UNIT I. KINEMATICS OF RIGID BODIES: RECTILINEAR MOTION
Overview
This unit will discuss the definition of kinematics quantities, derive the three
rectilinear motion equations with constant acceleration from the three basic concepts of
kinematics and also discuss Erratic motion. Knowledge in differential and integral calculus
is important for the discussion on this particular topic.
Learning Objectives
At the end of the module, I am able to:
1.) Define the kinematic quantities. The concept of the kinematic quantities should be
understood, and not confused.
2.) Derive the three equations for rectilinear motion with constant acceleration from
the three basic kinematics equation
3.) Study rectilinear erratic motion with piecewise function and motion graph.
4.) Explain the problem-solving strategy for problems involving constant acceleration
and erratic motion.
Topics
1.1. Kinematic Equations
1.2. Rectilinear Motion with Constant Acceleration
1.3. Rectilinear Erratic Motion
1
Conditioning Task
Name: ______________________________________
Section: _____________________________________
Date : _____________________________
Multiple Choice.
1. What is the acceleration of a car that maintains a constant velocity of 60km/hr for 15
seconds?
a.)
0
c.)
5km/s
b.)
4km/s
d.)
10km/s
2. If one object is 5-meter-high and is lifted 5 meters higher, how much potential energy
does it gain
a.)
The same
c.)
Four times as much
b.)
Twice as much
d.)
Six time as much
3. If a person mass on Earth is 80 kg, what would their mass be on the moon? (closest)
a.)
40kg
c.)
160 kg
b.)
80kg
d.)
100kg
4. If a person has a weight of 120N on Earth, what would their weight be on the moon?
(closest)
a.)
0 Newtons
c.)
20 Newtons
b.)
120 Newtons
d.)
60 Newtons
5. Which one of Newton’s Laws of Motion is “ for every action there is an equal and
opposite reaction”?
a.)
First Law
c.)
Third Law
b.)
Second Law
d.)
none of the above
6. A Zoo has 41 animals, if 6 is dead the how many animals remain?
a.)
35
c.)
41
b.)
30
d.)
47
7. A is the father of B. But B is not the son of A. How’s that possible?
a.)
A is the Father
c.)
A is the Daughter
b.)
B is the Father
d.)
B is the Daughter
8. If there are 6 apples and you take away 4, how many do you have?
a.)
7
c.)
5
b.)
6
d.)
4
9. If there are 12 fish and half of them drown, how many are there?
a.)
12
c.)
6
b.)
8
d.)
4
10. How many times can you subtract 10 from 100?
a.)
Ten times
c.)
twice
b.)
Five times
d.)
once
2
Lesson Proper
KINEMATICS OF RIGID BODIES: RECTILINEAR MOTION
Rectilinear Motion
Mechanics is a branch of the physical sciences concerned with the state of rest or
motion of bodies subjected to the action of forces. Engineering mechanics is divided into
two areas of study: statics and dynamics. In static, we look how a system performs under
the actions of balanced forces. In other words, we study systems when they are in
equilibrium. Example of systems we study under the condition of equilibrium include
everything from simple systems like tables and chairs, to more complex system like massive
bridges. And in dynamics, we look at bodies in motion under the action of unbalanced
forces. This means we are going to deal with systems that accelerate. When we go out for a
run from standing, we apply forces to our body so it can accelerate. When start driving, we
start accelerating our car. When a plane takes off, or lands, it is subjected to acceleration.
Dynamics has two distinct parts, kinematics and kinetics. Kinetics treats only the
motion of an object and does not consider the force that caused the motion and the kinetics
it is the kinetics that study the force that causing the motion.
Rectilinear Motion: Continuous Motion
Rectilinear Kinematics refers to straight motion. In kinematics we are going to deal
with displacement, velocity, acceleration and time. These are also called kinematic
quantities.
Position
It specifies the location of a point at any given instant of time, positive if the point is
located at the right of origin, negative if located at the left of the origin (by conventional
number line).
-s
+s
0
Displacement
It measures the change of position after some time from initial point. Positive if the
final position is on the right of the initial point and negative if the final postion is on left of
the initial point.
-Δs
0
3
+Δs
initial
point
Velocity
It is a physical vector quantity, velocity of an object is the rate of change of its
position with respect to a frame of reference, and is a function of time.
𝐼𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝑣 =
VELOCITY
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝑣𝑎𝑣𝑒 =
𝑑𝑠
𝑑𝑡
∆𝑠 𝑠 − 𝑠𝑜
=
∆𝑡 𝑡 − 𝑡𝑜
𝐼𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛, 𝑎 =
ACCELERATION
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛, 𝑣𝑎𝑣𝑒 =
𝑑𝑣
𝑑𝑡
∆𝑣 𝑣 − 𝑣𝑜
=
∆𝑡
𝑡 − 𝑡𝑜
Acceleration
It is the rate of change of velocity with respect to time in a particular direction.
Fundamental Equations:
Let
𝑠 =
𝑣 =
𝑎 =
𝑡 =
𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
𝑡𝑖𝑚𝑒
Δs 𝑑𝑠
=
Δt−0 Δt
𝑑𝑡
𝑣 = lim
𝑣=
𝑑𝑠
𝑑𝑡
Δv 𝑑𝑣
=
Δt−0 Δt
𝑑𝑡
𝑎 = lim
𝑎=
𝑑𝑣
𝑑𝑡
𝑑𝑠
𝑣 𝑑𝑡
=
𝑎 𝑑𝑣
𝑑𝑡
𝑣 𝑑𝑠
=
𝑎 𝑑𝑡
𝑣𝑑𝑣 = 𝑎𝑑𝑠
4
Rectilinear Kinematics with Constant Acceleration
Many motions that occur in real world, involves a constant acceleration, it happens
when a certain body or mass is acted upon by a constant force, such as free fall or when a
vehicle applies it brakes.
In deriving our equation, we are going to make “ a “ acceleration as a constant value.
𝑎=
𝑑𝑣
𝑑𝑡
𝑑𝑣 = 𝑎𝑑𝑡
𝑣
𝑡
∫ 𝑑𝑣 = ∫ 𝑎𝑑𝑡
𝑣𝑜
0
𝑣 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 𝑎𝑛𝑦 𝑡𝑖𝑚𝑒 “𝑡”
vo = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
[𝑣]𝑣𝑣𝑜 = 𝑎(𝑡)]𝑡0
𝑣 − 𝑣𝑜 = 𝑎𝑡
𝒗 = 𝒗𝒐 + 𝒂𝒕
𝑣=
𝑑𝑠
𝑑𝑡
𝑑𝑠 = 𝑣𝑑𝑡
Substituting;
𝑣 = 𝑣𝑜 + 𝑎𝑡
𝑑𝑠 = (𝑣𝑜 + 𝑎𝑡)𝑑𝑡
𝑠
𝑡
∫ 𝑑𝑠 = ∫ (𝑣𝑜 + 𝑎𝑡)𝑑𝑡
0
0
𝑡
𝑡
𝑠 = ∫ 𝑣𝑜 𝑑𝑡 + ∫ 𝑎𝑡𝑑𝑡
0
0
𝟏
𝒔 = 𝒗𝒐 𝒕 + 𝒂𝒕𝟐
𝟐
𝑣𝑑𝑣 = 𝑎𝑑𝑠
5
𝑣
𝑠
∫ 𝑣𝑑𝑣 = ∫ 𝑎𝑑𝑠
𝑣𝑜
0
𝑣2
[ ]𝑣𝑣𝑜 = 𝑎𝑠
2
𝑣 2 − 𝑣𝑜2
= 𝑎𝑠
2
𝒗𝟐 − 𝒗𝟐𝒐 = 𝟐𝒂𝒔
Examples:
1. The car on the left in the photo and in moves in a straight line such that for a short time
its velocity is defined by 𝑣 = (𝑡 2 + 2𝑡)𝑚/𝑠, where t is in seconds. Determine its position and
acceleration when 𝑡 = 4𝑠. When 𝑡 = 0, 𝑠 = 0.
Solution:
𝑑𝑥
𝑣=
= (𝑡 2 + 2𝑡)
𝑑𝑡
𝑥
𝑡
∫ 𝑑𝑥 = ∫ (𝑡 2 + 2𝑡)𝑑𝑡
0
0
𝑠 =
𝑡 3 2𝑡 2
+
3
2
When t = 4s
𝑠 =
43 2(4)2
+
= 𝟑𝟕 𝒎
3
2
𝑎=
𝑑𝑣
𝑑
= (𝑡 2 + 2𝑡)
𝑑𝑡 𝑑𝑡
𝑎 = 2𝑡 + 2
When t= 4s
a = 2(4) + 2 = 10 m/s2
6
2. Let’s consider the package we introduced at the beginning. At point A the package is
released from rest. It has constant acceleration (4.8𝑚/𝑠2) as it moves down section AB and
CD. The velocity between B and C is constant. The velocity at point D is 7.2 𝑚/𝑠. Determinde
the distance 𝑑 and the time required to reach at point D.
Solution:
For A – B and C – D we have
𝑣 2 = 𝑣𝑜2 + 2𝑎(𝑥 − 𝑥𝑜 )
2
𝑣𝐵𝐶
𝑣𝐵𝐶
Then, at B
= 0 + 2(4.8𝑚/𝑠 2 )(3𝑚 − 0)
= 5.366 𝑚/𝑠
And at D
=
+ 2𝑥𝐶𝐷 (𝑥𝐷 − 𝑥𝐶 ) ; 𝑑 = (𝑥𝐷 − 𝑥𝐶 )
𝑜𝑟
7.22 = 5.3662 + 2(4.8)(𝑑)
𝒅 = 𝟐. 𝟒𝟎𝒎
𝑣𝐷2
2
𝑣𝐵𝐶
For A – B and C – D we have
𝑣 = 𝑣𝑜 + 𝑎𝑡
Then A – B
5.366 = 0 + (4.8)(𝑡𝐴𝐵 )
𝑡𝐴𝐵 = 1.11804𝑠
And C – D
𝑚
𝑚
𝑚
7.2 = 5.366 + 4.8 ( 𝑡𝐶𝐷 )
𝑠
𝑠
𝑠2
𝑡𝐶𝐷 = 0.38196 𝑠
Now, for B – C, we have
𝑥𝑐 = 𝑥𝐵 + 𝑣𝐵𝐶 𝑡𝐵𝐶
or
3𝑚 = 5.366𝑡𝐵𝐶
or
𝑡𝐵𝐶 = 0.559 𝑠
7
Finally,
𝑡𝐷 = 𝑡𝐴𝐵 + 𝑡𝐵𝐶 + 𝑡𝐶𝐷
𝑡𝐷 = 1.11804 + 0.55901 + 0.38196
𝒕𝑫 = 𝟐. 𝟎𝟔𝒔
3. During a test a rocket travels upward at 75 m/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 downward
acceleration due to gravity, Neglect the effect of air resistance.
Solution:
𝑣 2 𝐵 = 𝑣 2 𝐴 + 2𝑎(𝑦𝐵 – 𝑦𝐴 )
0 = (75𝑚/𝑠)2 + 2(−9.81𝑚/𝑠2)( 𝑦𝐵 – 40𝑚)
𝑦𝐵 = 𝟑𝟐𝟕 𝒎
𝑣𝑐2 = 𝑣𝐵2 + 2𝑎(𝑦𝑐 – 𝑦𝐵 )
𝑣𝑐2 = (0)2 + 2(−9.81𝑚/𝑠2)( 0 – 327𝑚)
𝑣𝑐 = −80.1𝑚/𝑠 = 𝟖𝟎. 𝟏 𝒎/s
8
4. In adjacent highway lanes two automobile A and B are approaching each other as shown
in the fig. At t = 0, they are 3200ft apart, and their velocities are 𝑣𝐴 = 65 𝑚𝑖/ℎ and 𝑣𝐵 =
40 𝑚𝑖/ℎ, respectively. Knowing that A passes Point Q 40 s after B was there and that B
passes point P 42 s after A was there, determine a.) the constant acceleration of A and B, b.)
when the vehicle pass each other, c.) the speed of B at that time.
Solution:
𝑣𝐴𝑜 = 65
𝑚𝑖 5280𝑓𝑡
1ℎ𝑟
( 1𝑚𝑖 ) (3600𝑠)
ℎ
= 95.33
𝑓𝑡
𝑠
𝑣𝐵𝑜 = 40
𝑚𝑖 5280𝑓𝑡
1ℎ𝑟
( 1𝑚𝑖 ) (3600𝑠)
ℎ
= 58.667
𝑓𝑡
𝑠
a.) We have
1
𝑥𝐴 = 0 + (𝑣𝐴𝑜 )𝑡 + 2 𝑎𝐴 𝑡 2
x is positive
; origin at P.
At t = 40 s:
3200 𝑓𝑡 = (95.333
𝑓𝑡
1
) (40𝑠) + 𝑎𝐴 (40𝑠)2
𝑠
2
𝑎𝐴 = −𝟎. 𝟕𝟔𝟕 𝒇𝒕/𝒔𝟐
Also,
1
2
𝑥𝐵 = 0 + (𝑣𝐵𝑜 )𝑡 + 𝑎𝐵 𝑡 2
x is positive
; origin at Q.
