ch.12 kinematics of a particle

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12. Kinematics of a Particle

HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Engineering Mechanics – Dynamics 12.01 Kinematics of a Particle

Chapter Objectives

• To introduce the concepts of position, displacement, velocity,

and acceleration

• To study particle motion along a straight line and represent this

motion graphically

• To investigate particle motion along a curved path using

different coordinate systems

• To present an analysis of dependent motion of two particles.

• To examine the principles of relative motion of two particles

using translating axes


Engineering Mechanics – Statics 12.02 Kinematics of a Particle

§1. Introduction

- Mechanics: the study how body react to the forces acting on

them

- Branches of mechanics

Dynamics is concerned with the accelerated motion of bodies

• Kinematics: analysis only the geometric aspects of the motion

• Kinetics: analysis of the forces causing the motion



§1. Introduction

- Brief history of dynamics

Major contributors include

• Galileo Galilei (1564–1642): pendulums, falling bodies

• Sir Isaac Newton (1642–1727): laws of motion, law of

universal gravitation

• Others include: Kepler, Huygens, Euler, Lagrange, Laplace,

D’Alembert and many others



Galileo Galilei Sir Isaac Newton Leonard Euler Joseph Louis Lagrange

§2.Rectilinear Kinematics: Continuous Motion

- Assumptions

• The object is negligible size and shape (particle)

• The mass is not considered in the calculations

• Rotation of the object is neglected

- Rectilinear kinematics

• Kinematics of an object moving in a straight line

• The kinematics of a particle is characterized by specifying, at

any given instant, the particle’s position, velocity, and

acceleration




- Position

• Consider a particle in rectilinear motion from a fixed origin 𝑂in the 𝑠 direction

• For a given instant, 𝑠 is the position coordinate of the particle

• The magnitude of 𝑠 is the distance from the origin

• Note that the position coordinate would be negative if the

particle traveled in the opposite direction according to our

frame of reference

• Position is has a magnitude (distance from origin) and is

based on a specific direction. It is therefore a vector quantity



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- Displacement

• Displacement is defined as the change in position 𝑠

∆𝑠 = 𝑠′ − 𝑠

• Displacement is also a vector quantity characterized by a

magnitude and a direction

• Note that distance on the other hand is a scalar quantity

representing the length from an origin

- Distance

• Total length of the path over which object travelled (scalar)

• Distance ≠ displacement!

• Distance is path-dependentHCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien



- Velocity

• Average velocity

𝑣𝑎𝑣𝑔 =∆𝑠

∆𝑡(𝑚/𝑠)

• Instantaneous velocity

𝑣 = lim∆𝑡→0

∆𝑠

∆𝑡=

𝑑𝑠

𝑑𝑡(𝑚/𝑠)



∆𝑠: displacement, 𝑚

∆𝑡: time interval, 𝑠


- Speed

• Speed: the magnitude of velocity

• Average speed

(𝑣𝑠𝑝)𝑎𝑣𝑔=𝑠𝑇

∆𝑡(𝑚/𝑠)



𝑠𝑇: traveled distance, 𝑚

∆𝑡: elapsed time, 𝑠


- Acceleration

• Average acceleration

𝑎𝑎𝑣𝑔 =∆𝑣

∆𝑡(𝑚/𝑠2)

• Instantaneous acceleration

𝑎 = lim∆𝑡→0

∆𝑣

∆𝑡=

𝑑𝑣

𝑑𝑡=

𝑑2𝑠

𝑑𝑡2(𝑚/𝑠2)

• An important differential relation involving the displacement,

velocity, and acceleration along the path



∆𝑣:velocity, 𝑚/𝑠


𝑎𝑑𝑠 = 𝑣𝑑𝑣


- Constant acceleration

𝑎 = 𝑎𝑐 = 𝑐𝑜𝑛𝑠𝑡

• Velocity as a function of time

𝑎𝑐 =𝑑𝑣

𝑑𝑡⟹

𝑣0

𝑣

𝑑𝑣 =

𝑡0

𝑡

𝑎𝑐𝑑𝑡 ⟹ 𝑣 = 𝑣0 + 𝑎𝑐𝑡

• Position as a function of time

𝑣 =𝑑𝑠

𝑑𝑡⟹

𝑠0

𝑠

𝑑𝑠 =

𝑡0

𝑡

𝑣0 + 𝑎𝑐𝑡 𝑑𝑡 ⟹ 𝑠 = 𝑠0 + 𝑣0𝑡 +1

2𝑎𝑐𝑡

2

• Velocity as a function of position

𝑣𝑑𝑣 = 𝑎𝑐𝑑𝑠 ⟹

𝑣0

𝑣

𝑣𝑑𝑣 =

𝑠0

𝑠

𝑎𝑐𝑑𝑠 ⟹ 𝑣2 = 𝑣02 + 2𝑎𝑐(𝑠 − 𝑠0)




- Example 12.1 The car moves in a straight line such that for a

short time its velocity is defined by 𝑣 = 3𝑡2 + 2𝑡 (𝑚/𝑠) .

Determine its position and acceleration when 𝑠 = 0

Solution

Coordinate system

Position

𝑣 =𝑑𝑠

𝑑𝑡= 3𝑡2 + 2𝑡 ⟹

0

𝑠

𝑑𝑠 = 0

𝑡

3𝑡2 + 2𝑡 𝑑𝑡 ⟹ 𝑠 = 𝑡3 + 𝑡2(𝑚)

Acceleration

𝑎 =𝑑𝑣

𝑑𝑡=

𝑑

𝑑𝑡3𝑡2 + 2𝑡 = 6𝑡 + 2(𝑚/𝑠2)

At 𝑡 = 3𝑠, 𝑠(3) = 33 + 32 = 36𝑚

𝑎(3) = 6 × 3 + 2 = 20𝑚/𝑠2



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Solution

Coordinate system

Velocity

𝑎 =𝑑𝑣

𝑑𝑡= −0.4𝑣3 ⟹

60

𝑠 𝑑𝑣

−0.4𝑣3=

0

𝑡

𝑑𝑡 ⟹ 𝑣 =60

2880𝑡 + 1

Position

𝑣 =𝑑𝑠

𝑑𝑡⟹

0

𝑠

𝑑𝑠 = 0

𝑡 60

2880𝑡 +1𝑑𝑡 ⟹ 𝑠 =

1

0.4

60

2880𝑡 +1−

1

60

At 𝑡 = 4𝑠, ⟹ 𝑣 4 = 0.559𝑚/𝑠 ↓, 𝑠 4 = 4.43𝑚HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

- Example 12.2 A small projectile is fired

vertically downward into a fluid medium with

an initial velocity of 60𝑚/𝑠 and a deceleration

of 𝑎 = −0.4𝑣3(𝑚/𝑠2). Determine the projectile’s

velocity and position 4𝑠 after it is fired



- Example 12.3 During a test a rocket travels upward at 75𝑚/𝑠and when it is 40𝑚 from the ground its engine

fails. Determine the maximum height 𝑠𝐵

reached by the rocket and its speed just before

it hits the ground. While in motion the rocket is

subjected to a constant downward acceleration

of 9.81𝑚/𝑠2 due to gravity. Neglect the effect of

air resistance

Solution

Coordinate system

Position

𝑣𝐵2 = 𝑣𝐴

2 + 2𝑎𝑐 𝑠𝐵 − 𝑠𝐴

⟹ 𝑠𝐵 =𝑣𝐵

2 − 𝑣𝐴2

2𝑎𝑐+ 𝑠𝐴 =

0 − 752

2 −9.81+ 40 = 327𝑚




Coordinate system

Position

𝑣𝐵2 = 𝑣𝐴

2 + 2𝑎𝑐 𝑠𝐵 − 𝑠𝐴

⟹ 𝑠𝐵 =𝑣𝐵

2 − 𝑣𝐴2

2𝑎𝑐+ 𝑠𝐴 =

0 − 752

2 −9.81+ 40 = 327𝑚

Velocity

𝑣𝐶2 = 𝑣𝐵

2 + 2𝑎𝑐 𝑠𝐶 − 𝑠𝐵

⟹ 𝑣𝐶 = 𝑣𝐵2 + 2𝑎𝑐 𝑠𝐶 − 𝑠𝐵

= 0 + 2(−9.81)(0 − 327)

= −80.1𝑚/𝑠

= 80.1𝑚/𝑠 ↓




- Example 12.4 A metallic particle is subjected to the influence

of a magnetic field as it travels

downward through a fluid that extends

from plate 𝐴 to plate 𝐵. If the particle is

released from rest at the midpoint 𝐶, 𝑠 =100𝑚𝑚 and the acceleration is 𝑎 =4𝑠 (𝑚/𝑠2), determine the velocity of the

particle when it reaches plate 𝐵 , 𝑠 =200𝑚𝑚, and the time it takes to travel

from 𝐶 to 𝐵

Solution

Coordinate system




Velocity

𝑣𝑑𝑣 = 𝑎𝑑𝑠

⟹ 0

𝑣

𝑣𝑑𝑣 = 0.1

𝑠

4𝑠𝑑𝑠

⟹ 𝑣 = 2 𝑠2 − 0.01(𝑚/𝑠)

