Physics 1: Mechanics

Đào Ngọc Hạnh Tâm

WEEK2,LESSON 2

Review lesson 1: Motion in one Dimension

1. Displacement (m): Δx = xt – x0

2. Velocity (m/s): v = Δx/Δtspeed (m/s):

3. Acceleration (m/s2): a = Δv/Δt

4. Instantaneous velocity and acceleration:

5. Constant acceleration:

6. Free falling: affect by only gravity (g=9.8 m/s2)

Δtx)(totals =

dtdx(t)v(t) =

atvv 0 +=2

00 21tvxx at++=

2

2

dtxd

dtdx

dtd

dtdv(t)a(t) =⎟

⎠

⎞⎜⎝

⎛==

Chapter 1 Bases of Kinematics

1.1. Motion in One Dimension

1.2. Motion in Two Dimensions 1.2.1. The 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

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 componentnotation (ax and ay), we determine itin magnitude-angle notation (a and θ):

ax=acosθ a y=asinθ

a= ax2 + ay

2 tanθ =ayax

R3. Adding vectors:

R3.1. Adding vectors geometrically:

• Vector subtraction:

R3.2. Adding vectors by components:

!s = !a +!b θ

s2 = a2 + b2 − 2abcos(180−θ )

!d = !a −

!b = !a + (−

!b)

!a = axi + ay j;!b = bxi + by j

!s = !a +!b

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):

!a•!b = abcosφ

!a•!b = axbx + ayby

!c = !a×!b

c = absinφ

!a!b

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:

!c!a

!c

!b

!a = axi + ay j + azk!b = bxi + by j + bzk

!c = (aybz − byaz )i + (azbx − bzax ) j + (axby − bxay )k

!c = !a×!b

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 !!!−=Δ

)jyi(x)jyi(xrΔ 1122 +−+=!

jyixj)y-(yi)x-(xrΔ 1212 Δ+Δ=+=!

•Three dimensions: kzjyixr ++=!

kzjyixk)z-(zj)y-(yi)x-(xrΔ 121212 Δ+Δ+Δ=++=!

y

x

M

O X1

Y1

M

Δ

X2

Y2

Average Velocity and Instantaneous Velocity:

intervaltimentdisplacemevelocityaverage =

trvavg Δ

Δ=!

!

• Instantaneous Velocity, Δt→0:

dtrd

trlimv

0Δt

!!!

=ΔΔ

=→

•The direction of the instantaneous velocity of a particle is always tangent to the particle’s path at the particle position.

jvivv

jdtdyi

dtdx)jyi(x

dtdv

yx +=

+=+=

!

!

The scalar components of v!

dtdyv,

dtdxv yx ==

Three dimensions:

kvjvivv zyx ++=!

dtdzvz =

Average Acceleration and Instantaneous Acceleration:

intervaltimeyin velocit changeonacceleratiaverage =

tvaavg Δ

Δ=!!

• Instantaneous Acceleration, Δt→0:

dtvd

tvlima

0Δt

!!!=

ΔΔ

=→

a!The scalar components ofjaiaa yx +=

!

,,dtdv

adtdva y

yx

x ==

Three dimensions:

dtdva 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

!a = axi + ay j = constant

!r = !r0 +!v0t +

12!at2 = xi+yj

!v = !v0 +!at = vx i + vy j

Along the x axis:

vx =v0x + axt; x = x0 + v0xt + 12

axt2

Along the y axis:

vy = v0y + a yt; y = y0 + v0yt + 12

ayt2

Sample Problem:A particle with velocity (m/s) at t=0 undergoes a constant acceleration of magnitude a = 3.0 m/s2 at an angle θ= 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 must apply equations of straight-line motion separately for motion parallel

tavv;tavv y0yyx0xx +=+=v0x=-2.0 (m/s) and v0y=4.0 (m/s)

)(m/s2.30)sin(1303.0sinθaa

)(m/s-1.93)cos(1303.0cosθaa20

y

20x

=×==

=×==y

a!

θ• 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

y 127θ1.33vv

)θtan( =⇒−≈=

v!

Projectile motion!a = constantax = 0; ay = −g

θ0: launch angleR: horizontal range

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 + v0cosθ0t• Oy, vertical motion (free fall, ay = -g if the positive y direction is upward):

2000 gt21-tsinθvyy +=

vy = v0y + ayt = v0sinθ0 -gt

•The equation of the path:

( )( )200

2

0 θcosv2gx-xtanθy =

•Horizontal range:0

20 sin2θgvR =

vx = v0x = v0cosθ0 = 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 ==

g00sinθvt =

2000 gt21-tsinθvyy +=

( )max

200

2sinθvy Hg

==

Hmax

(m) 9.70max =H

Hmaxx

(b) its velocity:

(m/s) 25.53cosθvv 00x ==

yaΔ=− 2vv 20y

2y

(m/s) 28.37sinθvv 000y ==

)(m/s 8.9 2−=a(m) 1150 −=−=Δ yyy P

(m/s) 36.60vy −=

j (m/s) 36.60i (m/s)25.53v −=!

(c) Calculate X:g

y00y

v-sinvtatsinvv

θθ =⇒+=

(s) 96.9=t(m) 37.530cosθvvX 00x === tt

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 θ0from the horizontal must a ball be fired to hit the ship? (b) Howfar should the pirate ship be from the cannon if it is to be beyondthe maximum range of the cannon balls?

0

20 sin2θgvR =

20

1

vgRsin2θ −=

00 63θor 27θ ≈≈

(a)

(b)

(m)686)452sin(8.9

82sin2θgvR

2

0

20

max =×==

20

22

10

11

000

cosθvRt;

cosθvRt

tcosθvxx

==⇒

+=

R1 R2

è It is likely answer 4

However

gθsin2vt;

gθsin2vt

t θsinvv

202

101

00y

==⇒

−= g when the shells hit the ships, 00y θsinvv −=

→θ1>θ2 è t2 < t1:the answer is B

the farther ship gets hit first

VoVoθ1

θ2

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 circlev the speed of the particle

rva2

=

vr2T π

= (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 accelerationincludes radial and tangential components.

tr aaa !!!+=

ra!

a!

ta!

Tangential accelerationRadial (centripetal) acceleration

The path of a particle’s motion

ra!

a!ta!

The tangential acceleration causes the change in thespeed of the particle: parallel to the instantaneous velocity:

Td vadt

=

r

The radial acceleration arises from the change in direction of the velocity vector : perpendicular to the path

2

RvaR

= (R : radius of curvature of the path at the point)

Td vadt

=

r

The magnitude of the acceleration vector :

2

RvaR

=

2 2T Ra a a= +

Uniform circular motion (v is constant ) : aT = 0→ the acceleration is always completely radial

Homework: 6, 7, 11, 20, 27, 29, 54, 58, 66

From Page 78, in the Chapter 4, Principle of Physics

Acknowledgment:Some slides and images used in this lecture arereferred from Lectures of Prof. Phan Bao Ngocand Dr. Do Xuan Hoi in Department ofPhysics, HCM International University.