At t = 42 s:
3200𝑓𝑡 = (58.667
𝑓𝑡
1
)(42𝑠) + 𝑎𝐵 (42𝑠)2
𝑠
2
𝑎𝐵 = 𝟎. 𝟖𝟑𝟒 𝒇𝒕/𝒔𝟐
9
b.) When the cars pass each other
𝑥𝐴 + 𝑥𝐵 = 3200𝑓𝑡
Then
1
2
1
2
2
2
(𝑣𝐴𝑜 )𝑡 + 𝑎𝐴 𝑡𝐴𝐵
+ 0 + (𝑣𝐵𝑜 )𝑡 + 𝑎𝐵 𝑡𝐴𝐵
= 3200
1
1
2 )
2 )
(95.333)(𝑡𝐴𝐵 ) + (−0.767)(𝑡𝐴𝐵
+ (58.667)(𝑡𝐴𝐵 ) + (0.834)(𝑡𝐴𝐵
= 3200
2
2
2
0.034𝑡𝐴𝐵
+ 154𝑡𝐴𝐵 − 3200 = 0
Solving for t:
𝑡𝐴𝐵 = 20.685𝑠
and 𝑡𝐴𝐵 = −456𝑠
𝑡𝐴𝐵 > 0
𝒕𝑨𝑩 = 𝟐𝟎. 𝟔𝟖𝟓𝒔
c.) We have
𝑣𝐵 = 𝑣𝐵𝑜 + 𝑎𝐵 𝑡
at 𝑡𝐴𝐵 = 20.685𝑠
𝑣𝐵 = 58.667 + (0.834)(20.685)
𝒗𝑩 = 𝟕𝟓. 𝟗𝟐𝟕 𝒇𝒕/𝒔
10
Rectilinear Kinematics: Erratic Motion
Erratic Motion
What do we do when the motion is erratic?
Car accelerates for a short time, followed by a constant velocity for some period, and
ends with a deceleration (or negative acceleration) until it comes to a stop.
𝑣1 = 𝑣𝑜 + 𝑎𝑜−1 𝑡
1
𝑠1 = 𝑠𝑜 + 𝑣𝑜 𝑡 + 𝑎𝑜−1 𝑡 2
2
0
𝑣3 = 𝑣2 + 𝑎2 𝑡
𝑣2 = 𝑣1
1
𝑠𝑓2 = 𝑣𝑓1 𝑡 + 𝑠𝑓1
2
1
3
𝑠3 = 𝑠2 + 𝑣2 𝑡 + 𝑎2−3 𝑡 2
2
Graphs and calculus can be used to make these problems simpler in many cases.
Remember from calculus:
y
-
Derivative of a function = slope at a point
𝑑𝑦
𝑑𝑥
Small Change in y over distance, dx
x
Integral of a function = Area Under Curve
y
𝑏
∫ 𝑦𝑑𝑥
𝑎
a
b
Sum of rectangles dx wide by y tall
x
11
By definition:
𝑎=
𝑑𝑣
− − − ∫ 𝑎𝑑𝑡 = 𝑣
𝑑𝑡
𝑣=
𝑑𝑠
− − − ∫ 𝑣𝑑𝑡 = 𝑠
𝑑𝑡
Accel. (a)
Velocity (v)
Derivative
t (s)
v (m/s)
t (s)
a (m/s2)
Accel. (a)
Derivative
Area
Area
Integrate
Integrate
a (m/s2)
t (s)
Examples:
1. The motion of a particle starting from rest is governed by the a – t curve shown in figure.
Sketch the v – t and s – t curves. Determine the position at t = 15 sec.
a (ft/s2)
12
6
9
15
-4
12
t (s)
Solution:
Integrate
Area
Integrate
Area
𝐯 – 𝐭 𝐆𝐫𝐚𝐩𝐡. Integrating 𝑎 =
𝑑𝑣
𝑑𝑡
can determine the v − t function
13
For 0 ≤ t ≤ 6, a = 2t
𝑑𝑣 = 𝑎𝑑𝑡
𝑣
𝑡
∫ 𝑑𝑣 = ∫ 2𝑡𝑑𝑡
0
0
𝑡2
𝑣 − 0 = 2( )]𝑡0
2
𝑣 = 𝑡2
At t = 6 sec
𝒗 = 𝟔𝟐 = 𝟑𝟔
𝒇𝒕
𝒔
For 6 ≤ t ≤ 9, a = 12
𝑑𝑣 = 𝑎𝑑𝑡
𝑣
𝑡
∫ 𝑑𝑣 = ∫ 12𝑑𝑡
36
6
𝑣 − 36 = 12𝑡]𝑡6
𝑣 = 12𝑡 − 12(6) + 36
𝑣 = 12𝑡 − 36
At t = 9 sec.
𝒗 = 𝟏𝟐(𝟗) − 𝟑𝟔 = 𝟕𝟐 𝒇𝒕/𝒔
For 9 ≤ t ≤ 15, a = - 4
𝑑𝑣 = 𝑎𝑑𝑡
𝑣
𝑡
∫ 𝑑𝑣 = ∫ −4𝑑𝑡
72
9
𝑣 − 72 = −4𝑡]𝑡9
𝑣 − 72 = −4𝑡 − (−4)(9)
𝑣 = −4𝑡 + 108
At t = 15 sec.
𝒗 = −𝟒(𝟗) + 𝟏𝟎𝟖 = 𝟒𝟖 𝒇𝒕/𝒔
14
𝐬 – 𝐭 𝐆𝐫𝐚𝐩𝐡. Integrating 𝑣 =
𝑑𝑠
𝑑𝑡
can determine the s − t function
For 0 ≤ t ≤ 6, v = t2
𝑑𝑠 = 𝑣𝑑𝑡
𝑠
𝑡
∫ 𝑑𝑠 = ∫ 𝑡 2 𝑑𝑡
0
0
𝑠−0 =
𝑠=
𝑡3 𝑡
]
3 0
𝑡3
3
At t = 6 sec.
𝒔=
𝟔𝟑
𝟑
= 𝟕𝟐 𝒇𝒕
For 6 ≤ t ≤ 9, v = 12t – 36
𝑑𝑠 = 𝑣𝑑𝑡
𝑠
𝑡
∫ 𝑑𝑠 = ∫ (12𝑡 − 36)𝑑𝑡
72
6
𝑡2
𝑠 − 72 = 12 ( ) − 36𝑡]𝑡6
2
𝑡2
62
𝑠 − 72 = 12 ( ) − 36𝑡 − [12 ( ) − 36(6)]
2
2
At t = 9 sec.
92
62
𝑠 = 12 ( 2 ) − 36(9) − [12 ( 2 ) − 36(6)] + 72 = 𝟐𝟑𝟒 𝒇𝒕
For 9≤t≤15, v=-4t+108
𝑑𝑠 = 𝑣𝑑𝑡
𝑠
𝑡
∫ 𝑑𝑠 = ∫ (−4𝑡 + 108)𝑑𝑡
234
9
𝑡2
𝑠 − 234 = −4 ( ) + 108𝑡]𝑡9
2
𝑡2
92
𝑠 − 234 = −4 ( ) + 108𝑡 − [−4 ( ) + 108(9)]
2
2
15
Ans.
𝑠 = −2𝑡 2 + 108𝑡 − 576
At t = 15 sec.
𝑠 = −2(15)2 + 108(15) − 576 = 𝟓𝟗𝟒 𝒇𝒕
Alternative solution:
By Area Method:
𝐯 – 𝐭 𝐆𝐫𝐚𝐩𝐡
𝑣6 =
For 0 ≤ t ≤ 6, at t = 6sec
1
= (12)(6) = 𝟑𝟔 𝒇𝒕/𝒔
1
(𝑎6 )(𝑡0−6 )
2
2
For 6 ≤ t ≤ 9, at t = 9sec
𝑣9 = 𝑣6 + (𝑎9 )(𝑡6−9 ) = 36 + (12)(3) = 𝟕𝟐 𝒇𝒕/𝒔
For 9 ≤ t ≤ 15 at t = 15sec
𝑣15 = 𝑣9 + (𝑎15 )(𝑡9−15 ) = 72 + (−4)(6) = 𝟒𝟖 𝒇𝒕/𝒔
𝐬 – 𝐭 𝐆𝐫𝐚𝐩𝐡
For 0 ≤ t ≤ 6, at t = 6sec
1
1
𝑠6 = 3 (𝑣6 )(𝑡0−6 ) = 3 (36)(6) = 𝟕𝟐 𝒇𝒕
For 6 ≤ t ≤ 9, at t = 9sec
1
1
𝑠9 = 𝑠6 + 2 (𝑣6 + 𝑣9 )(𝑡6−9 ) = 72 + 2 (36 + 72)(3) = 𝟐𝟑𝟒 𝒇𝒕
For 9 ≤ t ≤ 15 at t = 15sec
1
1
𝑠15 = 𝑠9 + (𝑣9 + 𝑣15 )(𝑡9−15 ) = 234 + (72 + 48)(6) = 𝟓𝟗𝟒 𝒇𝒕
2
2
16
2. Let’s consider the car we introduced at the beginning.
- Accelerates from rest at a rate of 3m/s2
- Reaches a maximum velocity of 24m/s
- Maintain the Maximum velocity for 60 seconds
- Decelerates at a constant rate of 2 m/s2 until it stops
- How far did the vehicle travel?
- How long did it take?
-
m
Given:
m
vo =0 s vmax = v1 =24 s v2 =24
m
a0-1 =3 s2
m
a1-2 =0 s2
m
s
v3 =0
m
s
m
a2-3 =2 s2 𝑡1−2 = 60 𝑠
Required : s1, s2 , s3, and t2
Solution:
By Area Method:
The area (A1) under the acceleration curve will
provide us with a value for velocity. This area should
m
be equal to v1 =24 s or
A1a = v1
3(t1) = 24
t1 = 8s
Solve For s1
The area (A1) under the velocity curve will provide
us with a value for position (s1). This area should be
equal to s1
𝑆1 = 𝐴1𝑉.
1
𝑆1 = (𝑡1 )(𝑣1 )
2
1
𝑆1 = 2 (8)(24) = 𝟗𝟔 𝒎
Solve For t2
𝑡2 = 𝑡1 − 𝑡1−2
𝑡2 = 8 + 60 = 68 s
Solve for s2
The area (A1+A2) under the velocity curve will give us the next position value.
𝑠2 = 𝐴1𝑉. + 𝐴2𝑣
𝑠2 = 96 + 24(60) = 𝟏𝟓𝟑𝟔 𝐦
17
Solve for s3
The total area under the acceleration curve should give us the final velocity. Note
that the portion of the graph that is below the axis has a negative value and will provide a
negative area.
𝐴1𝑎 + 𝐴2𝑎 = 𝑣3
3(𝑡1 ) − 2(𝑡3 − 𝑡2 ) = 𝑣3
3(8) − 2(𝑡3 − 68) = 0
𝑡3 = 80𝑠
The total area under the velocity graph will provide us with the final position.
𝑠3 = 𝐴1𝑉. + 𝐴2𝑣 + 𝐴3𝑣
1
𝑠3 = 𝑠2 + (𝑡2−3 )(𝑣3 )
2
1
𝑠3 = 1536 + (12)(24) = 𝟏𝟔𝟖𝟎 𝒎
2
References
Hibbeler, R. C., 2016 – Engineering Mechanics: Dynamics Fourteenth Edition, Pearson
Prentice Hall Pearson Education, Inc.
18
Assessing Learning
Activity I
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
1. A man dropped a stone into a well. Four seconds later the sound of water splash is heard.
Assuming sound travels at a speed of 330m/s, Determine how deep is the well?
19
Name:
Date:
Course/Year/Section:
Direction: Solve the following problems.
2. Two objects fly toward each other. When they are 1200m apart, their velocities and
acceleration are v1 = 40m/s, v2 = 15m/s a1 = 0.50 m/s2, and a2 = 1.30 m/s2. A third object
left the first object and moves with constant velocity of 50 m/s toward the second object.
Upon reaching the second object, it flies back to the first object and so on until the first and
second object collide. Find the total distance travelled by the third object.
20
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
3. A ball is thrown vetically into the air at 44m/s, After t1 second another ball is thrown
vertically with the same velocity. Determine the time t1 and also determine the velocity of
the second ball relative to the first ball when they pass each other at 90m above the ground.
(hint the direction of the first ball must be downward, while the second ball is upward)
21
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
4. The accelearation of a particle moving along a straight line is a = 𝑎 = √𝑠ft/s2, where s is in
meter. If its position s = 0 velocity v = 0 when t = 0, determine its velocity when s = 60 ft.
what time is that?
22
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
5. The s – t graph for a sports car moving along a straight road. Construct the v – t graph and
a – t graph over the time interval shown. Determine the acceleration at t = 10s.
23
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
6. An object travels along a straight path. so = 0, vo = 10m/s and to = 0. Its acceleration (in
m/s2) function is given. Construct its v – s graph. How long does it take for it to reach s =
600m?
𝑎={
0.02𝑠
−0.01𝑠 + 6
0 ≤ 𝑠 ≤ 200 𝑚
200 ≤ 𝑠 ≤ 600𝑚
𝑒𝑞1
𝑒𝑞2
24
UNIT II. KINEMATICS OF RIGID BODIES: CURVILINEAR MOTION
Overview
In the preceding module, we discussed techniques for determining the velocity and
position of a particle from its acceleration. To keep things simple, we assumed that the
particle moved along a straight line. We will now relax this assumption, and solve problems
where the particle may travel along a curved path.
When dealing with particles that travel along a curved path, we need to describe
position, velocity and acceleration as vectors. This means we need to find a convenient basis
(a set of axes) to characterize our vectors.
Learning Objectives
At the end of this module, I am able to
1. Describe the motion of a particle traveling along a curved path,
2. Analyze curvilinear motion using normal and tangential coordinate system.
3. Analyze the free-flight motion of a projectile.
Topics
2.1.
2.2.