Time

𝑑𝑠 = 𝑣𝑑𝑡 = 2 𝑠2 − 0.01𝑑𝑡

⟹ 0.1

𝑠 𝑑𝑠

𝑠2 − 0.01=

0

𝑡

2𝑑𝑡

⟹ 𝑡 = 0.5 ln 𝑠2 − 0.01 + 𝑠 + 1.152

At 𝑠 = 200𝑚𝑚 = 0.2𝑚, 𝑣𝐵 = 0.346𝑚/𝑠 ↓

𝑡 = 0.658𝑠




- Example 12.5 A particle moves along a horizontal path with a

velocity of 𝑣 = 3𝑡2 − 6𝑡 (𝑚/𝑠) . If it is

initially located at the origin 𝑂 ,

determine the distance traveled in 3.5𝑠,

and the particle’s average velocity and

average speed during the time interval

Coordinate system

Distance traveled

𝑑𝑠 = 𝑣𝑑𝑡 = 3𝑡2 − 6𝑡 𝑑𝑡

⟹ 0

𝑠

𝑑𝑠 = 0

𝑡

3𝑡2 − 6𝑡 𝑑𝑡 ⟹ 𝑠 = 𝑡3 − 3𝑡2

Note: 0 < 𝑡 < 2 → 𝑣 < 0, 𝑡 > 2 → 𝑣 > 0

Distance traveled in 3.5𝑠

𝑠𝑇 = 4 + 4 + 6.125 = 14.125𝑚



Solution

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Distance traveled

𝑠 = 𝑡3 − 3𝑡2

Distance traveled in 3.5𝑠

𝑠𝑇 = 4 + 4 + 6.125 = 14.125𝑚

Velocity

Displacement from 𝑡 = 0 to 𝑡 = 3.5𝑠

∆𝑠 = 𝑠𝑡=3.5

− 𝑠𝑡=0

= 6.125 − 0 = 6.125𝑚

The average velocity

𝑣𝑎𝑣𝑔 =∆𝑠

∆𝑡=

6.125

3.5 − 0= 1.75𝑚/𝑠 →

The average speed

(𝑣𝑠𝑝)𝑎𝑣𝑔 =𝑠𝑇

∆𝑡=

14.125

3.5 − 0= 4.04𝑚/𝑠



Fundamental Problems

- F12.1 Initially, the car travels along a straight road with a

speed of 35𝑚/𝑠. If the brakes are applied and the speed of the

car is reduced to 10𝑚/𝑠 in 15𝑠 determine the constant

deceleration of the car




- F12.2 A ball is thrown vertically upward with a speed of

15𝑚/𝑠 . Determine the time of flight when it returns to its

original position




- F12.3 A particle travels along a straight line with a velocity

of 𝑣 = 4𝑡 − 3𝑡2. Determine the position of the particle when 𝑡 =4𝑠. 𝑠 = 0 when 𝑡 = 0




- F12.4 A particle travels along a straight line with a speed

𝑣 = 0.5𝑡3 − 8𝑡. Determine the acceleration of the particle when

𝑡 = 2𝑠




- F12.5 The position of the particle is given by 𝑠 = 2𝑡2 − 8𝑡 + 6.

Determine the time when the velocity of the particle is zero,

and the total distance traveled by the particle when 𝑡 = 3𝑠



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- F12.6 A particle travels along a straight line with an

acceleration of 𝑎 = 10 − 0.2𝑠(𝑚/𝑠2). Determine the velocity of

the particle when if 𝑣 = 5𝑚/𝑠 at 𝑠 = 0




- F12.7 A particle moves along a straight line such that its

acceleration is 𝑎 = 4𝑡2 − 2(𝑚/𝑠2). When 𝑡 = 0, the particle is

located 2𝑚 to the left of the origin, and when 𝑡 = 2𝑠, it is 20𝑚to the left of the origin. Determine the position of the particle

when 𝑡 = 4𝑠




- F12.8 A particle travels along a straight line with a velocity of

𝑣 = 20 − 0.05𝑠2(𝑚/𝑠) . Determine the acceleration of the

particle at 𝑠 = 15𝑚



§3.Rectilinear Kinematics: Erratic Motion

- The 𝑠 − 𝑡 graph

Plots of 𝑠 − 𝑡 can be used to find

the 𝑣 − 𝑡 curves by finding the

slope of the line tangent to the

motion curve at any point

𝑣(𝑡) =𝑑𝑠(𝑡)

𝑑𝑡

slope of 𝑠 − 𝑡 graph = velocity




- The 𝑣 − 𝑡 graph

Plots of 𝑣 − 𝑡 can be used to find

the 𝑎 − 𝑡 curves by finding the

slope of the line tangent to the

velocity curve at any point

𝑎(𝑡) =𝑑𝑣(𝑡)

𝑑𝑡

slope of 𝑣 − 𝑡 graph = acceleration




- The 𝑎 − 𝑡 graph

The 𝑣 − 𝑡 graph can be constructed

from an 𝑎 − 𝑡 graph if the initial

velocity of the particle is given

∆𝑣 = 0

𝑡1

𝑎𝑑𝑡

change in velocity = area under 𝑎 − 𝑡 graph



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- The 𝑎 − 𝑡 graph

The 𝑠 − 𝑡 graph can be constructed

from an 𝑣 − 𝑡 graph if the initial

position of the particle is given

∆𝑠 = 0

𝑡1

𝑣𝑑𝑡

displacement = area under 𝑣 − 𝑡 graph




- The 𝑣 − 𝑠 graph

Acceleration at one point can be

obtained by reading the velocity 𝑣at this point on the curve and

multiplying it by the slope of the

curve (𝑑𝑣/𝑑𝑠) at this same point

𝑎 = 𝑣 ×𝑑𝑣

𝑑𝑠

acceleration = velocity × slope of 𝑣 − 𝑠 graph




- The 𝑎 − 𝑠 graph

The area under the acceleration

versus position curve represents

the change in velocity

1

2𝑣1

2 − 𝑣02 =

𝑠0

𝑠1

𝑎𝑑𝑠

1

2𝑣1

2 − 𝑣02 = area under 𝑎 − 𝑠 graph




- Example 12.6 A bicycle moves along

a straight road such that its position is

described by the graph. Construct the

𝑣 − 𝑡 and 𝑎– 𝑡 graphs for 0 ≤ 𝑡 ≤ 30𝑠

Solution

𝑣 − 𝑡 graph

0 ≤ 𝑡 < 10: 𝑠 = 𝑡2(𝑚)

𝑣 =𝑑𝑠

𝑑𝑡= 2𝑡(𝑚/𝑠)

10 ≤ 𝑡 ≤ 30: 𝑠 = 20𝑡 − 100(𝑚)

𝑣 =𝑑𝑠

𝑑𝑡= 20(𝑚/𝑠)




𝑣 − 𝑡 graph

0 ≤ 𝑡 < 10: 𝑠 = 𝑡2(𝑚)

⟹ 𝑣 =𝑑𝑠


10 ≤ 𝑡 ≤ 30: 𝑠 = 20𝑡 − 100(𝑚)

⟹ 𝑣 =𝑑𝑠

𝑑𝑡= 20(𝑚/𝑠)

0 ≤ 𝑡 < 10: 𝑣 =𝑑𝑠


⟹ 𝑎 =𝑑𝑣

𝑑𝑡= 2(𝑚/𝑠2)

10 ≤ 𝑡 ≤ 30: 𝑠 = 20 (𝑚)

⟹ 𝑣 =𝑑𝑠

𝑑𝑡= 0(𝑚/𝑠)



𝑎 − 𝑡 graph


- Example 12.7 The car starts from

rest and travels along a straight track

such that it accelerates at 10𝑚/𝑠2 for

10𝑠, and then decelerates at 2𝑚/𝑠2.

Draw the 𝑣 − 𝑡 and 𝑠– 𝑡 graphs and

determine the time 𝑡′ needed to stop

the car. How far has the car traveled?

Solution

𝑣 − 𝑡 graph

0 ≤ 𝑡 < 10: 𝑎 = 10(𝑚/𝑠2)

𝑣 = 0

𝑣

𝑑𝑣 = 0

10

𝑑𝑡 = 10𝑡

𝑡 = 10: 𝑣 = 10 × 10 = 100𝑚/𝑠

10 < 𝑡 ≤ 𝑡′: 𝑎 = −2(𝑚/𝑠2)



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𝑣 − 𝑡 graph

0 ≤ 𝑡 ≤ 10: 𝑎 = 10(𝑚/𝑠2)

⟹ 𝑣 = 0

𝑣

𝑑𝑣

= 0

10

𝑑𝑡 = 10𝑡

𝑡 = 10: 𝑣 = 10 × 10 = 100(𝑚/𝑠)

10 ≤ 𝑡 ≤ 𝑡′: 𝑎 = −2(𝑚/𝑠2)

100

𝑣

𝑑𝑣 = 10

𝑡

−2𝑑𝑡

⟹ 𝑣 = −2𝑡 + 120(𝑚/𝑠)

𝑡 = 𝑡′: 𝑣 𝑡′ = 0

⟹ 𝑡′ = 60(𝑠)




𝑠 − 𝑡 graph

0 ≤ 𝑡 ≤ 10: 𝑣 = 10𝑡(𝑚/𝑠)

0

𝑠

𝑑𝑠 = 0

𝑡

10𝑡𝑑𝑡

⟹ 𝑠 = 5𝑡2(𝑚)

𝑡 = 10: 𝑠 = 5 × 102 = 500(𝑚)

10 ≤ 𝑡 ≤ 60:𝑣 = −2𝑡 + 120(𝑚/𝑠)

500

𝑠

𝑑𝑣 = 10

𝑡

(−2𝑡+120)𝑑𝑡

⟹ 𝑠 = −𝑡2 +120𝑡−600(𝑚)

𝑡 = 60: 𝑠 = − 60 2

+120×60−600

⟹ 𝑠 = 3000(𝑚)




- Example 12.8 Given the 𝑣– 𝑠 graph describing the motion of a

motorcycle. Construct the 𝑎– 𝑠 graph of the

motion and determine the time needed for the

motorcycle to reach the position 𝑠 = 400𝑚

Solution

𝑎 − 𝑠 graph

0 ≤ 𝑠 ≤ 200:𝑣 = 0.2𝑠 + 10(𝑚/𝑠)

⟹ 𝑎 = 𝑣𝑑𝑣

𝑑𝑠= (0.2𝑠 + 10) × 0.2

= 0.04𝑠 + 1

200 ≤ 𝑡 ≤ 400:

𝑣 = 50(𝑚/𝑠)

