Download - Ch.12 Kinematics of a Particle
5/26/2013
1
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.03 Kinematics of a Particle
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.04 Kinematics of a Particle
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.05 Kinematics of a Particle
§2.Rectilinear Kinematics: Continuous Motion
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.06 Kinematics of a Particle
5/26/2013
2
§2.Rectilinear Kinematics: Continuous Motion
- 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
Engineering Mechanics – Statics 12.07 Kinematics of a Particle
§2.Rectilinear Kinematics: Continuous Motion
- Velocity
• Average velocity
𝑣𝑎𝑣𝑔 =∆𝑠
∆𝑡(𝑚/𝑠)
• Instantaneous velocity
𝑣 = lim∆𝑡→0
∆𝑠
∆𝑡=
𝑑𝑠
𝑑𝑡(𝑚/𝑠)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.08 Kinematics of a Particle
∆𝑠: displacement, 𝑚
∆𝑡: time interval, 𝑠
§2.Rectilinear Kinematics: Continuous Motion
- Speed
• Speed: the magnitude of velocity
• Average speed
(𝑣𝑠𝑝)𝑎𝑣𝑔=𝑠𝑇
∆𝑡(𝑚/𝑠)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.09 Kinematics of a Particle
𝑠𝑇: traveled distance, 𝑚
∆𝑡: elapsed time, 𝑠
§2.Rectilinear Kinematics: Continuous Motion
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.10 Kinematics of a Particle
∆𝑣:velocity, 𝑚/𝑠
∆𝑡: time interval, 𝑠
𝑎𝑑𝑠 = 𝑣𝑑𝑣
§2.Rectilinear Kinematics: Continuous Motion
- 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)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.11 Kinematics of a Particle
§2.Rectilinear Kinematics: Continuous Motion
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.12 Kinematics of a Particle
5/26/2013
3
§2.Rectilinear Kinematics: Continuous Motion
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
Engineering Mechanics – Statics 12.13 Kinematics of a Particle
§2.Rectilinear Kinematics: Continuous Motion
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.14 Kinematics of a Particle
§2.Rectilinear Kinematics: Continuous Motion
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 ↓
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.15 Kinematics of a Particle
§2.Rectilinear Kinematics: Continuous Motion
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.16 Kinematics of a Particle
§2.Rectilinear Kinematics: Continuous Motion
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.17 Kinematics of a Particle
§2.Rectilinear Kinematics: Continuous Motion
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.18 Kinematics of a Particle
Solution
5/26/2013
4
§2.Rectilinear Kinematics: Continuous Motion
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.19 Kinematics of a Particle
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.20 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.21 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.22 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.23 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.24 Kinematics of a Particle
5/26/2013
5
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.25 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.26 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.27 Kinematics of a Particle
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.28 Kinematics of a Particle
§3.Rectilinear Kinematics: Erratic Motion
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.29 Kinematics of a Particle
§3.Rectilinear Kinematics: Erratic Motion
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.30 Kinematics of a Particle
5/26/2013
6
§3.Rectilinear Kinematics: Erratic Motion
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.31 Kinematics of a Particle
§3.Rectilinear Kinematics: Erratic Motion
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.32 Kinematics of a Particle
§3.Rectilinear Kinematics: Erratic Motion
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.33 Kinematics of a Particle
§3.Rectilinear Kinematics: Erratic Motion
- 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)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.34 Kinematics of a Particle
§3.Rectilinear Kinematics: Erratic Motion
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)
0 ≤ 𝑡 < 10: 𝑣 =𝑑𝑠
𝑑𝑡= 2𝑡(𝑚/𝑠)
⟹ 𝑎 =𝑑𝑣
𝑑𝑡= 2(𝑚/𝑠2)
10 ≤ 𝑡 ≤ 30: 𝑠 = 20 (𝑚)
⟹ 𝑣 =𝑑𝑠
𝑑𝑡= 0(𝑚/𝑠)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.35 Kinematics of a Particle
𝑎 − 𝑡 graph
§3.Rectilinear Kinematics: Erratic Motion
- 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)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.36 Kinematics of a Particle
5/26/2013
7
§3.Rectilinear Kinematics: Erratic Motion
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)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.37 Kinematics of a Particle
§3.Rectilinear Kinematics: Erratic Motion
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)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.38 Kinematics of a Particle