Acceleration in Circular motion
2.1.1. Tangent Acceleration
2.1.2. Normal Acceleration
Projectile Motion
2.2.1. Horizontal Component of Motion
2.2.1. Vertical Component of Motion
25
Conditioning Task
Name:
Course/Year/Section:
Date:
Direction: The following grid contains terms associated with curvilinear motion(as
enclosed in the box below). Find and encircle them. Look for them in all directions including
backwards and diagonally.
Q
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Lesson Proper
KINEMATICS OF RIGID BODIES: CURVILINEAR MOTION
Acceleration in Circular Motion
An object moves in a straight line if the net force on it acts in the direction of motion
or the net force is zero. If the net force act at an angle to the direction of motion at any
moment, then the object moves in a curved path.
Acceleration in circular motion can be broken down into two parts. One part is
tangent to the path of motion and the other part is perpendicular to the path of motion. The
tangent part of component is tangential component. The perpendicular part is the normal
component.
The tangential component represents the acceleration of the object in a straight line.
The normal component of acceleration will increase as the speed and curvature increase.
In this fig. the blue vector would be our acceleration vector (a),
The orange vector would be the tangent component of the
acceleration vector (at) and the yellow is the normal
component of acceleration (an) .
FUNDAMENTAL EQUATION:
Acceleration
𝒂 = √𝒂𝒏 𝟐 +𝒂𝒕 𝟐
Tangetial Acceleration
𝒂𝒕 =
𝒅𝒗 ∆𝒗
=
𝒅𝒕 ∆𝒕
Normal Acceleration or Centripetal Acceleration
𝒂𝒏 =
𝒗𝟐
𝒓
or
𝒂𝒏 =
Radius of Curvature
𝟑/𝟐
𝒅𝒚 𝟐
[𝟏 + ( ) ]
𝒅𝒙
𝝆=
𝟐
𝒅 𝒚
[ 𝟐]
𝒅𝒙
27
𝒗𝟐
𝝆
Examples:
1. A toboggan is traveling down along a curve which can be approximated by the parabla y =
0.01x2. Determine the magnitude of its acceleration when it reaches point A where its speed
vA = 10m/s and it is increasing at the rate of at = 3m/s2.
Solution:
y = 0.01x2
𝑑𝑦
𝑑𝑥
= 0.02𝑥
𝑑2 𝑦
𝑑𝑥 2
𝜌=
= 0.2
3/2
𝑑𝑦 2
) ]
𝑑𝑥
𝑑2 𝑦
[ 2]
𝑑𝑥
𝑣2
102
[1+(
𝑎𝑛 =
𝜌
= 190.57m
= 190.57 = 0.525𝑚/𝑠 2
𝑎 = √𝑎𝑛 2 +𝑎𝑡 2
𝑎 = √0.5252 +32 = 𝟑. 𝟎𝟓𝒎/𝒔𝟐
2. An object travels along a curved path as shown. If at the point shown its speed is 28.8 m/s
and the speed is increasing at 8m/s2, determine the direction of its velocity, and the
magnitude and direction of its acceleration at this point.
28
Solution:
𝑑𝑦
3
𝑠𝑙𝑜𝑝𝑒 = 𝑑𝑥 = 8 𝑥 1/2
At 𝑥 = 16𝑚
3
𝑠𝑙𝑜𝑝𝑒 = 8 (16)1/2 = 1.5
𝑡𝑎𝑛𝜃𝑣 = 1.5
𝜃𝑣 = 𝑡𝑎𝑛−1 1.5 = 𝟓𝟔. 𝟑𝑜
𝑎𝑡 =
𝑑𝑣
= 8𝑚/𝑠 2
𝑑𝑡
𝑑𝑦 3 1/2
= 𝑥
𝑑𝑥 8
𝑑2 𝑦
𝑑𝑥 2
3
= 16 𝑥 −1/2
3/2
𝑑𝑦 2
) ]
𝑑𝑥
𝑑2 𝑦
[ 2]
𝑑𝑥
[1 + (
𝜌=
𝑎𝑛 =
= 125 𝑚
𝑣 2 28.82
𝑚
=
= 6.64 2
𝜌
125
𝑠
𝑎 = √𝑎𝑛 2 +𝑎𝑡 2
𝑎 = √6.642 +82 = 𝟏𝟎. 𝟒𝒎/𝒔𝟐
𝜃𝑎 = 56.3𝑜 + 𝑡𝑎𝑛−1
6.64
= 𝟗𝟔. 𝟎𝑜
8
29
Projectile Motion
A projectile is an object that moves through space under the influence of the earth’s
gravitational force. Two coordinated must be used to describe the projectile’s motion, since
it moves horizontally as well as vertically.
FUNDAMENTAL EQUATIONS:
Horizontal Component of Motion:
The effect of gravity is vertical only, (ax = 0), hence the particle is just moving in a
contant velocity along the x direction.
𝑣𝑥 = 𝑣𝑥𝑜 = 𝑣𝑜 𝑐𝑜𝑠𝜃
𝑥 = 𝑣𝑥 𝑡
𝑥 = (𝑣𝑜 𝑐𝑜𝑠𝜃)𝑡
t = time after the projectile is launched
Vertical Component of Motion:
From the formula 𝑣 = 𝑣𝑜 + 𝑎𝑡, it follows that 𝑣𝑦 = 𝑣𝑦𝑜 − 𝑔𝑡. Also, from the formula
1
1
𝑠 − 𝑠𝑜 = 𝑣𝑜 𝑡 + 2 𝑎𝑡 2 , it follows that 𝑦 − 𝑦𝑜 = 𝑣𝑦𝑜 𝑡 − 2 𝑔𝑡 2
𝑣𝑦𝑜 = 𝑣𝑜 𝑠𝑖𝑛𝜃
Derived Formulars of Common Elements in Projectile
Maximum height occurs when the vertical component of the velocity becomes zero.
𝑣𝑦 = 𝑣𝑦𝑜 − 𝑔𝑡
Solving for time “t” so that the vertical component (vy) becomes zero.
0 = vyo – gt
30
t=
𝑣𝑦𝑜
𝑣𝑜 𝑠𝑖𝑛𝜃
𝑔
=
𝑔
From the formula: 𝑦𝑜 = 0
1
𝑦 − 0 = 𝑣𝑦𝑜 𝑡 − 𝑔𝑡 2
2
𝑣𝑜 𝑠𝑖𝑛𝜃
1
𝑣 𝑠𝑖𝑛𝜃
) − 2 𝑔( 𝑜 𝑔 )2
𝑔
y = hmax = (v0 sin𝜃)(
hmax =
(𝑣𝑜 𝑠𝑖𝑛𝜃)2 (𝑣𝑜 𝑠𝑖𝑛𝜃)2
𝑔
2𝑔
hmax=
(𝑣𝑜 𝑠𝑖𝑛𝜃)2
2𝑔
vyo2 = 2ghmax
Range
The horizontal displacement of projectile when the overall displacement in the Y
direction of the projectile is zero.
The time it takes to reach the maximum height is:
𝑣𝑦𝑜 𝑣𝑜 𝑠𝑖𝑛𝜃
=
𝑔
𝑔
The time it takes to reach the Range is:
t=
2𝑣𝑜 𝑠𝑖𝑛𝜃
𝑔
2𝑣𝑜 𝑠𝑖𝑛𝜃
)
𝑔
x = R = (vocos𝜃)(
R=
𝑣𝑜 2 (2 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃)
𝑔
R=
𝑉𝑂 2 sin 2𝜃
𝑔
Relationship between x and y
x = 𝑣𝑥𝑜 t
y = 𝑣𝑦𝑜 t – ½ gt2
𝑥
t=𝑣
𝑥𝑜
31
y = vyo (
𝑥
)–½
𝑣𝑥𝑜
𝑣
𝑥 2
)
𝑣𝑥𝑜
g(
𝑔𝑥 2
y = x (𝑣𝑦𝑜 ) - 2𝑣
𝑥𝑜
y= x tan𝜃 −
𝑥𝑜
2
𝑔𝑥 2
2(𝑣𝑜 𝑐𝑜𝑠𝜃)2
Examples:
1. A projectile is launched at an angle of 45 degrees relative to horizontal.The
projectileneeds to clear the top of a small mountain that is 8 km away.If the projectile
islaunched at A with an elevation of 600 m:
a.) detemine the minimum possible initial speed (vo)
b.) determing the height (H) of the mountain peak
c.) determine the range R of the projectile when it hits the water
Solution:
a.) When x = 8000m, we are at the mountain peak
𝑥 = (𝑣𝑥𝑜 )𝑡
𝑥 = (𝑣𝑜 𝑐𝑜𝑠𝜃)(𝑡𝑝𝑒𝑎𝑘 )
𝑡𝑝𝑒𝑎𝑘 = 𝑣
8000
𝑜 𝑐𝑜𝑠45
At point O, vy = 0 @ t = tpeak
𝑣𝑦 = 𝑣𝑦𝑜 − 𝑔𝑡
8000
0 = 𝑣𝑜 𝑠𝑖𝑛𝜃 − 9.81(
)
𝑣𝑜 𝑐𝑜𝑠45
𝑣𝑜 𝑠𝑖𝑛45(𝑣𝑜 𝑐𝑜𝑠45) = 9.81(8000)
𝒗𝒐 = 𝟑𝟗𝟔 𝒎/𝒔
32
b.) At point O, vy = 0 @ t = tpeak
1
𝑦 − 𝑦𝑜 = 𝑣𝑦𝑜 𝑡 − 𝑔𝑡 2
2
1
𝑦 − 0 = 𝑣𝑜 𝑠𝑖𝑛𝜃𝑡 − 𝑔𝑡 2
2
8000
1
8000 2
𝑦 − 0 = 396𝑠𝑖𝑛45(
) − (9.81)(
)
396𝑐𝑜𝑠45
2
396𝑐𝑜𝑠45
𝑦 = 4000𝑚
So
𝐻 = 600𝑚 + 4000𝑚 = 4600𝑚
c.) At point A y =- 600m
1
𝑦 − 𝑦𝑜 = 𝑣𝑦𝑜 𝑡 − 𝑔𝑡 2
2
1
−600 − 0 = 𝑣𝑜 𝑠𝑖𝑛𝜃𝑡 − 𝑔𝑡 2
2
1
−600 − 0 = 396𝑠𝑖𝑛45𝑡 − 9.81𝑡 2
2
Using quadratic formula
𝑡 = −2.07𝑠, 𝑡 = 59.2𝑠
𝑥 = 𝑣𝑥 𝑡
Where x=R
𝑅 = 𝑣𝑜 𝑐𝑜𝑠45𝑡
𝑅 = 396𝑐𝑜𝑠45(59.2) = 𝟏𝟔𝟔𝟎𝟎𝒎
2. The catapult is used to fire a ball so that it hits the building's wall at its trajectory’s
highest height. If moving from A to B takes 1.5 s, calculate the velocity vA which it was fired,
the release angle𝜃, and the height h.
33
Solution:
Horizontal Motion:
𝑠 = 𝑣𝐴𝑥 𝑡𝐴𝐵
; 𝑣𝐴𝑥 = 𝑣𝐴 𝑐0𝑠𝜃
18 = 𝑣𝐴 𝑐0𝑠𝜃(1.5)……..eq 1
Vertical Motion:
𝑣𝐵𝑦 = 𝑣𝐴𝑦 + 𝑎𝑡𝐴𝑩 ; 𝑣𝐴𝑦 = 𝑣𝐴 𝑠𝑖𝑛𝜃
0 = 𝑣𝐴 𝑠𝑖𝑛𝜃 – 32.2 (1.5)
48.3 = 𝑣𝐴 𝑠𝑖𝑛𝜃
. . . . . . . . . . . . . . . . . Eq. 2
𝑣𝐵𝑦 2 = 𝑣𝐴𝑦 2 + 2𝑎(𝑠 − 𝑠𝑜
0 = (𝑣𝐴 𝑠𝑖𝑛𝜃)2 + 2(−32.2)(ℎ − 3.5). . . . . . . . . Eq. 3
To solve, first divide Eq. 2 by Eq. 1 to get 𝜃. Then
a.) 𝜃 = 76.0o
b.) vA= 49.8 ft/s
c.) h = 39.7 ft
3. A projectile is launched with an initial velocity of 100m/s at an angle of 45 degrees with
respect to an inclined hill. If the hill makes an angle of 30 degrees with respect to the
horizontal how far down the hill D does the projectile land?
34
Solution:
Define initial values
𝑣𝑥𝑜 = 𝑣𝑜 cos(∝)
𝑣𝑦𝑜 = 𝑣𝑜 sin(∝)
𝑥𝑜 = 0
𝑥 = 𝐷𝑐𝑜𝑠(30)
𝑦𝑜 = 0
𝑦 = −𝐷𝑠𝑖𝑛(30)
Horizontal motion: a = 0
𝑣𝑥𝑜 = 𝑣𝑜 cos(15)
𝑥 = (𝑣𝑜 cos(15))𝑡
𝐷𝑐𝑜𝑠(30) = (100 cos(15))(𝑡)
𝐷𝑐𝑜𝑠(30)
t= 100 cos(25)
Vertical motion: 𝑎𝑦 = −𝑔
1
𝑦 − 0 = 𝑣𝑦𝑜 𝑡 − 2 𝑔𝑡 2
1
𝐷𝑐𝑜𝑠(30)
2
100 cos(15)
𝐷𝑐𝑜𝑠(30)
1
𝐷𝑐𝑜𝑠(30)
100𝑠𝑖𝑛15(100 cos(15)) − 2 9.81(100 cos(15))2
−𝐷𝑠𝑖𝑛30 = 𝑣𝑜 𝑠𝑖𝑛15𝑡 − 9.81𝑡 2 : substitute t=
−𝐷𝑠𝑖𝑛30 =
𝐷 = 𝟏𝟖𝟓𝟔. 𝟔𝟒𝒎
References
Hibbeler, R. C., 2016 – Engineering Mechanics: Dynamics Fourteenth Edition, Pearson
Prentice Hall Pearson Education, Inc.