⟹ 𝑎 = 𝑣𝑑𝑣

𝑑𝑠= 50 × 0 = 0(𝑚/𝑠)




Time

0 ≤ 𝑠 ≤ 200:

𝑣 = 0.2𝑠 + 10(𝑚/𝑠)

⟹ 𝑑𝑡 =𝑑𝑠

𝑣=

𝑑𝑠

0.2𝑠+10⟹

0

𝑡

𝑑𝑡 = 0

𝑠 𝑑𝑠

0.2𝑠+10

⟹ 𝑡 = 5 𝑙𝑛 0.2𝑠 + 10 − 5𝑙𝑛10(𝑠)

𝑠 = 200 ⟹ 𝑡 = 8.05(𝑠)

200 ≤ 𝑡 ≤ 400:

𝑣 = 50(𝑚/𝑠)

⟹ 𝑑𝑡 =𝑑𝑠

𝑣=

𝑑𝑠

50⟹

8.05

𝑡

𝑑𝑡 = 200

𝑠 𝑑𝑠

50

⟹ 𝑡 =𝑠

50+ 4.05 𝑠

𝑠 = 400 ⟹ 𝑡 = 12.0(𝑠)




- F12.9 The particle travels along a straight track such that its

position is described by the 𝑠 − 𝑡 graph. Construct the 𝑣 − 𝑡graph for the same time interval




- F12.10 A van travels along a straight road with a velocity

described by the graph. Construct the 𝑠 − 𝑡 and 𝑎 − 𝑡 graphs

during the same period. Take 𝑠 = 0 when 𝑡 = 0



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- F12.11 A bicycle travels along a straight road where its

velocity is described by the 𝑣 − 𝑠 graph. Construct the 𝑎 − 𝑠graph for the same time interval




- F12.12 The sports car travels along a straight road such that

its position is described by the graph. Construct the 𝑣 − 𝑡 and

𝑎 − 𝑡 graphs for the time interval 0 ≤ 𝑡 ≤ 10𝑠




- F12.13 The dragster starts from rest and has an acceleration

described by the graph. Construct the 𝑣 − 𝑡 graph for the time

interval 0 ≤ 𝑡 ≤ 𝑡′, where 𝑡′ is the time for the car to come to

rest




- F12.14 The dragster starts from rest and has a velocity

described by the graph. Construct the 𝑠 − 𝑡 graph during the

time interval 0 ≤ 𝑡 ≤ 15𝑠. Also, determine the total distance

traveled during this time interval



§4.General Curvilinear Motion

- Curvilinear motion: particle moves along a curved path

- Position 𝑟

• particle located at a point on a space curve defined by the

path function 𝑠(𝑡)

• position measured from a fixed point 𝑂

• determined by position vector

𝑟 = 𝑟(𝑡)

- Displacement ∆ 𝑟

• the change in the particle’s position

𝑟 → 𝑟′ = 𝑟 + ∆ 𝑟

• determined by vector subtraction

∆ 𝑟 = 𝑟′ − 𝑟




- Velocity

• average velocity

𝑣𝑎𝑣𝑔 =∆ 𝑟

∆𝑡(𝑚/𝑠)

• instantaneous velocity


∆ 𝑟

∆𝑡=

𝑑 𝑟

𝑑𝑡(𝑚/𝑠)

∆𝑡 → 0: ∆ 𝑟, 𝑑 𝑟 approaches the tangent to the curve path ⟹the direction of 𝑣 is also tangent to the curve path



∆ 𝑟: displacement, 𝑚


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- Speed

• speed: magnitude of 𝑣

∆𝑡 → 0: ∆ 𝑟 → ∆ 𝑠


∆𝑟

∆𝑡= lim

∆𝑡→0

∆𝑠

∆𝑡=

𝑑𝑠

𝑑𝑡(𝑚/𝑠)

• speed can be obtained by differentiating the path function 𝑠with respect to time




- Acceleration

• average acceleration

𝑎𝑎𝑣𝑔 =∆ 𝑣

∆𝑡=

𝑣′ − 𝑣

∆𝑡(𝑚/𝑠2)

• instantaneous acceleration

𝑎 = lim∆𝑡→0

∆ 𝑣

∆𝑡=

𝑑 𝑣

𝑑𝑡(𝑚/𝑠2)

𝑎 =𝑑 𝑣

𝑑𝑡=

𝑑

𝑑𝑡

𝑑 𝑟

𝑑𝑡=

𝑑2 𝑟

𝑑𝑡2(𝑚/𝑠2)

- Hodograph: the locus of points for the arrowhead of the

velocity vector in the same manner as the path 𝑠 describes the

locus of points for the arrowhead of the position vector

- The direction of 𝑎 is also tangent to the hodograph




- Note

• Acceleration acts tangential to the hodograph, but generally

not tangential to the path of motion 𝑠

• Velocity is always tangential to the path of motion, whereas

acceleration is always tangential to the hodograph



§5.Curvilinear Motion: Rectangular Components

- The motion of a particle can best be

described along a path that can be

expressed in terms of its 𝑥, 𝑦, 𝑧coordinates

- Position

• position vector 𝑟

𝑟 = 𝑥 𝑖 + 𝑦 𝑗 + 𝑧𝑘

• magnitude of 𝑟

𝑟 = | 𝑟| = 𝑥2 + 𝑦2 + 𝑧2

• unit vector 𝑢𝑟

𝑢𝑟 = 𝑟

| 𝑟|=

𝑥 𝑖 + 𝑦 𝑗 + 𝑧𝑘

𝑥2 + 𝑦2 + 𝑧2




- Velocity

• derivative of 𝑟 with respect to time

• velocity vector 𝑣

𝑣 =𝑑 𝑟

𝑑𝑡= 𝑣𝑥 𝑖 + 𝑣𝑦 𝑗 + 𝑣𝑧𝑘

𝑣𝑥 = 𝑥, 𝑣𝑦 = 𝑦, 𝑣𝑧 = 𝑧

• magnitude of 𝑣

𝑣 = | 𝑣| = 𝑥2 + 𝑦2 + 𝑧2

• unit vector 𝑢𝑣

𝑢𝑣 = 𝑣

| 𝑣|=

𝑣𝑥 𝑖 + 𝑣𝑦 𝑗 + 𝑣𝑧𝑘

𝑥2 + 𝑦2 + 𝑧2




- Acceleration

• derivative of 𝑣 with respect to time

• acceleration vector 𝑎

𝑎 =𝑑 𝑣

𝑑𝑡=

𝑑2 𝑟

𝑑𝑡2= 𝑎𝑥 𝑖 + 𝑎𝑦 𝑗 + 𝑎𝑧𝑘

𝑎𝑥 = 𝑣𝑥 = 𝑥,𝑎𝑦 = 𝑣𝑦 = 𝑦,𝑎𝑧 = 𝑣𝑧 = 𝑧

• magnitude of 𝑎

𝑎 = | 𝑎| = 𝑥2 + 𝑦2 + 𝑧2

• unit vector 𝑢𝑎

𝑢𝑎 = 𝑎

| 𝑎|=

𝑎𝑥 𝑖 + 𝑎𝑦 𝑗 + 𝑎𝑧𝑘

𝑥2 + 𝑦2 + 𝑧2



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- Example 12.9 The horizontal position of the weather balloon

is defined by 𝑥 = 8𝑡 (𝑚). If the equation of

the path is 𝑦 = 𝑥2/10, determine the velocity

and the acceleration when 𝑡 = 2𝑠

Solution

Velocity

𝑣𝑥 = 𝑥 =𝑑

𝑑𝑡8𝑡 = 8𝑚/𝑠

𝑣𝑦 = 𝑦 =𝑑

𝑑𝑡

𝑥2

10=

2

10𝑥 𝑥 =

2

10× 16 × 8 = 25.6𝑚/𝑠

⟹ 𝑣 = 𝑣𝑥2 + 𝑣𝑦

2 = 82 + 25.62 = 26.8𝑚/𝑠

𝜃𝑣 = 𝑡𝑎𝑛−1𝑣𝑦

𝑣𝑥= 𝑡𝑎𝑛−1

25.6

8= 72.60




Acceleration

𝑎𝑥 = 𝑣𝑥 = 𝑥 =𝑑

𝑑𝑡8 = 0𝑚/𝑠

𝑎𝑦 = 𝑣𝑦 = 𝑦 =𝑑

𝑑𝑡

2

10𝑥 𝑥 =

2

10 𝑥 𝑥 + 𝑥 𝑥

=2

10(8 × 8 + 16 × 0) = 12.8𝑚/𝑠2

⟹ 𝑎 = 𝑎𝑥2 + 𝑎𝑦

2 = 02 + 12.82 = 12.8𝑚/𝑠

𝜃𝑎 = 𝑡𝑎𝑛−1𝑎𝑦

𝑎𝑥= 𝑡𝑎𝑛−1

12.8

0= 900




- Example 12.10 For a short time, the path of the plane is

described by 𝑦 = 0.001𝑥2(𝑚). If the plane is rising with a

constant velocity of 10𝑚/𝑠, determine the magnitudes of the

velocity and acceleration of the plane when it is at 𝑦 = 100𝑚

Solution

𝑦 = 100𝑚 → 𝑥 = 1000𝑦 = 3.162𝑚

𝑣𝑦 = 10𝑚/𝑠 → 𝑡 = 𝑦/𝑣𝑦 = 100/10 = 10𝑠

Velocity

𝑣𝑦 = 𝑦 =𝑑

𝑑𝑡0.001𝑥2 = 0.002𝑥𝑣𝑥

⟹ 𝑣𝑥 = 500𝑣𝑦/𝑥 = 500 × 10/3.162 = 15.81𝑚/𝑠

The magnitude of the velocity

𝑣 = 𝑣𝑥2 + 𝑣𝑦

2 = 15.812 + 102 = 18.7𝑚/𝑠




𝑦 = 100𝑚 → 𝑥 = 1000𝑦 = 3.162𝑚

𝑣𝑦 = 10𝑚/𝑠 → 𝑡 = 𝑦/𝑣𝑦 = 100/10 = 10𝑠

𝑣𝑦 = 𝑦 = 0.002𝑥𝑣𝑥

Acceleration

𝑎𝑦 = 𝑣𝑦 =𝑑

𝑑𝑡0.002𝑥𝑣𝑥 = 0.002 𝑥𝑣𝑥 + 𝑥 𝑣𝑥 = 0.002(𝑣𝑥

2 + 𝑥𝑎𝑥)

with 𝑥 = 316.2𝑚, 𝑣𝑥 = 15.81𝑚/𝑠, 𝑣𝑦 = 𝑎𝑦 = 0

⟹ 𝑎𝑥 = −0.791𝑚/𝑠2

The magnitude of the acceleration

𝑎 = 𝑎𝑥2 + 𝑎𝑦

2 = (−0.791)2+02 = 0.791𝑚/𝑠2



§6.Motion of a Projectile

- What is a Projectile?

