specialized equations:

From Equations (1) & (2):

v 2 - v 0

2 = 2a(x - x 0 )

tvv )(2

1xx 00

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)

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

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

(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

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

(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:

(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

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

Homework: (1) Read Session 2-10, page 27

(2) Problems 1, 2, 3, 4, 7, 30, 34, 48, 50

(Page 30 - 35)

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

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

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

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

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

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

• 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

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.

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

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

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

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

θ0: launch angle R: horizontal range

Projectile motion

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

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

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)

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

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

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)

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

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

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:

B. In two dimensions:

BAPBPA rrr

BAPBPA vvv

dt

)vd(

dt

)vd(

dt

)vd( BAPBPA

PBPA aa

Note:

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

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

Homework: 6, 11, 20, 27, 29, 54, 66, 70, 76 From Page 78, in the Chapter 4, Principle of Physics

Review: Motion in two Dimensions

Position:

dt

vd

t

vlima

0Δt

Instantaneous velocity and acceleration:

Velocity:

Acceleration:

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:

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