35
Assessing Learning
Activity II
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
1. A man plans to jump over the hole as shown in the figure. Determine the smallest value of
the initial velocity so that he lands at point 0.
36
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
2. A policeman aimed his rifle at the bull’s eye of a target 50m away. If the speed of the
bullet is 1000ft/s, how far (cm) below the bull’s eye does the bullet strikes the target?
37
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
3. A baseball outfielder throws a ball at an initial speed of 100 ft/s. After the ball leaves the
player’s hand gravity starts to have an effect on the ball’s motion. Assume gravity is constant
at 32.2 ft/s2.
a.) determine the velocity after 2 seconds
b.) determine the distance traveled by the projectile
until it had reached the ground surface.
38
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
4. A projectile with a muzzle velocity of 500 m/s is fired from a gun on a top of a cliff 420 m
above sea level. If the projectile hits the water surface 48 second after being fired,
determine the horizontal range of the projectile.
39
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
5. An automobile travels along a horizontal circular curved road that has a radius of 600m.
If the speed is uniformly increased at a rate of 2000km/h2, determine the magnitude of the
acceleration at the instant speed of the car is 60km/h.
40
UNIT III. PLANE MOTION OF KINEMATICS OF RIGID BODIES
Overview
Within this unit we will analyze the planar kinematics of a rigid body. This study is
important for designing gears, cams, and mechanisms that are used in many mechanical
works. When the kinematics is completely understood, then we may apply the movement
equations, which link the forces on the body to the movement of the body.
Learning Objectives
At the end of the unit, I am able to:
1. Classify the various forms of rigid-body planar motion;
2. Analyze angular motion about a fixed axis and rigid-body translation; and
3. Test planar motion using measurement of absolute motion.
Topics
3.1.
3.2.
3.3.
3.4.
3.5.
Translation Motion
Plane Motion
Relative Motion Analysis: Velocity
Instantaneous Center of Zero Velocity
Absolute and Relative Reaction
41
Pre-Test
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
1. The angular velocity of the disk is defined by 𝜔 = (5𝑡 2 + 2) rad/s, where t is in seconds.
Assess the velocity and acceleration magnitudes of point A on the disk when t = 0.5 s.
42
2. The disk is originally rotating at 𝜔𝑂 = 12 rad/s. If it is subjected to a constant angular
acceleration of 𝛼 = 20 rad/s2, evaluate the velocity magnitudes and the acceleration n and t
components of point A at the moment t = 2s.
43
3. The disk is originally rotating at 𝜔𝑂 = 12 rad/s. If it is subjected to a constant angular
acceleration of 𝛼 = 20 rad/s2, evaluate the velocity magnitudes and the acceleration n and t
components of point A at the moment t = 2s.
44
Lesson Proper
PLANE MOTION OF KINEMATICS OF RIGID BODIES
Kinematics Planar Motion Equations
If all parts of the body move in parallel planes a rigid body performs plane motion.
For simplicity, we generally regard the motion plane as the plane that comprises the center
of mass, and we view the body as a thin slab whose motion is limited to the plane of the slab.
This idealization appropriately describes a very large category of rigid body movements
found in engineering
The plane motion of a rigid body may be divided into several categories, as
represented in Fig. 2-1. We note that in both of the two translation situations, the
movement of the body is entirely determined by the motion of any point in the body, as all
points have the same movement.
Translation is characterized as any movement in which every line in the body is
always parallel to its origin. In translation, no rotation of line in the body occurs. In
rectilinear translation, all body points travel in straight line. The curvilinear motion, by
the word itself, it moves in a curved direction.
Rotation about a fixed axis, it is the angular rotation of the axis. It says that every
object in a rigid body moves around the axis, circular in motion and all lines normal to the
axis of rotation rotate the same time around the same angle.
General plane motion is the combination of both translation and rotation.
45
Figure 2-1
Rotation
Figure 2-2 shows a rigid body that rotates in the direction of the image, as it
undergoes direction motion. The angular positions of any two lines 1 and 2 attached to the
body are specified by 𝜃1 and 𝜃2 measured from any convenient fixed reference direction.
Because the angle 𝛽 is invariant, the relation 𝜃2 = 𝜃1 + 𝛽 upon differentiation with respect
to time gives 𝜃2 = 𝜃1 and 𝜃2 = 𝜃1 or, during a finite interval, ∆𝜃2 = ∆𝜃1. Therefore, all
lines on a rigid body have the same angular displacement in its motion axis, the same
angular velocity and the same angular acceleration.
Figure 2-2
Note that a line 's angular motion depends only on its angular location with respect
to some arbitrary fixed reference, and on the displacement time derivatives. Angular motion
does not involve a fixed axis, natural to the motion plane on which the line and the body are
rotating.
46
Angular-Motion Relations
The angular velocity 𝜔 is the first derivative of the angular position 𝜃 and the
angular acceleration 𝛼 is the second derivative. These definitions give
Eqs. 2-1
For rotation with constant angular acceleration, the integrals of Eqs. 2-1 becomes
𝜔 = 𝜔𝑜 + 𝛼𝑡
𝜔2 = 𝜔𝑜 2 + 2𝛼(𝜃 − 𝜃𝑜 )
𝜃 = 𝜃𝑜 + 𝜔𝑜 𝑡 +
1 2
𝛼𝑡
2
Here, 𝜃𝑜 and 𝜔𝑜 are the values of the angular coordinate position and the angular
velocity at t = 0, respectively, and t is the length of the movement considered.
Rotation about a Fixed Axis
If a rigid body rotates around a fixed axis, all other points in concentric circles
except those on the axis travel around the fixed axis. So, for the rigid body in Fig. 2-3
rotating around the normal fixed axis to the plane of the figure through O, any point such as
A moving in the r-radius circle.
Figure 2-3
47
𝑣 = 𝑟𝜔
2
𝑎𝑛 = 𝑟𝜔2 = 𝑣 ⁄𝑟 = 𝑣𝜔
𝑎1 = 𝑟𝛼
Eqs. 2-2
Alternatively, those quantities can be expressed using the vector notation crossproduct relationship. The angular velocity can be expressed as 𝜔 as shown in Fig 2-4a,
normal to the rotational plane with a meaning regulated by the right-hand rule. From the
definition of the vector-cross product, we see that the vector v is obtained by crossing
somewhere in r. This cross product gives the right direction and magnitude for v and we
write
𝑣 =𝑟= 𝜔𝑥𝑟
We must retain the order of the vectors to be crossed. The reverse order gives 𝑟 𝑥 𝜔
= -v.
Figure 2-4
The acceleration of point A is obtained by differentiating the cross-product
expression for v, which
gives
48
Here 𝛼 = 𝜔, stands for the angular acceleration of the body. Thus, the vector
equivalents to Eqs. 2-2 are
𝑣= 𝜔 × 𝑟
𝑎𝑛 = 𝜔 × (𝜔 × 𝑟)
𝑎𝑡 = 𝛼 × 𝑟
and are shown figure 2-4b.
Examples:
1. A flywheel rotates freely at 1800 rev / min in clockwise direction is subjected to a
counterclockwise torque variable, which is first applied at t= 0. A counterclockwise angular
acceleration is generated by the torque α= 4 t rad / s2, where t is the time in seconds during
which the torque is applied. Solve for (a) the time needed by the flywheel to reduce its
angular velocity to 900 rev / min in the clockwise direction, (b) the time needed by the
flywheel to reverse its rotational direction and (c) the overall number of revolutions,
counterclockwise plus clockwise, turned by the flywheel within the first 14 seconds of
torque application.
Solution:
The counterclockwise path is arbitrarily acknowledged as positive.
We will integrate 𝛼 since 𝛼 is a known function of the time to obtain angular
velocity. With the initial angular velocity of -1800(2𝜋)/60 = -60𝜋 rad/s, we have
[𝑑𝜔 = 𝛼𝑑𝑡]
𝜔
𝑡
𝜔 = −60𝜋 + 2𝑡 2
∫−60𝜋 𝑑𝜔 = ∫0 4𝑡 𝑑𝑡
Substituting the clockwise angular speed of 900 rev/min or 𝜔 = −
𝑟𝑎𝑑
−30𝜋
𝑠
900(2𝜋)
60
=
gives
−30𝜋 = −60𝜋 + 2𝑡 2
𝑡 2 = 15𝜋
𝒕 = 𝟔. 𝟖𝟔 𝒔
The flywheel changes direction when its angular velocity is momentarily zero. Thus,
0 = −60𝜋 + 2𝑡 2
𝑡 2 = 30𝜋
𝒕 = 𝟗. 𝟕𝟏𝒔
49
The overall number of turns the wheel makes in 14 seconds is the number of
counterclockwise makes N1 in the first 9.71 seconds, plus the number of counterclockwise
turns N2 in the rest of the time.. Integrating the expression for 𝜔 in terms of 𝑡 gives us the
angular displacement in radians. Thus, for the first interval
𝜃1
[𝑑𝜃 = 𝜔𝑑𝑡]
0
𝜃1 = [−60𝜋𝑡 +
or 𝑁1 =
9.71
∫ 𝑑𝜃 = ∫
1220
2𝜋
(−60𝜋 + 2𝑡 2 ) 𝑑𝑡
0
2 3 9.71
𝑡 ]
= −1220 𝑟𝑎𝑑
3 0
= 194.2 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒.
For the second interval
𝜃2
14
∫ 𝑑𝜃 = ∫
0
(−60𝜋 + 2𝑡 2 )𝑑𝑡
9.71
𝜃2 = [−60𝜋𝑡 +
or 𝑁2 =
410
2𝜋
2 3 14
𝑡 ]
= 410 𝑟𝑎𝑑
3 9.71
= 65.3 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑐𝑜𝑢𝑛𝑡𝑒𝑟𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒.
The cumulative number of revolutions turned over in the 14 seconds is therefore
𝑁 = 𝑁1 + 𝑁2 = 194.2 + 65.3 = 𝟐𝟓𝟗 𝒓𝒆𝒗
2. The hoist motor pinion A drives gear B, which is fixed to the hoisting drum. The load L is
raised from its place of rest and with relentless acceleration, it acquires an upward speed of
3 ft / sec in a vertical rise of 4 feet. Calculate (a) the acceleration of point C on the cable in
contact with the drum as the charge passes this location and (b) the angular velocity and
angular acceleration of the pinion A.
50
Solution:
The acceleration of the load L will generally be the same as the tangential velocity v
and the tangential acceleration at C. The n- and t-components of the acceleration C become
with continuous acceleration for the rectilinear motion of L.
[𝑣 2 = 2𝑎𝑠]
[𝑎𝑛 =
𝑎 = 𝑎𝑡 =
𝑣2
]
𝑟
𝑎𝑛 =
𝑣2
32
=
= 1.125 𝑓𝑡/𝑠𝑒𝑐 2
2𝑠
2(4)
32
= 4.5 𝑓𝑡/𝑠𝑒𝑐 2
24
( )
12
𝑎𝑐 = √(4.5)2 + (1.125)2 = 𝟒. 𝟔𝟒 𝒇𝒕/𝒔𝒆𝒄𝟐
[𝑎 = √𝑎𝑛 2 + 𝑎𝑡 2 ]
The angular motion of gear A is determined by the velocity v1 and the tangential
acceleration a1 by their common point of contact from the angular motion of gear B. Firstly,
gear B angular movement is determined from point C motion on the attached drum. Thus,
[𝑣 = 𝑟𝜔]
[𝑎𝑡 = 𝑟𝛼]
𝜔𝐵 =
𝑣
𝑟
𝛼𝐵 =
3
𝑟𝑎𝑑
= 1.5
24
𝑠𝑒𝑐
(12)
𝑎𝑡 1.125
=
= 0.562 𝑟𝑎𝑑/𝑠𝑒𝑐 2
24
𝑟
(12)
Then from 𝑣1 = 𝑟𝐴 𝜔𝐴 = 𝑟𝐵 𝜔𝐵 and 𝛼1 = 𝑟𝐴 𝛼𝐴 = 𝑟𝐵 𝛼𝐵 , we have
18
𝑟𝐵
𝜔𝐴 =
𝜔𝐵 = 12 (1.5) = 𝟒. 𝟓 𝒓𝒂𝒅/ 𝐬𝐞𝐜 𝑪𝑾
6
𝑟𝐴
12
51
18
𝑟𝐵
𝛼𝐴 =
𝛼 = 12 (0.562) = 𝟏. 𝟔𝟖𝟖 𝒓𝒂𝒅/ 𝒔𝒆𝒄𝟐 𝑪𝑾
6
𝑟𝐴 𝐵
12
Absolute and Relative Velocity in Plane Motion
The velocity and acceleration of a point P undergoing rectilinear motion can be
correlated with the angular velocity and angular acceleration of a line embedded within a
body using the method below.
Position Coordinate Equation
Use the position coordinate s to determine the location of point P on the body which
is measured according from its fixed origin and is directed along the straight-line path of
point P motion.
Measure the angular position θ of a line lying in the body from a fixed reference axis.
From the body dimensions, relate 𝑠 𝑡𝑜 𝜃, 𝑠 = 𝑓(𝜃) using geometry and/or
trigonometry.