An object projected into the air at an angle, and once projected

continues in motion by its own inertia and is influenced only by

the downward force of gravity




- Example: release the balls and each picture in this sequence

is taken after the same time interval

• Red ball: ↓ falls from rest

• Yellow ball: → released with given

horizontal velocity

• Both balls accelerate downward

at the same rate, and so they

remain at the same elevation at

any instant

• The horizontal distance between

successive photos of the yellow

ball is constant since the velocity

in the horizontal direction remains

constant



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- History

• Niccolo Tartaglia (1500 –1557), realized that

projectiles actually follow a curved path

Yet no one knew what that path was

• Galileo (1564 –1642) accurately described

projectile motion by showing it could be

analyzed by separately considering the

horizontal and vertical components of motion

Galileo concluded that the path of any

projectile is a parabola




Analysis

- Consider a projectile launched at (𝑥0, 𝑦0) with an initial velocity

𝑣0[ 𝑣0 𝑥, 𝑣0 𝑦], constant downward acceleration 𝑎𝑐 =𝑔=9.81𝑚/𝑠2

- Horizontal motion 𝑎 = 𝑎𝑥 = 0

(+→) 𝑣 = 𝑣0 + 𝑎𝑥𝑡 𝑣𝑥 = (𝑣0)𝑥

(+→) 𝑥 = 𝑥0 + 𝑣0𝑡 + 1

2𝑎𝑥𝑡2 𝑥 = 𝑥0 + (𝑣0)𝑥𝑡

(+→) 𝑣2 = 𝑣02 + 2𝑎𝑥(𝑥 − 𝑥0) 𝑣𝑥 = (𝑣0)𝑥

The horizontal component of velocity always remains constant




- Vertical motion 𝑎 = 𝑎𝑐 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

(+↑) 𝑣 = 𝑣0 + 𝑎𝑐𝑡 𝑣𝑥 = (𝑣0)𝑦−𝑔𝑡

(+↑) 𝑥 = 𝑥0 + 𝑣0𝑡 + 1

2𝑎𝑐𝑡

2 𝑦 = 𝑦0 + (𝑣0)𝑦𝑡 − 1

2𝑔𝑡2

(+↑) 𝑣2 = 𝑣02 + 2𝑎𝑐(𝑥 − 𝑥0) 𝑣𝑦

2 = (𝑣0)𝑦2−2𝑔(𝑦 − 𝑦0)

The exponent of the time term confirms the parabolic shape of

the trajectory




- Example 12.11 A sack slides off the ramp with a horizontal

velocity of (𝑣𝐴)𝑥= 12𝑚/𝑠. If the height of the ramp is 6𝑚 from

the floor, determine the time needed for the sack to strike the

floor and the range 𝑅 where sacks begin to pile up

Solution

Coordinate system

Horizontal motion

(+↑) 𝑦𝐵 = 𝑦𝐴 + (𝑣𝐴)𝑦𝑡𝐴𝐵 + 1

2𝑎𝑦𝑡𝐴𝐵

2

⟹ 𝑡𝐴𝐵 =2𝑦𝐵

𝑎𝑦= 2 ×

−6

−9.81= 1.11𝑠

Vertical motion

(+→) 𝑥𝐵 = 𝑥𝐴 + (𝑣𝐴)𝑥𝑡𝐴𝐵

⟹ 𝑅 = 𝑥𝐵 = 0 + 12 × 1.11 = 13.3𝑚HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien



- Example 12.12 The chipping machine is

designed to eject wood chips

at 𝑣0 = 25𝑚/𝑠. If the tube is

oriented at 300 from the

horizontal, determine how

high, ℎ, the chips strike the

pile if at this instant they land

on the pile 20𝑚 from the tube

Solution

Coordinate system

(𝑣0)𝑥= 25𝑐𝑜𝑠300 = 21.65𝑚/𝑠 →

(𝑣0)𝑦= 25𝑠𝑖𝑛300 = 12.50𝑚/𝑠 ↑

(𝑣𝐴)𝑥= (𝑣0)𝑥= 21.65𝑚/𝑠

𝑎𝑦 = −32.2𝑚/𝑠2




(𝑣0)𝑥= 21.65𝑚/𝑠 →

(𝑣0)𝑦= 12.50𝑚/𝑠 ↑

(𝑣𝐴)𝑥= 21.65𝑚/𝑠

𝑎𝑦 = −32.2𝑚/𝑠2

Horizontal motion

(+↑) 𝑥𝐴 = 𝑥0 + (𝑣0)𝑥𝑡𝑂𝐴 = 0 + (𝑣0)𝑥𝑡𝑂𝐴

⟹ 𝑡𝑂𝐴 =𝑥𝐴

(𝑣0)𝑥=

20

21.65= 0.9238𝑠

Vertical motion

(+↑) 𝑦𝐴 = 𝑦𝑂 + (𝑣0)𝑦𝑡𝑂𝐴 + 1

2𝑎𝑐𝑡𝑂𝐴

2

⟹ ℎ − 4 = 0 + 12.5 × 0.9238 + 12(−32.2)(0.9238)2

⟹ ℎ = 1.81𝑚



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- Example 12.13 The track for this racing event was designed

so that riders jump off the slope at 300, from a height of 1𝑚.

During a race it was observed that the rider remained in mid

air for 1.5𝑠. Determine the speed at which he was traveling off

the ramp, the horizontal distance he travels before striking the

ground, and the maximum height he attains. Neglect the size

of the bike and rider

Solution

Coordinate system

Vertical motion

(+↑) 𝑦𝐵 = 𝑦𝐴 + (𝑣𝐴)𝑦𝑡𝐴𝐵 + 1

2𝑎𝑐𝑡𝐴𝐵

2

⟹ −1 = 0 + 𝑣𝐴𝑠𝑖𝑛300 × 1.5 + 12(−9.81)(1.5)2

⟹ 𝑣𝐴 = 13.38𝑚/𝑠 = 13.4𝑚/𝑠




Coordinate system

Vertical motion

(+↑) 𝑦𝐵 = 𝑦𝐴 + (𝑣𝐴)𝑦𝑡𝐴𝐵 + 1

2𝑎𝑐𝑡𝐴𝐵

2

⟹ −1 = 0 + 𝑣𝐴𝑠𝑖𝑛300 × 1.5 + 12(−9.81)(1.5)2

⟹ 𝑣𝐴 = 13.38𝑚/𝑠 = 13.4𝑚/𝑠

Horizontal motion

(+↑) 𝑥𝐵 = 𝑥𝐴 + (𝑣𝐴)𝑥𝑡𝐴𝐵 ⟹ 𝑅 = 0 + 13.38𝑐𝑜𝑠300 = 17.4𝑚/𝑠

(𝑣𝐶)𝑦2= (𝑣𝐴)𝑦

2+2𝑎𝐶(𝑦𝐶 − 𝑦𝐴)

⟹ 0 = (13.38𝑠𝑖𝑛300)2+2 × (−9.81) × [(ℎ − 1) − 0]

⟹ ℎ = 3.28𝑚




- F12.15 If the 𝑥 and 𝑦 components of a particle's velocity are

𝑣𝑥 = 32𝑡(𝑚/𝑠) and 𝑣𝑥 = 8(𝑚/𝑠), determine the equation of the

path 𝑦 = 𝑓(𝑥). 𝑥 = 0 and 𝑦 = 0 when 𝑡 = 0




- F12.16 A particle is traveling along the straight path. If its

position along the 𝑥 axis is 𝑥 = 8𝑡(𝑚), determine its speed

when 𝑡 = 2𝑠




- F12.17 A particle is constrained to travel along the path. If 𝑥 =4𝑡4(𝑚), determine the magnitude of the particle's velocity and

acceleration when 𝑡 = 0.5𝑠




- F12.18 A particle travels along a straight-line path 𝑦 = 0.5𝑥. If

the 𝑥 component of the particle's velocity is 𝑣𝑥 = 2𝑡2(𝑚/𝑠),determine the magnitude of the particle's velocity and

acceleration when 𝑡 = 4𝑠



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- F12.19 A particle is traveling along the parabolic path 𝑦 =0.25𝑥2 . If 𝑥 = 2𝑡2(𝑚) , determine the magnitude of the

particle's velocity and acceleration when 𝑡 = 2𝑠




- F12.20 The position of a box sliding down the spiral can be

described by 𝑟 = 2 sin 2𝑡 𝑖 + 2 cos 𝑡 𝑗 − 2𝑡2𝑘(𝑚) . Determine

the velocity and acceleration of the box when 𝑡 = 2𝑠




- F12.21 The ball is kicked from point 𝐴 with the initial velocity

𝑣𝐴 = 10𝑚/𝑠. Determine the maximum height ℎ it reaches




- F12.22 The ball is kicked from point 𝐴 with the initial velocity

𝑣𝐴 = 10𝑚/𝑠. Determine the range 𝑅, and the speed when the

ball strikes the ground




- F12.23 Determine the speed at which the basketball at 𝐴must be thrown at the angle of 300 so that it makes it to the

basket at 𝐵




- F12.24 Water is sprayed at an angle of 900 from the slope at

20𝑚/𝑠. Determine the range 𝑅



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- F12.25 A ball is thrown from 𝐴. If it is required to clear the wall

at 𝐵, determine the minimum magnitude of its initial velocity 𝑣𝐴




- F12.26 A projectile is fired with an initial velocity of 𝑣𝐴 =150𝑚/𝑠 off the roof of the building. Determine the range 𝑅where it strikes the ground at 𝐵