§3.Rectilinear Kinematics: Erratic Motion
- 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)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.39 Kinematics of a Particle
§3.Rectilinear Kinematics: Erratic Motion
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)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.40 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.41 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.42 Kinematics of a Particle
5/26/2013
8
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.43 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.44 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.45 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.46 Kinematics of a Particle
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.47 Kinematics of a Particle
§4.General Curvilinear Motion
- Velocity
• average velocity
𝑣𝑎𝑣𝑔 =∆ 𝑟
∆𝑡(𝑚/𝑠)
• instantaneous velocity
𝑣 = lim∆𝑡→0
∆ 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.48 Kinematics of a Particle
∆ 𝑟: displacement, 𝑚
∆𝑡: time interval, 𝑠
5/26/2013
9
§4.General Curvilinear Motion
- Speed
• speed: magnitude of 𝑣
∆𝑡 → 0: ∆ 𝑟 → ∆ 𝑠
𝑣 = lim∆𝑡→0
∆𝑟
∆𝑡= lim
∆𝑡→0
∆𝑠
∆𝑡=
𝑑𝑠
𝑑𝑡(𝑚/𝑠)
• speed can be obtained by differentiating the path function 𝑠with respect to time
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.49 Kinematics of a Particle
§4.General Curvilinear Motion
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.50 Kinematics of a Particle
§4.General Curvilinear Motion
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.51 Kinematics of a Particle
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.52 Kinematics of a Particle
§5.Curvilinear Motion: Rectangular Components
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.53 Kinematics of a Particle
§5.Curvilinear Motion: Rectangular Components
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.54 Kinematics of a Particle
5/26/2013
10
§5.Curvilinear Motion: Rectangular Components
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.55 Kinematics of a Particle
§5.Curvilinear Motion: Rectangular Components
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.55 Kinematics of a Particle
§5.Curvilinear Motion: Rectangular Components
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.57 Kinematics of a Particle
§5.Curvilinear Motion: Rectangular Components
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.58 Kinematics of a Particle
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.59 Kinematics of a Particle
§6.Motion of a Projectile
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.60 Kinematics of a Particle
5/26/2013
11
§6.Motion of a Projectile
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.61 Kinematics of a Particle
§6.Motion of a Projectile
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.62 Kinematics of a Particle
§6.Motion of a Projectile
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.63 Kinematics of a Particle
§6.Motion of a Projectile
- 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
Engineering Mechanics – Statics 12.64 Kinematics of a Particle
§6.Motion of a Projectile
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.65 Kinematics of a Particle
§6.Motion of a Projectile
(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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.66 Kinematics of a Particle
5/26/2013
12
§6.Motion of a Projectile
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.67 Kinematics of a Particle
§6.Motion of a Projectile
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.68 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.69 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.70 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.71 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.72 Kinematics of a Particle
5/26/2013
13
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.73 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.74 Kinematics of a Particle
Fundamental Problems
- F12.21 The ball is kicked from point 𝐴 with the initial velocity
𝑣𝐴 = 10𝑚/𝑠. Determine the maximum height ℎ it reaches
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.75 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.76 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.77 Kinematics of a Particle
Fundamental Problems
- F12.24 Water is sprayed at an angle of 900 from the slope at
20𝑚/𝑠. Determine the range 𝑅
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.78 Kinematics of a Particle
5/26/2013
14
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.79 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.80 Kinematics of a Particle
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.81 Kinematics of a Particle