Time Derivatives
Take the first derivative of s = f (𝜃) with respect to time to get a relation between
𝑣 𝑎𝑛𝑑 𝜔.
Take the second time derivative to get a relation between a and a.
For each case the chain rule of calculus must be used when the time derivatives of
the coordinate position equation are taken.
Examples:
1. The end of the rod R shown in the figure maintains contact with the cam through a spring.
Solve the speed and acceleration of the rod when the cam is in an arbitrary position 𝜃,
whether the cam rotates around an axis at point O with an angular acceleration of α and
angular velocity 𝜔.
52
Solution:
In order to relate the rotational motion of the line segment OA on the cam to the
rectilinear translation of the rod, coordinates θ and x are chosen. These coordinates are
measured from the fixed-point O and can be related to each other using trigonometry. Since
OC = CB = r cos 𝜃, then
𝑥 = 2𝑟𝑐𝑜𝑠𝜃
With the use of chain rule of calculus, we have
𝑑𝑥
𝑑𝜃
= −2𝑟(𝑠𝑖𝑛𝜃)
𝑑𝑡
𝑑𝑡
𝒗 = −𝟐𝒓𝝎𝒔𝒊𝒏𝜽
𝑑𝑣
𝑑𝑤
𝑑𝜃
= −2𝑟 ( ) 𝑠𝑖𝑛𝜃 − 2𝑟𝜔(𝑐𝑜𝑠𝜃)
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝒂 = −𝟐𝒓(𝜶𝒔𝒊𝒏𝜽 + 𝝎𝟐 𝒄𝒐𝒔𝜽)
Note: The negative signs mean that v and a are in contrary to the positive x
direction. If you imagine the motion, it seems normal.
2. At a given scenario, the cylinder of radius r, presented in the figure, has an angular
velocity V and angular acceleration A. Determine the velocity and acceleration of its center
G if the cylinder rolls without slipping.
53
Solution:
The cylinder is undergoing general plane motion, as it translates and rotates
simultaneously. By inspection, point G moves in a straight line to the left, from G to G’, as the
cylinder rolls. Consequently, its new position G’ will be specified by the horizontal position
coordinate 𝑠𝐺 , which is measured from G to G’. Also, as the cylinder rolls (without slipping),
the arc length A’B on the rim which was in contact with the ground from A to B, is equivalent
to 𝑠𝐺 . Consequently, the motion requires the radial line GA to rotate 𝜃 to the position G’A’.
Since the arc A’B = 𝑟𝜃, then G travels a distance 𝑠𝐺 = 𝑟𝜃.
𝑑𝜃
,
𝑑𝑡
Taking successive time derivatives of this equation, realizing that r is constant,𝜔 =
𝑑𝜔
and 𝛼 = 𝑑𝑡 , gives the necessary relationships:
𝑠𝐺 = 𝑟𝜃
𝒗𝑮 = 𝒓𝝎
𝒂𝑮 = 𝒓𝜶
Relative Motion Analysis: Velocity
The x, y system of coordinates is fixed and measures the absolute position of two
points A and B on the body, represented here as a bar, Fig. 2-5. The origin of the x, 'y'
coordinate system is attached to the "base point" A selected, which generally has a known
motion. The axes of this coordinate system move with respect to the fixed point, but do not
rotate with the bar.
Figure 2-5
The relative velocity equation can be implemented either using Cartesian vector
analysis, or directly writing the equations of the x and y scalar components. The following
method is recommended for use.
54
Vector Analysis
Kinematics Diagram
Set the fixed x, y coordinates directions and draw a body kinematic diagram.
Indicate the velocities 𝑣𝐴, 𝑣𝐵 of points A and B, and angular velocity 𝜔, and the
relative position vector Γ𝐵/𝐴
If the magnitudes of 𝑣𝐴, 𝑣𝐵 , or 𝜔 are unknown, the sense of direction of these
vectors can be assumed.
Velocity Equation
To apply 𝑣𝐵 = 𝑣𝐴 + 𝜔 𝑥 Γ𝐵/𝐴 , Express the Cartesian vector form of the
vectors and replace them with the equation. Assess the cross product and then
equate the respective i and j components to get two scalar equations.
If the solution provides an unknown magnitude with a negative response,
the direction of the vector is contrary to the direction shown in the kinematic
diagram.
Scalar Analysis
Kinematics Diagram
In scalar form, if the velocity equation is to be used, then the magnitude and
direction of the relative velocity 𝑣𝐵/𝐴 must be established. Draw a kinematic
diagram which shows the relative motion. Since the body is considered to be
“pinned” momentarily at the base point A, the magnitude of 𝑣𝐵/𝐴 is 𝑣𝐵/𝐴 = 𝜔𝑟𝐵/𝐴 .
The sense of direction of 𝑣𝐵/𝐴 is always perpendicular to 𝑟𝐵/𝐴 in accordance with
the rotational motion 𝜔 of the body.
Velocity Equation
Write the equation in symbolic form, 𝑣𝐵 = 𝑣𝐴 + 𝑣𝐵/𝐴 and underneath each
of the terms represent the vectors graphically by showing their magnitudes and
directions. The scalar equations of these vectors are calculated from the x and y
components.
Note: The notation 𝑣𝐵 = 𝑣𝐴 + 𝑣𝐵/𝐴(𝑝𝑖𝑛) may be helpful in recalling that “A”
is pinned.
55
Examples:
1. The link listed in Fig. 2-6a is driven by two blocks at A and B moving in the fixed slots. If
the speed of A is down 2 m / s, determine the speed of B at the instant 𝜃 = 45°.
Figure 2-6a
Solution:
Vector Analysis
Kinematic Diagram. Since points A and B are restricted to move along the fixed
slots and 𝑣𝐴 is directed downward, then velocity 𝑣𝐵 must be directed horizontally to the
right, Fig. 2-6b. This motion causes the link to rotate in the counterclockwise direction; that
is, the angular velocity ω is directed outward by the right-hand rule, perpendicular to the
motion plane.
Figure 2-6b
56
Velocity Equation. Expressing every single vector in Fig. 2-6b as to its components
i, j, k and applying the equation 𝑣𝐵 = 𝑣𝐴 + 𝜔 𝑥 Γ𝐵/𝐴 to A, the base point, and B, we have
𝑣𝐵 = 𝑣𝐴 + 𝜔 𝑥 Γ𝐵/𝐴
𝑣𝐵 𝒊 = −2𝒋 + [𝜔𝒌 𝑥 (0.2 sin 45°𝒊 − 0.2 cos 45°𝒋)]
𝑣𝐵 𝒊 = −2𝒋 + 0.2𝜔 sin 45°𝒋 + 0.2𝜔 cos 45°𝒊
Equating the i and j components gives
𝑣𝐵 = 0.2𝜔 cos 45°
0 = −2 + 0.2𝜔 sin 45°
Thus,
𝜔 = 14.1 𝑟𝑎𝑑/𝑠
𝒗𝑩 = 𝟐 𝒎/𝒔
Scalar Analysis
The kinematic diagram of the relative “circular motion: which produces 𝑣𝐵 is shown
𝐴
in Figure 2-6c. Here 𝑣𝐵 = 𝜔(0.2 𝑚)
𝐴
Figure 2-6c
Thus,
57
The solution produces the above results.
It should be emphasized that these results are valid only at the instant 𝜃 = 45°
2. In Fig 2-7a, collar C travels downwards at a pace of 2 m/s. Determine the angular speed
of CB at this moment.
Fig. 2-7a
Solution:
Vector Analysis
Kinematic Diagram. C's downward motion causes B to pass through a curved
direction to the right. CB and AB also rotate in counterclockwise direction.
Velocity Equation. Link CB (general plane motion): See Fig. 2-7b
Fig. 2-7b
𝑣𝐵 = 𝑣𝐶 + 𝜔𝐶𝐵 × 𝑟𝐵/𝐶
𝑣𝐵 𝑖 = −2𝑗 + 𝜔𝐶𝐵 × (0.2𝑖 − 0.2𝑗)
𝑣𝐵 𝑖 = −2𝑗 + 0.2𝜔𝐶𝐵 𝑗 + 0.2𝜔𝐶𝐵 𝑖
58
𝑣𝐵 = 0.2𝜔𝐶𝐵
(1)
0 = −2 + 0.2𝜔𝐶𝐵
(2)
𝜔𝐶𝐵 = 10 rad/s
𝒗𝑩 = 2 m/s
Scalar Analysis
The scalar component equations of vB = vC + vB/C can be obtained directly.
The kinematic diagram in Fig. 2-7c shows the relative “circular” motion which produces
vB/C. We have
Fig. 2-7c
vB = vC + vB/C
Resolving these vectors in the x and y directions yields
𝑣𝐵 = 0 + 𝜔𝐶𝐵 (0.2√2𝑐𝑜𝑠45°)
0 = −2 + 𝜔𝐶𝐵 (0.2√2𝑠𝑖𝑛45°
which is the same as Eqs. 1 and 2.
59
Instantaneous Center of Zero Velocity
The velocity of any point B situated on a rigid body can be obtained rather simply by
choosing the base point A as a point defined at the moment as having zero velocity. In this
case, vA=0, and hence, the equation of speed, 𝑣𝐵 = 𝑣𝐴 + 𝜔 𝑥 Γ𝐵/𝐴 , becomes 𝑣𝐵 = 𝜔 𝑥 Γ𝐵/𝐴 .
For a body with general plane motion, point A so chosen is called zero velocity
instantaneous center (IC), and it lies on the zero-velocity instantaneous axis.
The IC for bicycle wheel in Fig 6-8, for example is at ground contact point. There the
spokes are very obvious, while they are blurred at the top of the wheel. If one imagines that
the wheel is momentarily pinned at this point, the velocities of various points can be found
using 𝑣 = 𝜔𝑟. Here the radial distances shown in the photo, Fig. 6-8, must be determined
from the geometry of the wheel.
(© R.C. Hibbeler)
Figure 2-8
As shown on the kinematic diagram in Fig. 2-9, the body is imagined as “extended
and pinned” at the IC so that, at the instant considered, with its angular velocity 𝜔 it rotates
around this pin.
Using the equation 𝑣 = 𝜔𝑟 for each of the arbitrary points A, B, and C on the body,
the magnitude of velocity can be found, where r is the radial distance from the IC to each
point.
The action line of each vector v is perpendicular to its respective radial line r, and
the velocity has a sense of direction that appears to shift the point in a way compatible with
the radial line 's angular rotation V, Fig. 2-9.
60
Figure 2-9
Examples:
1. Block D shown in Fig. 2-10a moves with a velocity of 3 m/s. Determine the angular
velocities of connections BD and AB, at the situation show.
Figure 2-10a
Solution:
As D moves to the right, AB rotates in the clockwise direction over point A. Hence,
𝑣𝐵 is directed perpendicular to AB. The instantaneous center of zero velocity for BD is
located at the intersection of the line segments drawn perpendicular to 𝑣𝐵 and 𝑣𝐷 , Fig. 210b. From the geometry.
Figure 2-10b
61
𝑟𝐵/𝐶 = 0.4 tan 45° 𝑚 = 0.4𝑚
𝑟𝐷/𝐶 =
0.4 𝑚
= 0.5657𝑚
cos 45°
Since the magnitude of 𝑣𝐷 is known, the angular velocity of link BD is
𝜔𝐵𝐷 =
𝑣𝐷
𝑟𝐷/𝐼𝐶
=
3 𝑚/𝑠
= 𝟓. 𝟑𝟎 𝒓𝒂𝒅/𝒔 (𝑐𝑐𝑤)
0.5657 𝑚
The velocity of B is therefore
𝑣𝐵 = 𝜔𝐵𝐷 (𝑟 𝐵 ) = 5.30
𝐼𝐶
𝑟𝑎𝑑
(0.4𝑚) = 2.12 𝑚/𝑠
𝑠
From Fig. 2-10c, the angular velocity of AB is
𝜔𝐴𝐵 =
𝑣𝐵
2.12𝑚/𝑠
=
= 𝟓. 𝟑𝟎 𝒓𝒂𝒅/𝒔 (𝑐𝑤)
𝑟𝐵/𝐴
0.4 𝑚
2. The cylinder listed in Fig. 2-11a rolls between two moving plates E and D, without
slipping. Determine cylinder angular velocity and the velocity of its center C..
Figure 2-11a
Solution:
As no slipping occurs, the contact points A and B on the cylinder have the same
speeds as the plates E and D , respectively. In addition, the speeds 𝑣𝐴 and 𝑣𝐵 are parallel, so
that by the proportionality of right triangles the IC is located at a point on line AB, Fig. 211b. Assuming this point to be a distance x from B, we have
62
Figure 2-11b
𝑣𝐵 = 𝜔𝑋
𝑚
= 𝜔𝑋
𝑠
𝑚
0.25 = 𝜔(0.25𝑚 − 𝑋)
𝑠
0.4
𝑣𝐴 = 𝜔(0.25 𝑚 − 𝑋)
Dividing one equation into the other eliminates 𝜔 and yields
0.4(0.25 − 𝑋) = 0.25𝑋
𝑋=
0.1
= 0.1538 𝑚
0.65
Therefore, the angular velocity of the cylinder is
𝜔=
𝑣𝐵
0.4𝑚/𝑠
𝒓𝒂𝒅
=
= 𝟐. 𝟔𝟎
(𝑐𝑤)
𝑋
0.1538 𝑚
𝒔
The velocity of point C is therefore
𝑣𝐶 = 𝜔𝑟 𝐶 = 2.60
𝐼𝐶
𝑣𝐶 = 𝟎. 𝟎𝟕𝟓𝟎
𝑟𝑎𝑑
(0.1538 𝑚 − 0.125 𝑚)
𝑠
𝒎
𝒔
63
Absolute and Relative Acceleration
Velocity Analysis
Kinematic Diagram
Set the fixed x, y coordinates directions and draw the body's kinematic
diagram. Indicate on it𝑎𝐴 , 𝑎𝐵 , 𝜔, 𝛼, 𝑎𝑛𝑑 𝑟𝐵/𝐴 .