§7.Curvilinear Motion: Normal and Tangential Components

- Using for a particle moving along a known curvilinear path

- The position of the particle at the instant is took as the origin

- The coordinate axes: tangential (𝑡), and normal (𝑛) to the path

- Position

• Normal-Tangential Coordinate

• If 𝑦 = 𝑓(𝑥), 𝜌 = 1 +𝑑𝑦

𝑑𝑥

2 3

/𝑑2𝑦

𝑑𝑥2




- Velocity

• Direction: always tangential to the path 𝑠

• Magnitude

𝑠 = 𝑠 𝑡 ⟹ 𝑣 =𝑑𝑠

𝑑𝑡= 𝑠

𝑣 = 𝑣𝑢𝑡




- Acceleration

• The time rate of change of the velocity

𝑎 = 𝑣 = 𝑑 𝑣𝑢𝑡 /𝑑𝑡 = 𝑣𝑢𝑡 + 𝑣 𝑢𝑡

• Redraw the velocity unit vectors at the infinitesimal scale

𝑢𝑡′ = 𝑢𝑡 + 𝑑𝑢𝑡

𝑑𝑢𝑡 = 𝑑𝜃𝑢𝑡

𝑑𝑢𝑡 ↑↑ 𝑢𝑛

• The property of an arc

⟹ 𝑎 = 𝑣𝑢𝑡 + 𝑣 𝑢𝑡

𝑎 = 𝑣𝑢𝑡 +𝑣2

𝜌𝑢𝑛



𝑑𝑢𝑡 = 𝑑𝜃𝑢𝑛 ⟹ 𝑢𝑡 = 𝜃𝑢𝑛

𝑢𝑡 = 𝑠

𝜌𝑢𝑛 =

𝑣

𝜌𝑢𝑛

𝑑𝑠 = 𝜌𝑑𝜃 ⟹ 𝜃 = 𝑠

𝜌


• Acceleration = Tangential Acceleration + Normal Acceleration

𝑎 = 𝑣𝑢𝑡 +𝑣2

𝜌𝑢𝑛 = 𝑎𝑡𝑢𝑡 + 𝑎𝑛𝑢𝑛

• Magnitude of acceleration

𝑎 = 𝑎𝑡2 + 𝑎𝑛

2

• Normal acceleration ≡ Centripetal acceleration



𝑎𝑡 = 𝑣 or 𝑎𝑡𝑑𝑠 = 𝑣𝑑𝑣

𝑎𝑛 =𝑣2

𝜌

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- Special cases of motion

• The particle moves along a straight line

𝜌 → ∞: 𝑎𝑛 = 0, 𝑎 = 𝑎𝑡 = 𝑣

• The particle moves along a curve with a constant speed

𝑣 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡: 𝑎𝑛 = 𝑣 = 0, 𝑎 = 𝑎𝑛 = 𝑣2/𝜌




- Three dimensional motion

𝑡: tangent axis

𝑛: principal normal axis

𝑏: binormal axis

(𝑡, 𝑛):osculating plane (mật tiếp)

𝑢𝑏 = 𝑢𝑡 × 𝑢𝑛

Note: 𝑢𝑛is always on the concave side of the curve




- Example 12.14 When the skier reaches point 𝐴 along the

parabolic path, he has a speed of 6𝑚/𝑠 which is increasing at

2𝑚/𝑠2. Determine the direction of his velocity and the direction

and magnitude of his acceleration at this instant

Solution

Coordinate system

Velocity

𝑦 =1

20𝑥2 →

𝑑𝑦

𝑑𝑥=

1

10𝑥 →

𝑑2𝑦

𝑑𝑥2= 10

𝑥 = 10 →𝑑𝑦

𝑑𝑥= 1 → 𝜃 = 𝑡𝑎𝑛−11 = 450

Therefore

𝑣𝐴 = 6𝑚/𝑠 ↙ 2250




𝜌𝐴 = 1 +𝑑𝑦

𝑑𝑥

2 3𝑑2𝑦

𝑑𝑥2= 1 +

𝑥

10

2 31

10

⟹ 𝜌𝐴𝑥=10𝑚

= 28.18𝑚

𝑎𝐴 = 𝑣𝑢𝑡 +𝑣2

𝜌𝐴𝑢𝑛 = 2𝑢𝑡 +

62

28.28𝑢𝑛

⟹ 𝑎𝐴 = 2𝑢𝑡 + 1.273𝑢𝑛(𝑚/𝑠2)

𝑎 = 22 + 1.2732 = 2.37𝑚/𝑠2

𝜙 = 𝑡𝑎𝑛−12

1.273= 57.50

Therefore

𝑎 = 2.37𝑚/𝑠2 ↙ 192.50



Acceleration


- Example 12.15

A race car 𝐶 travels around the horizontal circular track that

has a radius of 300𝑚. If the car increases its speed at a

constant rate of 7𝑚/𝑠2, starting from rest, determine the time

needed for it to reach an acceleration of 8𝑚/𝑠2. What is its

speed at this instant?

Solution

Coordinate system




Acceleration

𝑣 = 𝑣0 + (𝑎𝑡)𝑐𝑡 = 0 + 7𝑡(𝑚/𝑠)

𝑎𝑛 =𝑣2

𝜌=

(7𝑡)2

300= 0.163𝑡2(𝑚/𝑠2)

𝑎𝑡 = 7(𝑚/𝑠2)

Velocity

𝑣 = 7𝑡 = 7 × 4.87 = 34.1(𝑚/𝑠)



𝑎 = 𝑎𝑡2 + 𝑎𝑛

2

⟹ 𝑡 =82 − 72

0.163= 4.87(𝑠)

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- Example 12.16 The boxes travel along the industrial

conveyor. If a box starts from rest at 𝐴 and increases its speed

such that 𝑎𝑡 = 0.2𝑡(𝑚/𝑠2) , determine the magnitude of its

acceleration when it arrives at point 𝐵

Solution

Coordinate system

Acceleration

𝑎𝑡 = 𝑣 = 0.2𝑡 → 0

𝑣

𝑑𝑣 = 0

𝑡

0.2𝑡𝑑𝑡

→ 𝑣 = 0.1𝑡2

𝑣 =𝑑𝑠

𝑑𝑡= 0.1𝑡2 →

0

6.142

𝑑𝑠 = 0

𝑡𝐵

0.1𝑡2𝑑𝑡

→ 𝑡𝐵 = 0.59𝑠




Acceleration

𝑎𝑡 = 𝑣 = 0.2𝑡 → 0

𝑣

𝑑𝑣 = 0

𝑡

0.2𝑡𝑑𝑡 → 𝑣 = 0.1𝑡2

𝑣 =𝑑𝑠

𝑑𝑡= 0.1𝑡2 →

0

6.142

𝑑𝑠 = 0

𝑡𝐵

0.1𝑡2𝑑𝑡 → 𝑡𝐵 = 0.59𝑠

(𝑎𝐵)𝑡= 𝑣𝐵 = 0.2 × 0.59 = 1.138𝑚/𝑠2

𝑣𝐵 = 0.1 × 5.692 = 3.238𝑚/𝑠

At 𝐵, 𝜌𝐵 = 2𝑚

(𝑎𝐵)𝑛=𝑣𝐵

2

𝜌𝐵=

3.238

2= 5.242𝑚/𝑠2

The magnitude of 𝑎𝐵

𝑎𝐵 = 1.1382 + 5.2422 = 5.36𝑚/𝑠2




- F12.27 The boat is traveling along the circular path with a

speed of 𝑣 = 0.0625𝑡2(𝑚/𝑠). Determine the magnitude of its

acceleration when 𝑡 = 10𝑠




- F12.28 The car is traveling along the road with a speed of 𝑣 =300/𝑠(𝑚/𝑠). Determine the magnitude of its acceleration when

𝑡 = 3𝑠 if 𝑡 = 0 at 𝑠 = 0




- F12.29 If the car decelerates uniformly along the curved road

from 25𝑚/𝑠 at 𝐴 to 15𝑚/𝑠 at 𝐶, determine the acceleration of

the car at 𝐵




- F12.30 When 𝑥 = 10𝑚, the crate has a speed of 20𝑚/𝑠 which

is increasing at 6𝑚/𝑠2. Determine the direction of the crate's

velocity and the magnitude of the crate's acceleration at this

instant



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- F12.31 If the motorcycle has a deceleration of 𝑎𝑡 =− 0.001𝑠(𝑚/𝑠2) and its speed at position 𝐴 is 25𝑚/𝑠 ,

determine the magnitude of its acceleration when it passes

point 𝐵




- F12.32 The car travels up the hill with a speed of 𝑣 =0.2𝑠(𝑚/𝑠), measured from 𝐴. Determine the magnitude of its