§7.Curvilinear Motion: Normal and Tangential Components
- Velocity
• Direction: always tangential to the path 𝑠
• Magnitude
𝑠 = 𝑠 𝑡 ⟹ 𝑣 =𝑑𝑠
𝑑𝑡= 𝑠
𝑣 = 𝑣𝑢𝑡
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.82 Kinematics of a Particle
§7.Curvilinear Motion: Normal and Tangential Components
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.83 Kinematics of a Particle
𝑑𝑢𝑡 = 𝑑𝜃𝑢𝑛 ⟹ 𝑢𝑡 = 𝜃𝑢𝑛
𝑢𝑡 = 𝑠
𝜌𝑢𝑛 =
𝑣
𝜌𝑢𝑛
𝑑𝑠 = 𝜌𝑑𝜃 ⟹ 𝜃 = 𝑠
𝜌
§7.Curvilinear Motion: Normal and Tangential Components
• Acceleration = Tangential Acceleration + Normal Acceleration
𝑎 = 𝑣𝑢𝑡 +𝑣2
𝜌𝑢𝑛 = 𝑎𝑡𝑢𝑡 + 𝑎𝑛𝑢𝑛
• Magnitude of acceleration
𝑎 = 𝑎𝑡2 + 𝑎𝑛
2
• Normal acceleration ≡ Centripetal acceleration
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.84 Kinematics of a Particle
𝑎𝑡 = 𝑣 or 𝑎𝑡𝑑𝑠 = 𝑣𝑑𝑣
𝑎𝑛 =𝑣2
𝜌
5/26/2013
15
§7.Curvilinear Motion: Normal and Tangential Components
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.85 Kinematics of a Particle
§7.Curvilinear Motion: Normal and Tangential Components
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.86 Kinematics of a Particle
§7.Curvilinear Motion: Normal and Tangential Components
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.87 Kinematics of a Particle
§7.Curvilinear Motion: Normal and Tangential Components
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.88 Kinematics of a Particle
Acceleration
§7.Curvilinear Motion: Normal and Tangential Components
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.89 Kinematics of a Particle
§7.Curvilinear Motion: Normal and Tangential Components
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)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.90 Kinematics of a Particle
𝑎 = 𝑎𝑡2 + 𝑎𝑛
2
⟹ 𝑡 =82 − 72
0.163= 4.87(𝑠)
5/26/2013
16
§7.Curvilinear Motion: Normal and Tangential Components
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.91 Kinematics of a Particle
§7.Curvilinear Motion: Normal and Tangential Components
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.92 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.93 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.94 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.95 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.96 Kinematics of a Particle
5/26/2013
17
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.97 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.98 Kinematics of a Particle
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.99 Kinematics of a Particle
§8.Curvilinear Motion: Cylindrical Components
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.100 Kinematics of a Particle
§8.Curvilinear Motion: Cylindrical Components
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.101 Kinematics of a Particle
∆𝑢𝑟 = ∆𝜃𝑢𝜃 ⟹ 𝑢𝑟 = lim∆𝑡→0
∆𝑢𝑟
∆𝑡= 𝜃𝑢𝜃
§8.Curvilinear Motion: Cylindrical Components
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.104 Kinematics of a Particle
𝑣 = 𝑟𝑢𝑟 + 𝑟 𝜃𝑢𝜃
5/26/2013
18
§8.Curvilinear Motion: Cylindrical Components
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.103 Kinematics of a Particle
∆𝑢𝜃 = −∆𝜃𝑢𝑟 ⟹ 𝑢𝜃 = lim∆𝑡→0
∆ 𝑢𝜃
∆𝑡= − 𝜃 𝑢𝑟
§8.Curvilinear Motion: Cylindrical Components
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.104 Kinematics of a Particle
𝑎 = 𝑟 − 𝑟 𝜃2 𝑢𝑟
+(𝑟 𝜃 + 2 𝑟 𝜃)𝑢𝜃
§8.Curvilinear Motion: Cylindrical Components
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.105 Kinematics of a Particle
§8.Curvilinear Motion: Cylindrical Components
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.106 Kinematics of a Particle
§8.Curvilinear Motion: Cylindrical Components
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.107 Kinematics of a Particle
Acceleration
𝑎𝑟 = 𝑟 − 𝑟 𝜃2 = −𝑟 𝜃2
𝑎𝜃 = 𝑟 𝜃 + 2 𝑟 𝜃 = 𝑟 𝜃
§8.Curvilinear Motion: Cylindrical Components
- 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
Coordinate system: polar coordinate ( 𝑟, 𝜃)
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.108 Kinematics of a Particle
5/26/2013
19
§8.Curvilinear Motion: Cylindrical Components
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.109 Kinematics of a Particle
§8.Curvilinear Motion: Cylindrical Components
- 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
Coordinate system: polar coordinate ( 𝑟, 𝜃)
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
Engineering Mechanics – Statics 12.110 Kinematics of a Particle
§8.Curvilinear Motion: Cylindrical Components
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.111 Kinematics of a Particle
§8.Curvilinear Motion: Cylindrical Components
- 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
Coordinate system: polar coordinate ( 𝑟, 𝜃)
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.112 Kinematics of a Particle
§8.Curvilinear Motion: Cylindrical Components