If points A and B travel along curved paths, their accelerations in terms of
their tangential and normal components should be indicated, i.e., 𝑎𝐴 = (𝑎𝐴 )𝑡 +
(𝑎𝐴 )𝑛 and 𝑎𝐵 = (𝑎𝐵 )𝑡 + (𝑎𝐵 )𝑛 .
Acceleration Equation
To apply 𝑎𝐵 = 𝑎𝐴 + 𝛼 𝑥 𝑟𝐵/𝐴 − 𝜔2 𝑟𝐵/𝐴 , express the vectors in Cartesian
vector form and substitute them into the equation. Evaluate the cross product and
then equate the respective i and j components to obtain two scalar equations.
𝑎𝐵 = acceleration of point 𝐵
𝑎𝐴 = acceleration of the base point 𝐴
𝛼 = angular acceleration of the body
𝜔 = angular velocity of the body
𝑟𝐵/𝐴 = position vector directed from 𝐴 to 𝐵
If the solution gives a negative answer for an unknown magnitude, it means
that the vector's sense of direction is contrary to that shown on the kinematic
diagram.
Scalar Analysis
Kinematic Diagram
If the acceleration equation is applied in scalar form, then the magnitudes
and directions of the relative-acceleration components (𝑎𝐵/𝐴 )𝑡 and (𝑎𝐵/𝐴 )𝑛 must be
established. To do this, draw a kinematic diagram such as shown in Fig. 2-12. Since
the body is considered to be momentarily “pinned” at the base point A, the
magnitudes of these components are (𝑎𝐵/𝐴 )𝑡 = 𝛼𝑟𝐵/𝐴 and (𝑎𝐵/𝐴 )𝑛 = 𝜔2 𝑟𝐵/𝐴 . Their
sense of direction is established from the diagram such that (𝑎𝐵/𝐴 )𝑡 acts
perpendicular to 𝑟𝐵/𝐴 , in accordance with the rotational motion 𝛼 of the body, and
(𝑎𝐵/𝐴 )𝑛 is directed from B toward A.
Figure 2-12
64
Acceleration Equation
Represent the vectors in 𝑎𝐵 = 𝑎𝐴 + (𝑎𝐵/𝐴 )𝑡 + (𝑎𝐵/𝐴 )𝑛 graphically by
showing their magnitudes and directions underneath each term. The scalar
equations of these vectors are calculated from the x and y components.
Consider a disk that rolls without slipping as shown in Figure 2-12a. As a
result, 𝑣𝐴 = 0 and so from the kinematic diagram in Figure 2-12b, the velocity of the
mass center G is
𝑣𝐺 = 𝑣𝐴 + 𝜔 × 𝑟𝐺/𝐴 = 0 + (−𝜔𝐤) × (𝑟𝐣)
(𝑒𝑞 2 − 3)
Figure 2-12a
Figure 2-12b
So that
𝑣𝐺 = 𝜔𝑟
Since G moves along a straight line, its acceleration in this case can be determined
from the time derivative of its velocity.
𝑑𝑣𝐺 𝑑𝜔
=
𝑟
𝑑𝑡
𝑑𝑡
𝑎𝐺 = 𝛼𝑟
(𝑒𝑞 2 − 4)
65
Examples:
1. The rod AB shown in Figure 2-13a is confined to travel along the tilted planes at A and B.
If point A has an acceleration of 3 m/s2 and a speed of 2 m / s, both of which are guided
down the plane at the moment when the rod is horizontal, evaluate the angular acceleration
of the rod at this moment.
Figure 2-13a
Solution:
Vector Analysis. We'll apply the equation of acceleration to points A and B on the
rod. For this to happen, the angular velocity of the rod must first be calculated. Show that is
is 𝜔 = 0.283 rad/s ↶ using either the velocity equation or the method of instantaneous
centers.
Kinematic Diagram. Since both points A and B travel along straight paths, they do
not have acceleration components normal to the paths. There are two unknowns in Figure
2-13b, namely, 𝑎𝐵 and 𝛼.
Figure 2-13b
Acceleration Equation
𝑎𝐵 = 𝑎𝐴 + 𝛼 × 𝑟𝐵/𝐴 − 𝜔2 𝑟𝐵/𝐴
𝑎𝐵 𝑐𝑜𝑠45°𝑖 + 𝑎𝐵 𝑠𝑖𝑛45°𝑗 = 3𝑐𝑜𝑠45°𝑖 − 3𝑠𝑖𝑛45°𝑗 + (𝛼𝑘) × (10𝑖) − (0.283)2 (10𝑖)
Carrying out the cross product and equating the i and j components yields
𝑎𝐵 𝑐𝑜𝑠45° = 3𝑐𝑜𝑠45° − (0.283)2 (10)
(1)
𝑎𝐵 𝑠𝑖𝑛45° = −3𝑠𝑖𝑛45° + 𝛼(10)
(2)
Solving, we have
66
𝑎𝐵 = 1.87𝑚/𝑠 2 ∡45°
𝜶 = 𝟎. 𝟑𝟒𝟒 𝒓𝒂𝒅/𝒔𝟐 ↶
2. The disk rolls without slipping and is shown in Figure 2-14a has an angular motion.
Determine point A's acceleration at this moment.
Figure 2-14a
Solution:
Vector Analysis
Kinematic Diagram. Since no slipping occurs, applying Eq. 2-4
𝑎𝐺 = 𝛼𝑟 = (4
𝑟𝑎𝑑
) (0.5𝑓𝑡) = 2𝑓𝑡/𝑠 2
𝑠2
Acceleration Equation
We will apply the acceleration equation to points G and A, Figure 2-14b
67
Figure 2-14b
𝑎𝐴 = 𝑎𝐺 + 𝛼 × 𝑟𝐴/𝐺 − 𝜔2 𝑟𝐴/𝐺
𝑎𝐴 = −2𝐢 + (4𝐤) × (−0.5𝐣) − (6)2 (−0.5𝐣)
𝑎𝐴 = 18𝒋 𝑓𝑡/𝑠2
Scalar Analysis
Using the result for 𝑎𝐺 = 2 ft/s2 determined above, and from the kinematic diagram,
showing the relative motion 𝑎𝐴/𝐺 , Figure 2-14c, we have
Figure 2-14c
𝑎𝐴 = 𝑎𝐺 + (𝑎𝐴⁄𝐺 )𝑥 + (𝑎𝐴⁄𝐺 )𝑦
Therefore,
𝑎𝐴 = √(0)2 + (18𝑓𝑡/𝑠 2 )2 = 18 ft/s
Reference
Hibbeler, R. C. (2016), Planar Kinematics of a Rigid Body, Engineering Mechanics Dynamics,
14th Edition
Meriam, J.L., Kraige, L.G., Plane Kinematics of Rigid Bodies, Engineering Mechanics
Dynamics, 6th Edition
68
Assessing Learning
Activity III
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
1. If the block at C falls down at 4 𝑓𝑡/𝑠, evaluate the angular velocity of bar AB at the
specified instant.
69
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
2. The disk is originally rotating at𝜔𝑂 = 12 𝑟𝑎𝑑/𝑠. If it is subjected to a constant angular
acceleration of 𝛼 = 20 𝑟𝑎𝑑/𝑠2, calculate the velocity magnitude and the acceleration n and
t components of point B when the disk undergoes 2 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠.
70
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
3. Determine in this instant the angular acceleration and angular velocity of the link AB.
Note: The Guide's upward movement is in the negative y direction. At the instant 𝜃 = 50°,
the slotted guide is moving upward with an acceleration of 3 𝑚/𝑠2 and a velocity of 2 𝑚/𝑠.
71
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
4. At the given instant shown, 𝜃 = 50°, and rod AB faces a deceleration of 16𝑚/𝑠2 when the
velocity is 10 𝑚/𝑠. Evaluate the angular velocity at this instant and the angular acceleration
of the link CD.
72
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
5. Calculate the acceleration of the ladder's bottom A and the angular acceleration of the
ladder at this moment At a given instant, the top B of the ladder has an acceleration 𝑎𝐵 =
2 𝑓𝑡/𝑠2 and a velocity of 𝑣𝐵 = 4 𝑓𝑡/𝑠, both acting downward.
73
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
6. Determine the angular velocity of connection AB at the moment shown if block C moves
upward at 12 𝑖𝑛 / 𝑠.
74
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
7. Calculate the velocity of the gear rack C. The pinion gear A rolls on the fixed gear rack B
with an angular velocity 𝜔 = 4 𝑟𝑎𝑑/s.
75
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
8. At a given moment, the bottom A of the ladder has an acceleration 𝑎𝐴 = 4 𝑓𝑡/𝑠2 and
velocity 𝑣𝐴 = 6 𝑓𝑡/𝑠, they both behave to the left. Calculate the acceleration of the top of the
ladder, B, and the ladder’s angular acceleration at this same instant.
76
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
9. The mechanism of the shaper is designed to give a slow cutting stroke and fast return to a
blade attached to the slider at C. Calculat the angular velocity of the CB link if the AB link
rotates at 4 𝑟𝑎𝑑 / 𝑠.
77
UNIT III. PLANE MOTION OF KINETICS OF RIGID BODIES: FORCE
AND ACCELERATION
Overview
This unit will give you a deeper understanding about different equations of motion,
how to use it, and clarify some misconceptions regarding to this topic.
Learning Objectives
At the end of the unit, I am able to:
1. Understand different equations of motion and the concept behind it;
2. Clarify some misconceptions about motion; and
3. Use these equations and solve motion problems easier.
Topics
3.1 Equations of Motion
78
Pre-Test
Name: _________________________________________________
Date: ___________________
Course/Year/Section: _______________________________
Directions: Solve the following questions as neatly as possible. Apply the basic concept of
Newton’s 2nd Law of Motion.
1. A 65 – kg woman descends in an elevator that briefly accelerates at 0.20g downward
when leaving the floor. She stands on a scale that reads in kg. (a) During this acceleration,
what is her weight and what does the scale read? (b) What does the scale read when the
elevator descends at a constant speed of 2.0 m/s?
79
Name: _________________________________________________
Date: ___________________
Course/Year/Section: _______________________________
Directions: Solve the following questions as neatly as possible. Apply the basic concept of
Newton’s 2nd Law of Motion.
2. Suppose a friend asks to examine the 10-kg box you were given, hoping to guess what is
inside. She then pulls the box by the attached cord, as shown in the figure, along the smooth
surface of the table. The magnitude of the force exerted by the person is Fp = 40 N, and it is
exerted at a 30° angle as shown. Calculate (a) the acceleration of the box, and (b) the
magnitude of the upward force FN exerted by the table on the box. Assume that friction can
be neglected.
80
Name: _________________________________________________
Date: ___________________
Course/Year/Section: _______________________________
Directions: Solve the following questions as neatly as possible. Apply the basic concept of
Newton’s 2nd Law of Motion.
3. Two boxes, A and B, are connected by a lightweight cord and are resting on a smooth
(frictionless) table. The boxes have masses of 12.0 kg and 10.0 kg. A horizontal force FP of
40.0 N is applied to the 10 kg box, as shown in the figure. Find (a) the acceleration of each
box (b) the tension in the cord connecting the boxes.
81
Name: _________________________________________________
Date: ___________________
Course/Year/Section: _______________________________
Directions: Solve the following questions as neatly as possible. Apply the basic concept of
Newton’s 2nd Law of Motion.
4. A 20-kg box rests on a table. (a) What is the weight of the box and the normal force acting
on it? (b) A 10-kg box is placed on the top of the 20-kg box, as shown in the figure.
Determine the normal force that the table exerts on the 20-kg box and the normal force that
the 20-kg box exerts on the 10-kg box.
82
Name: _________________________________________________
Date: ___________________
Course/Year/Section: _______________________________
Directions: Solve the following questions as neatly as possible. Apply the basic concept of
Newton’s 2nd Law of Motion.
5. The cable supporting a 2125-kg elevator has a maximum strength of 21,750 N. What
maximum upward acceleration can it give the elevator without breaking?
83
Lesson Proper
PLANE MOTION OF KINETICS OF RIGID BODIES: FORCE AND ACCELERATION
Equations of Motion
Kinetics is a branch of dynamics which deals with study of bodies in motion
particularly the force involved in the motion.
When a body accelerates Newton’s 2nd law relates motion of body to forces acting on
it.
Newton’s Second law the accelerationS (a) of a body is directly proportional to the
net force (Fnet) acting on it and inversely proportional to its mass (m).
𝑎∝
𝐹𝑛𝑒𝑡
𝑚
𝑎=
𝐹𝑛𝑒𝑡
𝑚
∑ 𝐹 = 𝐹𝑛𝑒𝑡 = 𝑚𝑎
This shows that the force, F causes translational motion.
The mass, m, is the resistance to the translational motion.
Friction, if a moving particle contacts a rough surface, it may be necessary to use the
frictional equation, which relates the frictional and normal forces Ffr and N acting at the
surface of contact by using the coefficient of kinetic friction 𝐹𝑓𝑟 = 𝜇𝑘 𝑁. Friction always act
on the free-body diagram such that it opposes the motion of the particle relative to the
surface it contracts. If the particle in on the verge of relative motion, the the coefficient of
static friction should be used.