acceleration when it is at point 𝑠 = 50𝑚, where 𝜌 = 500𝑚



§8.Curvilinear Motion: Cylindrical Components

- In some cases the motion of a

particle is constrained on a path

amenable to analysis using

cylindrical coordinates

- If the motion is restricted to a

plane, then we can use polar

coordinates




- Polar Coordinates

• A particle is located by

+ a radial coordinate 𝑟, which extends from an origin 𝑂

+ an angle 𝜃 measured counterclockwise form a fixed

reference line to the axis of 𝑟

• 𝑢𝜃, 𝑢𝑟: unit vectors in the directions of increasing 𝜃 and 𝑟

- Position

At any instant, the position vector of the particle

𝑟 = 𝑟𝑢𝑟




- Velocity

• The instantaneous velocity 𝑣

𝑣 = 𝑟 = 𝑟𝑢𝑟 + 𝑟 𝑢𝑟

• During the time interval ∆𝑡

a change ∆𝑟 will not cause a chance in the direction of 𝑢𝑟

a change Δ𝜃 will cause 𝑢𝑟 to become 𝑢𝑟′ : 𝑢𝑟

′ = 𝑢𝑟 + ∆𝑢𝑟

∆𝑢𝑟 ≈ 𝑢𝑟 ∆𝜃 = ∆𝜃

∆𝑢𝑟 ↑↑ 𝑢𝜃



∆𝑢𝑟 = ∆𝜃𝑢𝜃 ⟹ 𝑢𝑟 = lim∆𝑡→0

∆𝑢𝑟

∆𝑡= 𝜃𝑢𝜃


𝑣 = 𝑟𝑢𝑟 + 𝑟 𝑢𝑟

𝑢𝑟 = 𝜃𝑢𝜃

⟹ 𝑣 = 𝑣𝑟𝑢𝑟 + 𝑣𝜃𝑢𝜃

𝑣𝑟 = 𝑟: radial component, a measure of the rate of increase

or decrease in the length of the radial coordinate, 𝑟

𝑣𝜃 = 𝑟 𝜃: transverse component, the rate of motion along

the circumference of a circle having a radius 𝑟

• Magnitude of velocity 𝑣 = 𝑣𝑟2 + 𝑣𝜃

2 = ( 𝑟)2+(𝑟 𝜃)2



𝑣 = 𝑟𝑢𝑟 + 𝑟 𝜃𝑢𝜃

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- Acceleration

• 𝑎 = 𝑣 = 𝑑 𝑟𝑢𝑟 + 𝑟 𝜃𝑢𝜃 /𝑑𝑡

⟹ 𝑎 = 𝑟𝑢𝑟 + 𝑟 𝑢𝑟 + 𝑟 𝜃𝑢𝜃 + 𝑟 𝜃𝑢𝜃 + 𝑟 𝜃 𝑢𝜃

• During the time interval ∆𝑡

a change ∆𝑟 will not cause a chance in the direction of 𝑢𝜃

a change Δ𝜃 will cause 𝑢𝜃 to become 𝑢𝜃′ : 𝑢𝜃

′ = 𝑢𝜃 + ∆𝑢𝜃

∆𝑢𝜃 ≈ 𝑢𝜃 ∆𝜃 = ∆𝜃

∆𝑢𝜃 ↑↓ 𝑢𝑟



∆𝑢𝜃 = −∆𝜃𝑢𝑟 ⟹ 𝑢𝜃 = lim∆𝑡→0

∆ 𝑢𝜃

∆𝑡= − 𝜃 𝑢𝑟


𝑎 = 𝑟𝑢𝑟 + 𝑟 𝑢𝑟 + 𝑟 𝜃𝑢𝜃 + 𝑟 𝜃𝑢𝜃 + 𝑟 𝜃 𝑢𝜃

𝑢𝜃 = − 𝜃𝑢𝑟

𝑢𝑟 = 𝜃𝑢𝜃

⟹ 𝑎 = 𝑎𝑟𝑢𝑟 + 𝑎𝜃𝑢𝜃

𝑎𝑟 = 𝑟 − 𝑟 𝜃2: radial component of acceleration

𝑎𝜃 = 𝑟 𝜃 + 2 𝑟 𝜃: transverse component of acceleration

• Magnitude of acceleration 𝑎 = ( 𝑟 − 𝑟 𝜃2)2+(𝑟 𝜃 + 2 𝑟 𝜃)2



𝑎 = 𝑟 − 𝑟 𝜃2 𝑢𝑟

+(𝑟 𝜃 + 2 𝑟 𝜃)𝑢𝜃


- Cylindrical Coordinates

• If the particle moves along a space curve, then its location

may be specified by the three cylindrical coordinates 𝑟,𝜃,𝑧

𝑟𝑃 = 𝑟𝑢𝑟 + 𝑧𝑢𝑧

𝑣 = 𝑟𝑢𝑟 + 𝑟 𝜃𝑢𝜃 + 𝑧𝑢𝑧

𝑎 = ( 𝑟 − 𝑟 𝜃2)𝑢𝑟 + (𝑟 𝜃 + 2 𝑟 𝜃)𝑢𝜃 + 𝑧𝑢𝑧




𝑟𝑃 = 𝑟𝑢𝑟 + 𝑧𝑢𝑧

𝑣 = 𝑟𝑢𝑟 + 𝑟 𝜃𝑢𝜃 + 𝑧𝑢𝑧

𝑎 = ( 𝑟 − 𝑟 𝜃2)𝑢𝑟 + (𝑟 𝜃 + 2 𝑟 𝜃)𝑢𝜃 + 𝑧𝑢𝑧

• Time derivative

The above equations require 𝑟, 𝑟, 𝜃, 𝜃. Two types of problems

generally occur

+ Given the time parametric equations: 𝑟 = 𝑟(𝑡), 𝜃 = 𝜃(𝑡)

⟹ the time derivatives can be found directly

+ Given the path: 𝑟 = 𝑟(𝑡)

⟹ using the chain rule of calculus to find the relation

between 𝑟 and 𝜃 and between 𝑟 and 𝜃




- Example 12.17 The amusement park ride consists of a chair

that is rotating in a horizontal circular

path of radius 𝑟 such that the arm 𝑂𝐵

has an angular velocity 𝜃 and angular

acceleration 𝜃. Determine the radial and

transverse components of velocity and

acceleration of the passenger

Solution

Coordinate system: polar coordinate ( 𝑟, 𝜃)

𝑟 = 𝑟 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 → 𝑟 = 0, 𝑟 = 0

Velocity

𝑣𝑟 = 𝑟 = 0

𝑣𝜃 = 𝑟 𝜃



Acceleration

𝑎𝑟 = 𝑟 − 𝑟 𝜃2 = −𝑟 𝜃2

𝑎𝜃 = 𝑟 𝜃 + 2 𝑟 𝜃 = 𝑟 𝜃


- Example 12.18 The rod 𝑂𝐴 rotates in the horizontal plane

such that 𝜃 = 𝑡3(𝑟𝑎𝑑). At the same time,

the collar 𝐵 is sliding outward along 𝑂𝐴so that 𝑟 = 100𝑡2(𝑚𝑚). Determine the

velocity and acceleration of the collar

when 𝑡 = 1𝑠

Solution


Time derivative of 𝑟 and 𝜃 at 𝑡 = 1𝑠

𝑟 𝑡 = 100𝑡2 → 𝑟 𝑡 = 200𝑡 𝑟 𝑡 = 200

𝜃 𝑡 = 𝑡3 → 𝜃 𝑡 = 3𝑡2 𝜃 𝑡 = 6𝑡

⟹ 𝑟 1 = 100 𝑟 1 = 200 𝑟 1 = 200

𝜃 1 = 1 𝜃 1 = 3 𝜃 1 = 6



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Velocity

𝑣 = 𝑟𝑢𝑟 + 𝑟 𝜃𝑢𝜃 = 200𝑢𝑟 + 300𝑢𝜃

𝑣 = 2002 + 3002 = 361𝑚𝑚/𝑠

𝛿 = 𝑡𝑎𝑛−1 300

200= 56.30, 𝛿 + 57.30 = 1140

Acceleration

𝑎 = ( 𝑟 − 𝑟 𝜃2)𝑢𝑟 + (𝑟 𝜃 + 2 𝑟 𝜃)𝑢𝜃

= 200 − 100 × 32 𝑢𝑟

+(100×6+2×200×3) 𝑢𝜃(𝑚𝑚/𝑠)

= −700𝑢𝑟 + 1800𝑢𝜃(𝑚𝑚/𝑠2)

𝑎 = 7002 + 18002 = 1930𝑚𝑚/𝑠2

𝜙 = 𝑡𝑎𝑛−1 1800

700= 68.70

(1800 − 𝜙) + 57.30 = 1690




- Example 12.19 The searchlight casts a spot of light along the

face of a wall that is located 100𝑚 from the searchlight.