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.113 Kinematics of a Particle
Fundamental Problems
- F12.33 The car has a speed of 55𝑚/𝑠. Determine the angular
velocity 𝜃 of the radial line 𝑂𝐴 at this instant
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.114 Kinematics of a Particle
5/26/2013
20
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.115 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.116 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.117 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.118 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.119 Kinematics of a Particle
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.120 Kinematics of a Particle
5/26/2013
21
§9.Absolute Dependent Motion Analysis of Two Particles
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.121 Kinematics of a Particle
§9.Absolute Dependent Motion Analysis of Two Particles
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.122 Kinematics of a Particle
§9.Absolute Dependent Motion Analysis of Two Particles
- 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 ↓
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.123 Kinematics of a Particle
§9.Absolute Dependent Motion Analysis of Two Particles
- 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 ↓
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.124 Kinematics of a Particle
§9.Absolute Dependent Motion Analysis of Two Particles
- 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 ↑
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.125 Kinematics of a Particle
§9.Absolute Dependent Motion Analysis of Two Particles
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.126 Kinematics of a Particle
5/26/2013
22
§9.Absolute Dependent Motion Analysis of Two Particles
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 ↑
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.127 Kinematics of a Particle
§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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.128 Kinematics of a Particle
𝑟𝐴, 𝑟𝐵: absolute position vectors
𝑟𝐵/𝐴: relative position vector
§10. Relative Motion of Two Particles Using Translating Axes
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.129 Kinematics of a Particle
§10. Relative Motion of Two Particles Using Translating Axes
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.130 Kinematics of a Particle
𝑡𝑎𝑛𝜃 =𝑣𝑇/𝐴
𝑦
𝑣𝑇/𝐴𝑥 =
31.8
28.2⟹ 𝜃 = 48.50 ↘
§10. Relative Motion of Two Particles Using Translating Axes
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)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.131 Kinematics of a Particle
§10. Relative Motion of Two Particles Using Translating Axes
- 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 ↓
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.132 Kinematics of a Particle
5/26/2013
23
§10. Relative Motion of Two Particles Using Translating Axes
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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.133 Kinematics of a Particle
§10. Relative Motion of Two Particles Using Translating Axes
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.134 Kinematics of a Particle
§10. Relative Motion of Two Particles Using Translating Axes
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.135 Kinematics of a Particle
Fundamental Problems
- F12.39 Determine the speed of block 𝐷 if end 𝐴 of the rope is
pulled down with a speed of 𝑣𝐴 = 3𝑚/𝑠
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.136 Kinematics of a Particle
Fundamental Problems
- F12.40 Determine the speed of block 𝐴 if end 𝐵 of the rope is
pulled down with a speed of 6𝑚/𝑠
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.137 Kinematics of a Particle
Fundamental Problems
- F12.41 Determine the speed of block 𝐴 if end 𝐵 of the rope is
pulled down with a speed of 1.5𝑚/𝑠
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.138 Kinematics of a Particle
5/26/2013
24
Fundamental Problems
- F12.42 Determine the speed of block 𝐴 if end 𝐹 of the rope is
pulled down with a speed of 𝑣𝐹 = 3𝑚/𝑠
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.139 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.140 Kinematics of a Particle
Fundamental Problems
- F12.44 Determine the speed of cylinder 𝐵 if cylinder 𝐴 moves
downward with a speed of 𝑣𝐴 = 4𝑚/𝑠
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.141 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.142 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.143 Kinematics of a Particle
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.144 Kinematics of a Particle
5/26/2013
25
Fundamental Problems
- 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
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics – Statics 12.145 Kinematics of a Particle