Module
Dynamics of Rigid Bodies
Module
USMKCC-COL-F-050
Dynamics
of Math
Rigidfor
Bodies
Advanced
IE
Page 1 of 8
Introduction to Dynamics
Chapter 1
Intended Learning Outcomes: At the end of this chapter, the students are expected to:
1. Define the dynamics and enumerate its different branches.
2. Explain the various principles that govern the study of dynamics
3. Familiarize the various types of motion
1.1 Definition of Dynamics
•
•
•
•
Dynamics is the branch of mechanics which deals with the accelerated motion of
a body.
Motion is the progressive change of position of the body.
Particle usually denotes an object of point size.
Body denotes a system of particles which form an object of appreciable size.
1.2 Branches of Dynamics
1. Kinematics is the geometry of motion. It is the branch of dynamics which describes
the motion of bodies without reference to the forces that either cause the motion
or are generated as a result of motion.
2. Kinetics relates to force acting on a body to its mass and acceleration.\
1.3 Basic Units of Dynamics
Kinds of units necessary in the foundation of dynamics
1. Force
2. Length
3. Time
Force is something which changes the state of motion of the body.
Length is the measured of distance or dimension.
Time is a measure of succession of events and is considered an absolute quantity in
Newtonian mechanics.
Quantity
Force (𝐹)
Length (𝐿)
Time (𝑡)
SI Units
Newton (𝑁)
Meter (𝑚)
Second (𝑠𝑒𝑐)
English
Pound (𝑙𝑏)
Feet (𝑓𝑡)
Second (𝑠𝑒𝑐)
Note:
𝑚
1 𝑁 = (1 𝑘𝑔) (1 2 )
𝑠
1 𝑙𝑏 = (1 𝑠𝑙𝑢𝑔) (1
𝑆𝑙𝑢𝑔 = 𝑙𝑏.
Dynamics of Rigid Bodies
Module
USMKCC-COL-F-050
𝑓𝑡
𝑠𝑒𝑐 2
)
sec2
𝑓𝑡
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1.4 Important Principles Governing the Study of Dynamics
Velocity (𝑣) is defined as the time rate of change of displacement.
𝑣 =
𝑑𝑠
𝑑𝑡
Acceleration (𝑎) is defined as the time rate of change of velocity.
𝑎 =
𝑑2 𝑠
𝑑𝑣
𝑎 = 𝑑𝑡 2
𝑑𝑡
𝑑𝑣
𝑎 = 𝑣 𝑑𝑠
Resultant Force (𝐹) = 𝑚𝑎𝑠𝑠 x 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
𝑑𝑣
𝐹 = 𝑚𝑎
𝐹 = 𝑚 𝑑𝑡
1.5 Types of Motion
1. Rectilinear Motion – if the particle moves along a straight path or is moving in the
direction parallel to its displacement.
2. Curvilinear Motion – if the particle moves along a curve path.
Translation of Rigid Bodies. Motion of a rigid body in which a straight line passing through
any two of its particles always remain parallel to its initial position.
A’
’
A
Parallel (//)
Same motion
B’
B
Represent motion of translating body
A
Not parallel (//)
B
Not same
motion
Represent motion of non-translating body
Dynamics of Rigid Bodies
Module
USMKCC-COL-F-050
Page 3 of 8
Rectilinear Motion
1. Uniform Motion (𝒂 = 𝟎). A motion with constant speed or velocity.
When 𝑣 = constant (𝑐); 𝑎 = 0
1
2
𝑆
𝑺 = 𝒗𝒕
where:
𝑆 = distance - 𝑓𝑡, 𝑚
𝑣 = velocity – 𝑓𝑡/𝑠𝑒𝑐; 𝑚/𝑠𝑒𝑐
𝑡 = time – 𝑠𝑒𝑐, 𝑚𝑖𝑛
𝑎 = acceleration
2. Uniformly Accelerated Motion (𝒂 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕). A motion with constant change
in velocity or of uniform acceleration.