Free Body Diagram:
To identify external force acting on the rigid body.
84
Translation Motion Equation
∑ 𝐹𝑥 = 𝑚(𝑎𝐺 )𝑥
∑ 𝐹𝑦 = 𝑚(𝑎𝐺 )𝑦
∑ 𝑀𝐺 = 0
Examples:
1. The car shown has a mass of 2000 kg and a center of mass at G. Determine the
acceleration if the rear “driving” wheels are always slipping, whereas are free to rotate.
Neglect the mass of the wheels. The coefficient of kinetic friction between the wheels and
the road is 𝜇 = 0.25.
Solution:
Solve for acceleration using three equation
⃗⃗ ∑ 𝐹𝑥 = 𝑚(𝑎𝐺 )𝑥
+
eq. 1
−0.25𝑁𝐵 = −(2000𝑘𝑔)𝑎𝐺
∑ 𝐹𝑦 = 𝑚(𝑎𝐺 )𝑦
eq. 2
𝑁𝐴 + 𝑁𝐵 − 2000(9.81) = 0
∑ 𝑀𝐺 = 0
−𝑁𝐴 (1.25𝑚) − 𝐹𝐵 (0.3𝑚) + 𝑁𝐵 (0.75𝑚) = 0
85
−𝑁𝐴 (1.25𝑚) − 0.25𝑁𝐵 (0.3𝑚) + 𝑁𝐵 (0.75𝑚) = 0
eq. 3
−𝑁𝐴 (1.25𝑚) + 0.675𝑁𝐵 = 0
Eliminate NA to solve NB in eq 1 and 2
−𝑁𝐴 (1.25𝑚) + 0.675𝑁𝐵 = 0
[𝑁𝐴 + 𝑁𝐵 − 2000(9.81) = 0](1.25)
𝑁𝐵 = 12.7 𝐾𝑁
𝑁𝐴 = 6.68𝑘𝑁
Solve for aG in equation 1 : 𝑁𝐵 = 12.7 𝐾𝑁
−0.25(12.7) = −(2000𝑘𝑔)𝑎𝐺
𝒂𝑮 = 𝟏. 𝟓𝟗𝒎/𝒔𝟐
2. If a 80 lb force is applied to the 100 lb uniform crates as shown, determine the linear
acceleration aG of this crate. The coefficient of kinetic friction between the crate and the
surface is 𝜇 = 0.20.
Solution:
From Static we learned that through equilibrium, if x is calculated to be b/2 or
above, that means that point A is right there at the corner, and that indicates the crate is
going to be tipping over.
Assume sliding, not tipping, occurs:
𝐹𝑓𝑟 = 𝜇𝑘 𝑁 = 0.2𝑁
⃗+
⃗ ∑ 𝐹𝑥 = 𝑚(𝑎𝐺 )𝑥
100
80 − 𝐹𝑓𝑟 = 32.2 𝑎𝐺
80 − 0.2𝑁 =
100
𝑎
32.2 𝐺
eq. 1
86
∑ 𝐹𝑦 = 𝑚(𝑎𝐺 )𝑦
100 − 𝑁 = 0
𝑁 = 100
∑ 𝑀𝐺 = 0
−80(1) + 𝑁(𝑥) − 𝐹𝑓𝑟 (2) = 0
−80(1) + 𝑁(𝑥) − 0.2𝑁(2) = 0
𝑥 = 1.2𝐹𝑇 < 1.5 𝑓𝑡
The assumption is correct
Substitute N = 100 to equation 1
80 − 0.2(100) =
100
𝑎
32.2 𝐺
𝑎𝐺 = 19.3 𝑓𝑡/𝑠 2
General Plane Motion Equations
A body may have a variety of forces acting on it including gravity. These forces and
moment will cause the body to moved and rotate, tending to accelerate in x and y and rotate
about its CG.
∑ 𝐹𝑥 = 𝑚(𝑎𝐺 )𝑥 – moves in a straight line
∑ 𝐹𝑥 = 𝑚(𝑎𝐺 )𝑥 – still moves in a straight line but also
has a tendency to rotate about the center of gravity
The force applied at a distance, d, from the mass center will cause a moment which
tries to rotate the mass center, G, such that
𝐹𝑛𝑒𝑡 (𝑑) = 𝐼𝐺 ∝
87
I = mass moment of inertia
𝑀𝐺 = 𝐼𝐺 ∝
(𝐹𝑛𝑒𝑡 (𝑑)) − 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑓𝑜𝑟𝑐𝑒
𝑚(𝑑) − 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑚𝑎𝑠𝑠
𝐴(𝑑) − 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑎𝑟𝑒𝑎
From this we develop a set of equations that defined 2D motion. These are general
equation which apply in all circumstances.
∑ 𝐹𝑥 = 𝑚(𝑎𝐺 )𝑥
∑ 𝐹𝑦 = 𝑚(𝑎𝐺 )𝑦
𝑀𝐺 = 𝐼𝐺 ∝
If the moment about G are hard to find, we can choose another point of convenience
to sum moment about and the moment equation becomes.
∑ 𝑀𝑝 = 𝑚(𝑎𝐺 )𝑦 𝑥̅ + 𝑚(𝑎𝐺 )𝑥 𝑦̅
Where 𝑥̅ and 𝑦̅ are the distance of P from G. These terms
𝑚(𝑎𝐺 )𝑦 and 𝑚(𝑎𝐺 )𝑥 𝑦̅ are the factors in the “kinetics
equation”. Similar to our angular momentum equation
having moment of momentum, these are the ‘ moments of
the mass x acceleration’’
Examples:
1. How fast does a bicycle and rider have to accelerate to just bring the front wheel off the
ground?
Given: 𝑊𝑅 = 165𝑙𝑏
Required 𝑎 =?
𝑊𝐵 = 30𝑙𝑏
𝑑1 = 1.5𝑓𝑡
𝑑2 = 2𝑓𝑡
ℎ1 = 2.5𝑓𝑡
ℎ2 = 4 𝑓𝑡
88
Solution:
Using FBD.
∑ 𝑀𝐵 = 𝑚(𝑎𝐺 )𝑦 𝑥̅ + 𝑚(𝑎𝐺 )𝑥 𝑦̅
𝑊𝑅 𝑑1 + 𝑊𝐵 𝑑2 = 𝑚𝑅 (𝑎)(ℎ1 ) + 𝑚𝐵 (𝑎)(ℎ2 )
165
30
165(1.5) + 30(2) = 32.2 (𝑎)(4) + 32.2 (𝑎)(2.25) ; a = 13.5 ft/s2
2. The 18- kg wheel is rolling under the constant moment of 80 N.m. If the wheel has mass
center at point G and the radius of gyration is rg = 0.30m, determine its angular acceleration
and the linear acceleration of its mass center. The kinetic coefficient of friction between the
wheel is 0.30.
Solution:
Using FBD
∑ 𝐹𝑥 = 𝑚(𝑎𝐺 )𝑥
𝐹𝑓𝑟 = 𝑚(𝑎𝐺 )𝑥 = 18(𝑎𝐺 )𝑥
𝜇𝑘 𝑁 = 18(𝑎𝐺 )𝑥
0.3(177) = 18(𝑎𝐺 )𝑥
(𝑎𝐺 )𝑥 = 𝟐. 𝟗𝟓𝒎/𝒔𝟐
89
∑ 𝐹𝑦 = 𝑚(𝑎𝐺 )𝑦
𝑁−𝑊 =0
𝑁 = 𝑊 = 18(9.81) = 177𝑁
∑ 𝑀𝐺 = 𝐼𝐺 ∝
80 − 𝐹𝑓𝑟 (0.5) = 𝐼𝐺 ∝
where IG = mrg2
80 − 𝐹𝑓𝑟 (0.5) = m𝑟𝑔2 ∝
80 − 𝐹𝑓𝑟 (0.5) = 18(0.3)2 ∝
80 − 𝜇𝑘 𝑁(0.5) = 18(0.3)2 ∝
80 − 0.3(177)(0.5) = 18(0.3)2 ∝
∝ = 33.0 rad/s2
90
Assessing Learning
Activity IV
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
1. The 20 – kg wheel is rolling while slipping under the constant force of 300 N through the
handle as shown. If the wheel has mass center at a point G and the radius of Gyration is rg =
0.35, determine the linear acceleration of its mass center. The coefficient of kinetic friction
between the wheel and the ground 𝜇𝑘 = 0.32. Hint: what is the direction of the friction
force?
91
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
2. The assembly has a mass of 6 Mg and is hoisted using the boo and pulley system. If the
winch at B draws in the cable with an acceleration of 2 m/s2, Determine the compressive
force in the hydraulic cylinder needed to support the boom. The boom has a mass of 1.8 Mg
and mass center at G.
92
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
3. The uniform 200-lb beam is initially at rest when the forces are applied to the cable.
Determine the magnitude of the acceleration of the mass center and the angular
acceleration of the beam at this instant.
93
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
4. For a 90- kg crate simply standing on the moving cart, what is the maximum acceleration
a the cart can have without the crate tipping over or sliding relative to the cart? Take the
coefficient of static friction 𝜇𝑠 = 0.5 between the crate and the cart.
94
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
5. The 18 – kg wheel is rolling while slipping under the constant force of 300 N through the
handle as shown. If the wheel has mass center at a point G and the radius of Gyration is rg =
0.35, determine the angular acceleration of its mass center. The coefficient of kinetic friction
between the wheel and the ground 𝜇𝑘 = 0.32. Hint: what is the direction of the friction
force?
95
UNIT V. PLANE MOTION OF KINETICS OF RIGID BODIES: WORK AND
ENERGY
Overview
This unit provides you a broad discussion about the principles and relationships of
work and energy of rigid bodies, how the forces and moments do work, how the
conservation of energy can be utilized to solve the kinetic problems of rigid bodies in plane
motion.
Learning Objectives
At the end of the unit, I am able to:
1. Formulate solutions for the kinetic energy of rigid bodies;
2. Define and solve the different ways of the work of forces and moments;
3. Apply the principle of work and energy in solving kinetic problems of rigid
bodies in plane motion involving force, velocity, and position; and
4. Apply the principle of the energy conservation in solving kinetic problems of
rigid bodies also in plane motion.
Topics
5.1
5.2
5.3
5.4
Kinetic Energy of Rigid Bodies
The Work of Forces and Couple Moments
The Principle of Work and Energy of Rigid Bodies
The Conservation of Energy
96
Pre-Test
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
1. A spring is to be placed at the bottom of the elevator shaft. Considering a total mass of 𝑀
of the passengers and the elevator, the cable of the elevator breaks when the elevator is the
height h above the top of the spring. Find the value of the spring constant k so that the
passengers accelerate at 4 g.
97
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
2. A steel cable must lift a precast concrete with a mass of 300 𝑘𝑔. If the concrete is lifted at
a height of 15 𝑚 with an acceleration of 2 𝑚/𝑠 2 , find the tension at the cable, the work done
on the precast concrete, the work done by steel cable on the concrete, and the final velocity
of the concrete. The concrete accelerated from rest.
98
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
3. A 30 𝑘𝑔 circular disk is subjected to a couple moment of 35 𝑁 − 𝑚. The disk rotates from
rest, if the disk made 3 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠, find the angular velocity.
99
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
4. A sphere with a mass 𝑚 is released from rest on an incline with an angle of 𝜃. Find the
velocity of the sphere after it has rolled through a distance ℎ corresponding to a change of
elevation of its center.
100
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
5. A 1.65 𝑚 long rod with a mass of 20 𝑘𝑔 is pin-connected about a point half meter from
the right end of the rod. The left end is pressed against a vertical spring of spring constant
𝑘 = 305 𝑁/𝑚. When the rod is in exact horizontal position, the spring is compressed 5 𝑐𝑚.
If the rod is suddenly released from the horizontal position, the spring will push it upward.
Find the angular velocity of the rod as it passes through a vertical position.
101
Lesson Proper
PLANE MOTION OF KINETICS OF RIGID BODIES: WORK AND ENERGY
Kinetic Energy of Rigid Bodies
In this part, the basic methods of work and energy that have been discussed in
Physics will be applied in solving problems involving force, velocity and object positions,
but this time in a more complex type of situations, in planar motion problems. Before diving
into the deeper approach of kinetic energy, it will be essential to distinguish first if the body
is subjected to translation, rotation about a fixed axis or general plane motion.
Translation
Translation can be defined in two types: it can be subjected to either rectilinear or
curvilinear motion. If the rigid body is in translation only, therefore, the kinetic energy of
the rigid body is
𝑻=
𝟏
𝒎𝒗𝟐𝑮
𝟐
The kinetic energy due to the rotation is already zero since there is no rotation
happened in the body.
Rotation about a Fixed Axis
102
If the rigid body rotates about a fixed axis that passes through a single point, the
rigid body has a rotational kinetic energy and the already existing translational kinetic
energy due to the motion. The kinetic energy can now be defined as
𝑻=
𝟏
𝟏
𝒎𝒗𝟐𝑮 + 𝑰𝑮 𝝎𝟐
𝟐
𝟐
The kinetic energy of the body in translational and rotational motion can also be
solved by
𝑻=
𝟏
𝑰 𝝎𝟐
𝟐 𝑶
Note that the rotational and translational motion have a relationship in terms of
1
velocity, 𝑣 = 𝑟𝜔, substituting the value of the translational velocity, we have 𝑇 = (𝐼𝐺 +
2
𝑚𝑟𝐺2 )𝜔2 . The quantity inside the parenthesis is equivalent to the moment of inertia, 𝐼𝑂 , of
the rigid body rotating about an axis perpendicular to the plane of motion and passing
through a single point.