Determine the magnitudes of the velocity and acceleration at

which the spot appears to travel across the wall at the instant

𝜃 = 450. The searchlight rotates at a constant rate of 𝜃 = 4𝑟𝑎𝑑/𝑠

Solution


𝑟 = 100/𝑐𝑜𝑠𝜃 = 100𝑠𝑒𝑐𝜃

Time derivative of 𝑟 and 𝜃

𝑑 𝑠𝑒𝑐𝜃 = 𝑠𝑒𝑐𝜃𝑡𝑎𝑛𝜃𝑑𝜃 𝑑 𝑡𝑎𝑛𝜃 = 𝑠𝑒𝑐2𝜃𝑑𝜃

𝑟 = 100𝑠𝑒𝑐𝜃 → 𝑟 = 100𝑠𝑒𝑐𝜃𝑡𝑎𝑛𝜃 𝜃

→ 𝑟 = 100𝑠𝑒𝑐𝜃𝑡𝑎𝑛2𝜃 𝜃2 +100𝑠𝑒𝑐3𝜃 𝜃2 +100𝑠𝑒𝑐𝜃𝑡𝑎𝑛𝜃 𝜃

𝜃 = 450 𝜃 = 4𝑟𝑎𝑑/𝑠

→ 𝜃 = 0HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien



At 𝜃 = 450, 𝑟 = 100𝑠𝑒𝑐450 = 141.4

𝑟 = 400𝑠𝑒𝑐450𝑡𝑎𝑛450 = 564.7

𝑟 = 1600 𝑠𝑒𝑐450𝑡𝑎𝑛2450 + 𝑠𝑒𝑐3450 = 6788.2

Velocity

𝑣 = 𝑟𝑢𝑟 + 𝑟 𝜃𝑢𝜃

= 565.7𝑢𝑟 + 565.7𝑢𝜃

𝑣 = 565.72 + 565.72 = 800𝑚/𝑠

Acceleration

𝑎 = 𝑟 − 𝑟 𝜃2 𝑢𝑟 + 𝑟 𝜃 + 2 𝑟 𝜃 𝑢𝜃

= 4525.5𝑢𝑟 + 4525.5𝑢𝜃

𝑎 = 4525.52 + 4525.52 = 6400𝑚/𝑠2




- Example 12.20 Due to the rotation of the forked rod, the ball

travels around the slotted path, a portion

of which is in the shape of a cardioid,

𝑟 = 0.5(1 − 𝑐𝑜𝑠𝜃). If the ball’s velocity is

4𝑚/𝑠 and its acceleration is 30𝑚/𝑠2 at

the instant 𝜃 = 1800 , determine the

angular velocity 𝜃 and angular

acceleration 𝜃 of the fork

Solution


Time derivative of 𝑟 and 𝜃

𝑟 𝑡 = 0.5 1 − 𝑐𝑜𝑠𝜃 → 𝑟 𝑡 = 0.5𝑠𝑖𝑛𝜃 𝜃

→ 𝑟 𝑡 = 0.5𝑐𝑜𝑠𝜃 𝜃2 + 0.5𝑠𝑖𝑛𝜃 𝜃

At 𝜃 = 1800: 𝑟 = 1, 𝑟 = 0, 𝑟 = −0.5 𝜃2




𝑣 = ( 𝑟)2+(𝑟 𝜃)2

⟹ 𝜃 =𝑣2 − 𝑟2

𝑟

=42 − 02

1= 4𝑟𝑎𝑑/𝑠

𝑎 = ( 𝑟 − 𝑟 𝜃2)2+(𝑟 𝜃 + 2 𝑟 𝜃)2

⟹ 𝜃 =𝑎2 − 𝑟 − 𝑟 𝜃2 2

− 2 𝑟 𝜃

𝑟

=302 − −0.5 × 42 − 1 × 42 2 − 2 × 0 × 4

1= 18𝑟𝑎𝑑/𝑠2




- F12.33 The car has a speed of 55𝑚/𝑠. Determine the angular

velocity 𝜃 of the radial line 𝑂𝐴 at this instant



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- F12.34 The platform is rotating about the vertical axis such

that at any instant its angular position is 𝜃 = 4 𝑡3. A ball rolls

outward along the radial groove so that its position is 𝑟 = 0.1𝑡3.

Determine the magnitudes of the velocity and acceleration of

the ball when 𝑡 = 1.5𝑠




- F12.35 Peg 𝑃 is driven by the fork link 𝑂𝐴 along the curved

path described by 𝑟 = 2𝜃(𝑚) . At the instant 𝜃 = 𝜋/4 , the

angular velocity and angular acceleration of the link are 𝜃 =3𝑟𝑎𝑑/𝑠 and 𝜃 = 1𝑟𝑎𝑑/𝑠2 . Determine the magnitude of the

peg’s acceleration at this instant




- F12.36 Peg 𝑃 is driven by the forked link 𝑂𝐴 along the path

described by 𝑟 = 𝑒𝜃. When 𝜃 = 𝜋/4, the link has an angular

velocity and angular acceleration of 𝜃 = 2𝑟𝑎𝑑/𝑠 and 𝜃 =4𝑟𝑎𝑑/𝑠2. Determine the radial and transverse components of

the peg’s acceleration at this instant




- F12.37 The collars are pin-connected at 𝐵 and are free to

move along rod 𝑂𝐴 and the curved guide 𝑂𝐶 having the shape

of a cardioid, 𝑟 = 0.2 1 + 𝑐𝑜𝑠𝜃 . At 𝜃 = 300 , the angular

velocity of 𝑂𝐴 is 𝜃 = 3𝑟𝑎𝑑/𝑠. Determine the magnitudes of the

velocity of the collars at this point




- F12.38 At the instant 𝜃 = 450, the athlete is running with a

constant speed of 2𝑚/𝑠. Determine the angular velocity at

which the camera must turn in order to follow the motion



§9.Absolute Dependent Motion Analysis of Two Particles

• In some cases the motion of one particle depends on the

motion of another

• Consider two objects physically interconnected by inextensible

chords of a pulley system. Choose the coordinate system

+ measured from a fixed point (𝑂 ) or

fixed datum line

+ measured along each inclined plane in

the direction of motion of each block

+ has a positive sense from 𝐶 → 𝐴,𝐷 → 𝐵

The total cord length 𝑙𝑇 = 𝑠𝐴 + 𝑙𝑐 + 𝑠𝐵 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

⟹𝑑𝑠𝐴

𝑑𝑡+

𝑑𝑠𝐵

𝑑𝑡= 𝑣𝐴 + 𝑣𝐵 = 0 ⟹ 𝑣𝐴 = −𝑣𝐵

⟹𝑑𝑣𝐴

𝑑𝑡+

𝑑𝑣𝐵

𝑑𝑡= 𝑎𝐴 + 𝑎𝐵 = 0 ⟹ 𝑎𝐴 = −𝑎𝐵



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Another example

𝑙𝑇 = 𝑠𝐴 + ℎ + 𝑙 + 2𝑠𝐵 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑙: total length of curved cords

⟹𝑑𝑠𝐴

𝑑𝑡+

𝑑𝑠𝐵

𝑑𝑡= 𝑣𝐴 + 2𝑣𝐵 = 0 ⟹ 𝑣𝐴 = −2𝑣𝐵

⟹𝑑𝑣𝐴

𝑑𝑡+ 2

𝑑𝑣𝐵

𝑑𝑡= 𝑎𝐴 + 2𝑎𝐵 = 0 ⟹ 𝑎𝐴 = −2𝑎𝐵




By choosing the different coordinate

𝑙𝑇 = 𝑠𝐴 + ℎ + 𝑙 + 2𝑠𝐵 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑙: total length of curved cords

⟹𝑑𝑠𝐴

𝑑𝑡+

𝑑𝑠𝐵

𝑑𝑡= 𝑣𝐴 + 2𝑣𝐵 = 0 ⟹ 𝑣𝐴 = −2𝑣𝐵

⟹𝑑𝑣𝐴

𝑑𝑡+ 2

𝑑𝑣𝐵

𝑑𝑡= 𝑎𝐴 + 2𝑎𝐵 = 0 ⟹ 𝑎𝐴 = −2𝑎𝐵




- Example 12.21 Determine the speed of block 𝐴 if block 𝐵 has

an upward speed of 6𝑚/𝑠

Solution

Position coordinate equation

𝑠𝐴 + 3𝑠𝐵 + 𝑙𝑐 = 𝑙

Time derivative

𝑣𝐴 + 3𝑣𝐵 = 0

⟹ 𝑣𝐴 = −3𝑣𝐵

= − −6𝑚/𝑠

= 6𝑚/𝑠 ↓




- Example 12.22 Determine the speed of block 𝐴 if block 𝐵 has

an upward speed of 6𝑚/𝑠

Solution

Position coordinate equations

𝑠𝐴 + 2𝑠𝐵 + 𝑙𝑐1= 𝑙1

𝑠𝐵 + (𝑠𝐵 − 𝑠𝑐) + 𝑙𝑐2= 𝑙2

Time derivatives

𝑣𝐴 + 2𝑣𝐵 = 0

2𝑣𝐵 − 𝑣𝐶 = 0

⟹ 𝑣𝐴 = −4𝑣𝐵

= −4 −6𝑚/𝑠

= 24𝑚/𝑠 ↓




- Example 12.23 Determine the speed of block 𝐵 if the end of the

cord at 𝐴 is pulled down with a speed of 2𝑚/𝑠

Solution

Position coordinate equations

𝑠𝐶 + 𝑠𝐵 + 𝑙𝑐1= 𝑙1

𝑠𝐴 − 𝑠𝐶 + 𝑠𝐵 − 𝑠𝑐 + 𝑠𝐵 + 𝑙𝑐2= 𝑙2

Time derivatives

𝑣𝐶 + 𝑣𝐵 = 0

𝑣𝐴 + 2𝑣𝐵 − 2𝑣𝐶 = 0

⟹ 𝑣𝐵 = −𝑣𝐵/4

= − −2 /4

= 0.5𝑚/𝑠 ↑




- Example 12.24 A man at 𝐴 is hoisting a safe 𝑆 by walking to

the right with a constant velocity 𝑣𝐴 = 0.5𝑚/𝑠 .