When a = c
1
2
𝑣1 ≠ 𝑣2
𝑣1
𝑡0
𝑡𝑡
dv
𝑎=
from:
dt
dv = adt
𝑣 = 𝑎𝑡 + 𝑣1
𝑑𝑠
ʃdv = aʃdt
𝑑𝑡
𝑣 = 𝑎𝑡 + 𝑐
@ 𝑡 = 0; 𝑣 = 𝑣1
𝑣1 = 𝑎(0) + 𝑐
𝐶 = 𝑣1
Then:
𝑣 = 𝑎𝑡 + 𝑣1
@ 𝑡 = 𝑡; 𝑣 = 𝑣2
𝑣2 = 𝑎𝑡 + 𝑣1
𝒗𝟐 – 𝒗𝟏 = 𝒂𝒕
v=
ds
v
;a =
dt
ds
dt =
ds
𝑣2
=
where:
𝑣1 = initial velocity
𝑣2 = final velocity
𝑡0 = initial time
𝑡𝑡 = final time
v
dv
= 𝑎𝑡 + 𝑣1
𝑑𝑠 = 𝑎𝑡 𝑑𝑡 + 𝑣1 𝑑𝑡
ʃ𝑑𝑠 = 𝑎ʃ𝑡𝑑𝑡 + 𝑣1 ʃ𝑑𝑡
𝑠 = ½ 𝑎𝑡2 + 𝑣1 𝑡 + 𝑐2
@ 𝑡 = 0; 𝑠 = 0;
0 = ½ (𝑎) (0)2 + 𝑣1 (0) + 𝑐2
𝑐2 = 0
then:
𝑆 = ½ 𝑎𝑡2 + 𝑣1 𝑡
@ 𝑡 = 𝑡; 𝑠 = 𝑠
𝑺 = 𝒗𝟏 𝒕 + ½ 𝒂𝒕𝟐
dv
dt
; dt =
dv
a
a
𝑣ʃ𝑑𝑣 = 𝑎ʃ𝑑𝑠
v2
2
= as + c3
@ 𝑣 = 𝑣1 ; 𝑠 = 0
Dynamics of Rigid Bodies
Module
USMKCC-COL-F-050
@ 𝑣 = 𝑣2 ; 𝑠 = 𝑠
v22
2
= as +
v21
2
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(v1 )2
v22
= a(0) + c3
2
C3 =
2
v21
−
v21
2
= as
𝒗𝟐𝟐 – 𝒗𝟐𝟏 = 𝟐𝒂𝒔
2
3. Rectilinear Motion with variable acceleration (a not constant)
- when bodies are acted upon by variable forces, they move with variable
acceleration.
4. Free Falling Body
-When a body is subjected to free fall, the body is acted by a single force known as
its weight. The weight of the body can also be considered as the resultant force
acting on the body.
𝑡 = 0, 𝑣 = 𝑣0 , ℎ = 0
𝑊
[T
ℎ
𝐹 = 𝑚𝑎
Since: 𝐹 = 𝑊
Then: 𝑊 = 𝑚𝑎
𝑚𝑔 = 𝑚𝑎
𝑔 = 𝑎
𝑡 = 𝑡, 𝑣 = 𝑣, ℎ
=h
Therefore: the body under free fall is moving w/ constant acceleration 𝑔.
𝑔 = 9.81𝑚/𝑠 2 ; = 32.2𝑓𝑡/𝑠 2
where:
𝑔 is (+) when going down
𝑔 is (-) when going up (against the gravity)
𝑣𝑜 = 0 (when it is dropped, and the one that drops is
not moving; this must also be the case for free fall)
𝑣𝑓 = final velocity
ℎ = height
𝑡 = time
1
ℎ = 𝑣0 𝑡 + 2 𝑔𝑡 2
𝑣𝑓 − 𝑣𝑜 = 𝑔𝑡
𝑣𝑓2 − 𝑣𝑜2 = 2𝑔ℎ
Illustrative Examples:
1. The car moves in a straight line such that for a short time its velocity is defines by
𝜐 = (3𝑡 2 + 2𝑡)
𝑓𝑡
𝑠
, where 𝑡 is in seconds. Determine its position and acceleration
when 𝑡 = 4 𝑠𝑒𝑐. when 𝑡 = 0, 𝑠 = 0.