Determine the velocity and acceleration of the safe

when it reaches the elevation of 10𝑚. The rope is

30𝑚 long and passes over a small pulley at 𝐷

Solution

Position coordinate equation

𝑙𝐴𝐷 + 𝑙𝐷𝐶 + 𝑙𝐶 = 𝑙

⟹ 152 + 𝑥2 + 15 − 𝑦 = 30

⟹ 𝑦 = 225 + 𝑥2 − 15

Time derivatives

𝑣𝑆 =𝑑𝑦

𝑑𝑡=

1

2

2𝑥

225 + 𝑥2

𝑑𝑥

𝑑𝑡=

𝑥

225 + 𝑥2𝑣𝐴



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𝑣𝑆 =𝑥

225 + 𝑥2𝑣𝐴

At 𝑦 = 10𝑚 → 𝑥 = 20𝑚 with 𝑣𝐴 = 0.5𝑚/𝑠

𝑣𝑆 =20

225 + 202× 0.5 = 0.4𝑚/𝑠 ↑

𝑎𝑆 =𝑑2𝑦

𝑑𝑡2

=−𝑥(𝑑𝑥/𝑑𝑡)𝑥𝑣𝐴

(225 + 𝑥2)3+

(𝑑𝑥/𝑑𝑡)𝑣𝐴

225 + 𝑥2

+𝑑𝑣𝐴/𝑑𝑡 𝑥

225 + 𝑥2

At 𝑥 = 20𝑚, 𝑣𝐴 = 0.5𝑚/𝑠

𝑎𝑆 =225 × 0.5

(225 + 202)3= 0.00360𝑚/𝑠2 = 3.6𝑚𝑚/𝑠2 ↑



§10. Relative Motion of Two Particles Using Translating Axes

In some cases it is easier to analyze

the motion using two or ` more

frames of reference

⟹ Analysis translating frames of

reference to change our point of

view of the object(s) in motion

- Position

• Consider particles 𝐴 and 𝐵, which move along the arbitrary

paths

• Fixed reference frame: 𝑂, 𝑥, 𝑦, 𝑧

• Second reference frame: 𝐴, 𝑥′, 𝑦′, 𝑧′

• Relation

𝑟𝐵 = 𝑟𝐴 + 𝑟𝐵/𝐴



𝑟𝐴, 𝑟𝐵: absolute position vectors

𝑟𝐵/𝐴: relative position vector


- Velocity

𝑑

𝑑𝑡 𝑟𝐵 =

𝑑

𝑑𝑡 𝑟𝐴 +

𝑑

𝑑𝑡 𝑟𝐵/𝐴

⟹ 𝑣𝐵 = 𝑣𝐴 + 𝑣𝐵/𝐴

𝑣𝐴, 𝑣𝐵:absolute velocity

𝑣𝐵/𝐴: relative velocity

- Acceleration

𝑑

𝑑𝑡 𝑣𝐵 =

𝑑

𝑑𝑡 𝑣𝐴 +

𝑑

𝑑𝑡 𝑣𝐵/𝐴

⟹ 𝑎𝐵 = 𝑎𝐴 + 𝑎𝐵/𝐴

𝑎𝐴, 𝑎𝐵:absolute acceleration

𝑎𝐵/𝐴: relative acceleration




- Example 12.25 A train travels at a constant speed of 60𝑘𝑚/ℎcrosses over a road. If the automobile 𝐴is traveling at 45𝑘𝑚/ℎ along the road,

determine the magnitude and direction of the

velocity of the train relative to the automobile

Solution

Vector analysis

𝑣𝑇 = 𝑣𝐴 + 𝑣𝑇/𝐴

⟹ 60 𝑖 = 45𝑐𝑜𝑠450 𝑖 + 45𝑠𝑖𝑛450 𝑗 + 𝑣𝑇/𝐴

⟹ 𝑣𝑇/𝐴 = 28.2 𝑖 − 31.8 𝑗(𝑘𝑚/ℎ)

⟹ 𝑣𝑇/𝐴 = 28.2 2 + −31.8 2 = 42.5 (𝑘𝑚/ℎ)

The direction of 𝑣𝑇/𝐴



𝑡𝑎𝑛𝜃 =𝑣𝑇/𝐴

𝑦

𝑣𝑇/𝐴𝑥 =

31.8

28.2⟹ 𝜃 = 48.50 ↘


Scalar analysis

𝑣𝑇 = 𝑣𝐴 + 𝑣𝑇/𝐴 ⟹ 𝑣𝑇 = 𝑣𝐴 + 𝑣𝑇/𝐴𝑥 + 𝑣𝑇/𝐴

𝑦

𝑣𝑇 = 𝑣𝐴 + 𝑣𝑇/𝐴𝑥 + 𝑣𝑇/𝐴

𝑦

→ ↗ 450 → ↑60𝑘𝑚/ℎ 45𝑘𝑚/ℎ ? ?

Resolving each vector into its 𝑥 and 𝑦components

(+⟶): 60 = 45𝑐𝑜𝑠450 + 𝑣𝑇/𝐴𝑥 + 0

(+ ↑): 0 = 45𝑠𝑖𝑛450 + 0 + 𝑣𝑇/𝐴𝑦

⟹ 𝑣𝑇/𝐴𝑥 = 28.2𝑘𝑚/ℎ = 28.2𝑘𝑚/ℎ →

𝑣𝑇/𝐴𝑦

= −31.8𝑘𝑚/ℎ = 31.8𝑘𝑚/ℎ ↓

⟹ 𝑣𝑇/𝐴 = 28.2 2 + −31.8 2 = 42.5 (𝑘𝑚/ℎ)




- Example 12.26 Plane 𝐴 is flying along a straight-line path,

here as plane 𝐵 is flying along a circular

path having a radius of curvature of

𝜌𝐵 = 400𝑘𝑚 . Determine the velocity

and acceleration of 𝐵 as measured by

the pilot of 𝐴

Solution

Velocity

𝑣𝐵 = 𝑣𝐴 + 𝑣𝐵/𝐴

↑ ↑ ↑600𝑘𝑚/ℎ 700𝑘𝑚/ℎ ?

(+↑): 𝑣𝐵 = 𝑣𝐴 + 𝑣𝐵/𝐴

⟹ 𝑣𝐵/𝐴 = 𝑣𝐵 − 𝑣𝐴 = 600 − 700 = −100𝑘𝑚/ℎ = 100𝑘𝑚/ℎ ↓



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Acceleration

𝑎𝐵𝑛 =

𝑣𝐵2

𝜌𝐵=

6002

400= 900𝑘𝑚/ℎ2

Relative acceleration equation

𝑎𝐵𝑛 + 𝑎𝐵

𝜏 = 𝑎𝐴 + 𝑎𝐵/𝐴

→ ↓ ↑ ?900 100 50 ?

⟹ 900 𝑖 − 100 𝑗 = 50 𝑗 + 𝑎𝐵/𝐴

⟹ 𝑎𝐵/𝐴 = 900 𝑖 − 150 𝑗

The magnitude and direction of 𝑎𝐵/𝐴

𝑎𝐵/𝐴 = 9002 + (−150)2= 912𝑘𝑚/ℎ2

𝜃 = 𝑡𝑎𝑛−1150

900= 9.460




- Example 12.27 At the instant, cars 𝐴and 𝐵 are traveling with speeds of

18𝑚/𝑠and 12𝑚/𝑠 respectively. Also at

this instant, 𝐴 has a decrease in speed

of 2𝑚/𝑠2 and 𝐵 has an increase in

speed of 3𝑚/𝑠2. Determine the velocity

and acceleration of 𝐵 with respect to 𝐴

Solution

Relative velocity equation

𝑣𝐵 = 𝑣𝐴 + 𝑣𝐵/𝐴

↓ ↙ 2400 ?12𝑚/𝑠 18𝑚/𝑠 ?

⟹ −12 𝑗 = (−18𝑐𝑜𝑠600 𝑖−18𝑠𝑖𝑛600 𝑗)+ 𝑣𝐵/𝐴 ⟹ 𝑣𝐵/𝐴 = −9 𝑖+ 3.588 𝑗

𝑣𝐵/𝐴 = 92 + 3.5882 = 9.69𝑚/𝑠, 𝜃 = 𝑡𝑎𝑛−1(3.588/9) = 21.60




- Acceleration

𝑎𝐵𝑛 =

𝑣𝐵2

𝜌=

122

100= 1.440𝑚/𝑠2

Relative acceleration equation

𝑎𝐵 = 𝑎𝐴 + 𝑎𝐵/𝐴

↓ ↗ 600 ?3𝑚/𝑠2 2𝑚/𝑠2 ?

⟹ −1.44 𝑖−3 𝑗 = 2𝑐𝑜𝑠600 𝑖−4.732 𝑗 + 𝑎𝐵/𝐴

⟹ 𝑎𝐵/𝐴 = −2.440 𝑖 − 4.732 𝑗(𝑚/𝑠2)

The magnitude and direction of 𝑎𝐵/𝐴

𝑎𝐵/𝐴 = 2.4402 + 4.7322 = 5.32𝑚/𝑠2

𝜃 = 𝑡𝑎𝑛−14.732

2.440= 62.70




- F12.39 Determine the speed of block 𝐷 if end 𝐴 of the rope is

pulled down with a speed of 𝑣𝐴 = 3𝑚/𝑠




- F12.40 Determine the speed of block 𝐴 if end 𝐵 of the rope is

pulled down with a speed of 6𝑚/𝑠




- F12.41 Determine the speed of block 𝐴 if end 𝐵 of the rope is

pulled down with a speed of 1.5𝑚/𝑠



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- F12.42 Determine the speed of block 𝐴 if end 𝐹 of the rope is

pulled down with a speed of 𝑣𝐹 = 3𝑚/𝑠




- F12.43 Determine the speed of car 𝐴 if point 𝑃 on the cable

has a speed of 4𝑚/𝑠 when the motor 𝑀 winds the cable in




- F12.44 Determine the speed of cylinder 𝐵 if cylinder 𝐴 moves

downward with a speed of 𝑣𝐴 = 4𝑚/𝑠




- F12.45 Car 𝐴 is traveling with a constant speed of 80𝑘𝑚/ℎdue north, while car 𝐵 is traveling with a constant speed of

100𝑘𝑚/ℎ due east. Determine the velocity of car 𝐵 relative to

car 𝐴




- F12.46 Two planes 𝐴 and 𝐵 are traveling with the constant

velocities shown. Determine the magnitude and direction of the

velocity of plane 𝐵 relative to plane 𝐴




- F12.47 The boats 𝐴 and 𝐵 travel with constant speeds of 𝑣𝐴 =15𝑚/𝑠 and 𝑣𝐵 = 10𝑚/𝑠 when they leave the pier at 𝑂 at the

same time. Determine the distance between them when 𝑡 = 4𝑠



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- F12.48 At the instant shown, cars 𝐴 and 𝐵 are traveling at the

speeds shown. If 𝐵 is accelerating at 1200𝑘𝑚/ℎ2 while 𝐴maintains a constant speed, determine the velocity and

acceleration of 𝐴 with respect to 𝐵



ch.12 kinematics of a particle

Documents

rectilinear motion

continuous motion position

study particle motion

laws of motion

motion kinetics

continuous motion assumptions

analysis of dependent

position coordinate