Dynamics of Rigid Bodies
Module
USMKCC-COL-F-050
Page 5 of 8
Solution:
Position
From: 𝜐 = 𝑑𝑠/𝑑𝑡, since this equation relates 𝑣, 𝑠, and 𝑡.
when 𝒕 = 𝟎, 𝒔 = 𝟎
v=
ds
𝒔
dt
= (3t2 + 2t)
𝒕
∫𝟎 𝒅𝒔 = ∫𝟎 (𝟑𝒕𝟐 + 𝟐𝒕)𝒅𝒕
𝒔 = 𝒕𝟑 + 𝒕𝟐
when 𝑡 = 4 𝑠𝑒𝑐
𝒔 = (𝟒)𝟑 + (𝟒)𝟐 = 𝟖𝟎 𝒇𝒕 ------answer
Acceleration
From: 𝒂 = 𝒅𝒗/𝒅𝒕, since this equation relates v, s, and t.
𝑎=
𝑑𝑣
𝑑
= 𝑑𝑡 (3𝑡 2 + 2𝑡)
𝑑𝑡
= 6𝑡 + 2
when 𝑡 = 4 𝑠𝑒𝑐
𝑎 = 6(4) + 2 = 26 𝑓𝑡/𝑠 2 -------answer
2. The position coordinate of a particle which is confined to move along a straight line
is given by 𝑠 = 2𝑡3 − 24𝑡 + 6, where 𝑠 is measured in meters from a
convenient origin and 𝑡 is in seconds. Determine (a) the time required for the
particle to reach a velocity of 72
𝑚
𝑠
from its initial condition at 𝑡 = 0, (b) the
𝑚
acceleration of the particle when 𝑣 = 30 𝑠 , and (c) the net displacement of the
particle during the interval from 𝑡 = 1 𝑠𝑒𝑐 to 𝑡 = 4 𝑠𝑒𝑐.
Solution:
From: 𝑆 = 2𝑡 3 − 24𝑡 + 6
Since: 𝑣 =
𝑑𝑠
𝑑𝑡
𝑣 = 6𝑡 2 – 24
dv
=a
dt
𝑚
𝑎 = 12 𝑡 2
𝑠
a. @ 𝑣 = 72
𝑚
𝑠
and
𝑚
𝑠
Then: 72 = 6𝑡 2 – 24
𝑡 = 4 𝑠𝑒𝑐 ----- answer
b. @ 𝑣 = 30
𝑚
𝑠
Then: 30 = 6𝑡 2 – 24
Dynamics of Rigid Bodies
Module
USMKCC-COL-F-050
Page 6 of 8
𝑡 = 3 𝑠𝑒𝑐
then:
𝑎 = (12) (3)
𝑎 = 36 𝑚/sec2 ----- answer
c. Net displacement at t = 1 sec to t = 4 sec
∆𝑆 = 𝑆4 − 𝑆1
∆𝑆 = [2(43 ) − 24(4) + 6] − [2(13 ) − 24(1) + 6 ]
𝛥𝑆 = 54 𝑚 ----- answer
3. If a particle moves along a straight path in such a way that 𝑆 = 2𝑡 2 𝑓𝑒𝑒𝑡, where 𝑡 is
in seconds, what kind of acceleration has it? Find the acceleration when 𝑡 = 4 𝑠𝑒𝑐𝑠.
Find 𝑣 when 𝑡 = 10 𝑠𝑒𝑐.
Solution:
𝑑𝑠
but 𝑆 = 2𝑡 2 since: 𝑎 =
From: 𝑣 = 𝑑𝑡
𝑑
𝑣 = 𝑑𝑡 (2𝑡
2)
𝑑𝑡
𝑑
𝑎 = 𝑑𝑡 (4𝑡)
𝑓𝑡
𝑣 = 4𝑡 𝑓𝑡/𝑠𝑒𝑐
when 𝑡
when 𝑡
Then: 𝑣
𝑣
𝑑𝑣
𝑎 = 4 𝑠 2---- (𝑎 = constant)
= 4 sec, still 𝑎 = 4 ft/sec2
= 10 secs
= 4 (10)
= 40 ft/sec ----- answer
4. Same problem in 3, except that 𝑆 = 2𝑡 3 + 𝑡 2
Solution:
𝑑𝑠
From: 𝑣 = 𝑑𝑡
𝑣 =
𝑑
𝑑𝑡
𝑎 =
3
2
(2𝑡 + 𝑡 )
𝑎 =
𝑣 = 6𝑡2 + 2𝑡, ft/sec
𝑑𝑣
𝑑𝑡
𝑑
𝑑𝑡
(6𝑡 2 + 2𝑡)
𝑎 = 12𝑡 + 2----- (variable acceleration)
when 𝑡 = 4 𝑠𝑒𝑐, 𝑎 = 12 (4) + 2
𝒂 = 𝟓𝟎 𝒇𝒕/𝒔𝒆𝒄𝟐
when 𝑡 = 10 𝑠𝑒𝑐, 𝑣 = 6 (10)2 + 2 (10)
𝒗 = 𝟔𝟐𝟎 𝒇𝒕/𝒔𝒆𝒄
5. A ball is dropped down a well and 5 seconds later the sound of the splash is heard. If
the velocity of sound is 330 𝑚/𝑠𝑒𝑐 , what is the depth of the well?
Solution:
Let: 𝑡1 = time for the ball to travel a distance 𝑆
𝑡2 = time for the sound to travel a distance 𝑆
𝑡1 + 𝑡2 = 5---------------------1
1
From: 𝑆 or ℎ = 𝑣0 𝑡 + 2 𝑔𝑡 2 ---------2
330
S
𝑚
𝑠
Since 𝑣0 = 0
𝑆 = ½ 𝑔𝑡 2
2S
t1 = √ g
Dynamics of Rigid Bodies
Module
USMKCC-COL-F-050
Page 7 of 8
when: 𝑆 = 330 𝑡2
S
t2 = 330
Substitute in eqn. 1
2S
√
9.81
From Quadratic formula:
S
+ 330 = 5
1
2
𝑦=
𝑆
0.452 𝑆 + 330 − 5 = 0
2
𝑦1 = 10.35
1
2
𝑆 + 149 𝑆 − 1650 = 0
Let: 𝑦 = 𝑆
−149±√(149)2 −4(1)(−1650)
𝑦2 = −159.35 (neglected)
1
2
1
𝑦2 = 𝑆
𝑦 2 + 149 𝑦 – 1650 = 0
𝑆 2 = 10.35
𝑺 = 𝟏𝟎𝟕. 𝟐 𝒎 ----- answer
𝑚
6. During a test a rocket travels upward at 75 𝑠 , and when it is 40 𝑚 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
𝑚
downward acceleration of 9.81 𝑠 2 due to gravity. Neglect the effect of air resistance.
Solution:
Letting 𝑂 as origin and going up as positive
𝑚
𝑎 = −9. 81 𝑠 2, since downward acceleration
Solving for Maximum height
From: 𝑣𝐵2 − 𝑣𝐴2 = 2𝑎𝑠
𝟎 = (𝟕𝟓𝒎/𝒔)𝟐 + 𝟐(−𝟗. 𝟖𝟏 𝒎/𝒔𝟐 )(𝒔𝑩 − 𝟒𝟎𝒎)
𝒔𝑩 = 𝟑𝟐𝟕 𝒎 ----- answer
Solving for the velocity of the rocket just before it
hits the ground
𝑣𝐶2 − 𝑣𝐵2 = 2𝑎(𝑠𝐶 − 𝑠𝐵 )
= 0 + 2(−9.81m/𝑠 2 )(0 − 327𝑚)
𝑣𝐶 = −80.1 𝑚/𝑠 = 80.1 𝑚/𝑠 𝑔𝑜𝑖𝑛𝑔 𝑑𝑜𝑤𝑛
Solving for the velocity of the rocket between points A
and C
𝑣𝐶2 − 𝑣𝐴2 = 2𝑎(𝑠𝐶 − 𝑠𝐴)
= (75𝑚/𝑠)2 + 2(−9.81𝑚/𝑠 2 )(0 − 40𝑚)
𝑣𝐶 = −80.1 𝑚/𝑠 = 80.1 𝑚/𝑠 𝑔𝑜𝑖𝑛𝑔 𝑑𝑜𝑤𝑛
Note: The rocket is subjected to a deceleration from 𝐴 to 𝐵 of 9.81 𝑚/𝑠 2 , and then from
𝐵 to 𝐶 it is accelerated at this rate.
Dynamics of Rigid Bodies
Module
USMKCC-COL-F-050
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