General Plane Motion
If the rigid body is in general plane motion, the body has an angular velocity 𝜔 and
the center of mass has a translational velocity, 𝑣𝐺 . General plane motion has the same
kinetic energy as the rotation.
𝑻=
𝟏
𝟏
𝒎𝒗𝟐𝑮 + 𝑰𝑮 𝝎𝟐
𝟐
𝟐
Considering the body’s motion about its instantaneous center of zero velocity, the
kinetic energy can be defined as
𝑻=
𝟏
𝑰 𝝎𝟐
𝟐 𝑰𝑪
103
The Work of Forces
In Physics, the work of forces has several types referring to the different ways a
body does a motion, the work done by a constant force, the work done by a varying force
and the work done equivalent to the potential energies such as the gravity (weight) and the
spring elasticity.
Work done by a Constant Force
When a force acts on a rigid body externally and the force doesn’t change its
magnitude in a constant direction moving in a translation, then
𝑾 = 𝑭𝒅
𝑾 = (𝑭 𝐜𝐨𝐬 𝜽) 𝒅
Work done by a Varying Force
When a force acts on a rigid body externally, the work done by the force
when the body moves along the path of the motion is defined as
104
𝑾 = ∫ 𝑭𝒅𝒓
𝑾 = ∫ (𝑭 𝐜𝐨𝐬 𝜽) 𝒅𝒔
𝒔
Work done by Gravity
The gravitational work or the work done by the Weight is occurred when the
center of mass of the body takes a displacement in a vertical manner, considering
downward as negative, therefore
𝑾 = −𝑷𝑬𝒈 = −𝒎𝒈∆𝒚
Work done by Spring Force
When a body is attached to a spring, the spring exerts a force to the body
and creating a work as it compresses or stretches from on
𝑾 = 𝑷𝑬𝒆𝒍 =
105
𝟏 𝟐
𝒌𝒔
𝟐
The Work of Couple Moments
The figure above is subjected to a couple moment, 𝑀 = 𝐹𝑟. When the body
experiences a differential displacement, we can calculate the work done by couple forces by
the displacement of the body as it is equivalent to the sum of its translation and rotation. In
translation of a body, the work done by the forces can only be found by multiplying the
components of the displacement to the forces applied to the body considering one line of
motion, 𝑑𝑠𝑡 .
The negative work of one force is cancelled by the positive work of the opposite
force. When the body experiences a differential rotation 𝑑𝜃 about an arbitrary point, a
𝑟
2
displacement will be experienced by each force, 𝑑𝑠𝜃 = ( ) 𝑑𝜃, same with the direction of
the force.
Thus, the total work done by the couple moment can be defined by
𝒓
𝒓
𝒅𝑾𝑴 = 𝑭 ( ) 𝒅𝜽 + 𝑭 ( ) 𝒅𝜽 = (𝑭𝒓)𝒅𝜽 = 𝑴𝒅𝜽
𝟐
𝟐
Take note that the direction will affect the magnitude of the work, if 𝑀 and 𝑑𝜃 have
the same direction, the work is positive and negative if the direction is opposite.
The work of a couple moments can also be defined if the rigid body rotates in planar
motion through a finite angle (in radians from 𝜃1 to 𝜃2 ).
𝜃2
𝑾𝑴 = ∫ 𝑴𝒅𝜽
𝜃1
106
The Principle of Work and Energy of Rigid Bodies
We can recall the basic principle of the work and energy relationship with regards
to the changes of the body’s velocity, external force and position as
𝑲𝑬𝟏 + 𝑷𝑬𝟏 = 𝑲𝑬𝟐 + 𝑷𝑬𝟐
The principle of this equation is still applicable even if the rotational energy is to be
considered, the final translational and rotational kinetic energy of the body is equivalent to
the sum of the initial translational and rotational kinetic energy of the body and the total
work done by the external forces and couple moments applied to the body being analyzed.
As we study rigid bodies, no relative movement between internal forces occur, therefore,
internal forces cannot produced an internal work. Take note that some rigid bodies are
connected by pin, cable or in mesh with another body, in these types of connected body
systems, the work and energy principle can still be applied. Internal forces are ignored.
Examples:
1. A man created a wheel by assembling a 1 − 𝑚 diameter ring and a pair of rods, the ring
weighs 10 𝑘𝑔 and each rod weigh 1.5 𝑘𝑔. He attached a spring at the center of the wheel
with a stiffness of 2.5 𝑁 − 𝑚/𝑟𝑎𝑑. Find the maximum angular velocity of the wheel after
being released from rest if he rotated it until a torque of 30 𝑁 − 𝑚 is reached.
Solution:
The mass moment of inertia of the entire wheel assembled about the center is 𝐼𝑐 .
The quantity 𝑚𝑅 represents the mass of the ring, 𝑟 represents the radius, 𝑚𝑟 represents the
mass of the rod and 𝑙 represents the length of the rod.
1
𝐼𝑐 = 𝑚𝑅 𝑟 2 + 2 ( 𝑚𝑟 𝑙 2 )
2
1
𝐼𝑐 = (10 𝑘𝑔)(0.5 𝑚)2 + 2 [( ) (1.5 𝑘𝑔)(1 𝑚)2 ]
2
𝑰𝒄 = 𝟐. 𝟕𝟓 𝒌𝒈 − 𝒎𝟐
Thus, the kinetic energy of the wheel is
1
1
𝐾𝐸2 = 𝐼𝑐 𝜔2 = (2.75 𝑘𝑔 − 𝑚2 )𝜔2
2
2
107
𝐾𝐸2 = 𝟏. 𝟑𝟕𝟓𝝎𝟐
Since, the wheel is released from rest, 𝐾𝐸1 = 0. The torque is 𝑀 = 𝑘𝜃 = 2.5𝜃. Using
this equation, we can solve for the angle of rotation needed to produce the 30 𝑁 − 𝑚 torque.
𝑀 = 2.5𝜃
30 𝑁 − 𝑚 = 2.5𝜃
𝜽 = 𝟏𝟐 𝒓𝒂𝒅
Thus, the work done
𝜃2
𝑊𝑀 = ∫ 𝑀𝑑𝜃
𝜃1
𝜃2
𝑊𝑀 = ∫ 2.5𝜃𝑑𝜃
𝜃1
12
𝑊𝑀 = ∫ 2.5𝜃𝑑𝜃 =
0
2.5 2 12
𝜃 |
2
0
𝑾𝑴 = 𝟏𝟖𝟎 𝑱
From the principle of work and energy
𝐾𝐸1 + 𝑊𝑀 = 𝐾𝐸2
0 + 180 𝐽 = 1.375𝜔2
𝝎 = 𝟏𝟏. 𝟒𝟒 𝒓𝒂𝒅/𝒔
The angular velocity of the wheel is 11.44 𝑟𝑎𝑑/𝑠.
2. A 2.5 diameter reel weighing 250 kg is resting on two rollers 1 m apart from each other
until a horizontal pulling force of 85 N is applied using a cable causing the reel to turn. If the
turn made 2.5 revolutions from rest, find the angular velocity assuming the radius of
gyration of the reel about its center remains the same, 𝑘𝑂 = 0.7 𝑚. Ignore the mass of
rollers and the cable.
108
Solution:
Since, the reel turned from rest, therefore, 𝐾𝐸1 = 0.The mass moment of inertia of
the reel about the center is
𝐼𝐶 = 𝑚𝑘𝑂2 = (250 𝑘𝑔)(0.7𝑚)2
𝑰𝑪 = 𝟏𝟐𝟐. 𝟓 𝒌𝒈 − 𝒎𝟐
Thus, the final kinetic energy is
1
1
𝐾𝐸2 = 𝐼𝐶 𝜔2 = (122.5 𝑘𝑔 − 𝑚2 )𝜔2
2
2
𝑲𝑬𝟐 = 𝟔𝟏. 𝟐𝟓𝝎𝟐
From the free-body diagram, the pulling force from the cable is the only one that
created a positive work. To calculate the work done by the force at the cable, the total
displacement is needed. When the cable is pulled, the reel rotates 2.5 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠,
therefore, the force displaces 𝑑 = 𝑆 = 𝑟𝜃 = 0.80𝑚(2.5𝑟𝑒𝑣)(2𝜋) = 4𝜋 𝑚.
𝑊𝑃 = 𝐹𝑑 = (85 𝑁)(4𝜋 𝑚)
𝑾𝑷 = 𝟑𝟒𝟎𝝅 𝑱
From the principle of work and energy
𝐾𝐸1 + 𝑊𝑃 = 𝐾𝐸2
0 + 340𝜋 𝐽 = 61.25𝜔2
𝝎 = 𝟐. 𝟑𝟓𝟔 𝒓𝒂𝒅/𝒔
The angular velocity of the reel is 2.356 𝑟𝑎𝑑/𝑠.
109
3. A 4 𝑚 uniform slender rod with a mass of 15 𝑘𝑔 is suspended and resting until a
horizontal force is applied to its end. If the force of 175 𝑁 made the rod to rotate a 180°
clockwise from the initial position, find the angular velocity of the rod. Take note that the
force is always perpendicular to the rod.
Solution:
The rod starts from rest, therefore, 𝐾𝐸1 = 0. The mass moment of inertia of the
slender rod about its center is
𝐼𝐶 =
1
1 2
𝑚𝑙 2 + 𝑚 ( 𝑙)
12
2
𝐼𝐶 =
2
1
1
(15 𝑘𝑔)(4 𝑚)2 + (15 𝑘𝑔) [( ) (4 𝑚)]
12
2
𝑰𝑪 = 𝟖𝟎 𝒌𝒈 − 𝒎𝟐
1
1
𝐾𝐸2 = 𝐼𝐶 𝜔2 = (80 𝑘𝑔 − 𝑚2 )𝜔2
2
2
𝑲𝑬𝟐 = 𝟒𝟎𝝎𝟐
From the free-body diagram, the horizontal force creates a positive work as it
pushes the rod clockwise making an angular displacement 𝜃 whereas the weight of the rod
itself creates a negative work. The rod rotates an angle 𝜃 = 180°, therefore, 𝑑𝐹 = 𝑆𝐹 = 𝑟𝜃 =
(4 𝑚)(𝜋) = 4𝜋 𝑚 and the displacement of the weight, 𝑑𝑊 = 𝑆𝑊 = 4 𝑚.
The work of the force is
𝑊𝐹 = 𝐹𝑑𝐹 = (175 𝑁)(4𝜋 𝑚)
𝑾𝑭 = 𝟕𝟎𝟎𝝅 𝑱
The work of the weight is
𝑊𝑊 = 𝐹𝑑𝑊 = −(15 𝑘𝑔) (9.81
𝑚
) (4 𝑚)
𝑠2
𝑾𝑾 = −𝟓𝟖𝟖. 𝟔 𝑱
The total work of the force and the weight is
𝑊𝑇 = 𝑊𝐹 + 𝑊𝑊 = 700𝜋 𝐽 + (−588.6 𝐽)
𝑾𝑻 = 𝟏𝟔𝟏𝟎. 𝟓𝟐 𝑱
From the principle of work and energy
110
𝐾𝐸1 + 𝑊𝑇 = 𝐾𝐸2
0 + 1610.52 𝐽 = 40𝜔2
𝝎 = 𝟔. 𝟑𝟒𝟓 𝒓𝒂𝒅/𝒔
The angular velocity of the reel is 2.356 𝑟𝑎𝑑/𝑠.
4. A 0.5 𝑚 diameter disk with a mass of 45 𝑘𝑔 attached to two steel rods with its center is
being rotated. The disk is in contact with a belt, so when the disk rotates, the belt moves. If
the belt moves with a speed of 1.75 𝑚/𝑠, how many revolutions does the disk turn before it
reaches its constant angular velocity? The coefficient of friction between the disk and belt is
𝜇 = 0.25.
Solution:
To determine the work at the disk, the force of friction between the disk and the belt
must be calculated first, and for that we solve for the normal force 𝑭𝑵 .
∑ 𝐹𝑦 = 𝑚𝑎𝑦
∑ 𝐹𝑦 = 𝐹𝑁 + 𝑊
Since, there is no vertical motion, vertical acceleration is zero.
0 = 𝐹𝑁 − (45 𝑘𝑔) (9.81
𝑚
)
𝑠2
𝑭𝑵 = 𝟒𝟒𝟏. 𝟒𝟓 𝑵
For the frictional force
𝐹𝑓𝑟 = 𝜇𝑘 𝐹𝑁 = (0.25)(441.45 𝑁)
𝑭𝒇𝒓 = 𝟏𝟏𝟎. 𝟑𝟔 𝑵
As the disk rotates, the frictional force produces a constant couple moment about
the center of the disk, 𝑀 = 𝐹𝑓𝑟 𝑟 = (110.36 𝑁)(0.25 𝑚) = 27.59 𝑁 − 𝑚. Tis couple moment
creates a positive work with the disk’s angular displacement.
𝑊𝑀 = 𝑀𝜃
𝑾𝑴 = 𝟐𝟕. 𝟓𝟗𝜽
111
Take note that whenever a force does not take a displacement, there is no work
created, therefore, the normal force, the weight and the force at the steel rod produced zero
work.
The constant angular velocity of the disk is reached when the point of contact of the
𝑚
disk and belt reaches its linear velocity, 𝑣 = 1.75 𝑚/𝑠. From 𝑣 = 𝑟𝜔, 1.75 𝑠 = (0.25 𝑚)𝜔,
𝜔 = 7 𝑟𝑎𝑑/𝑠.
The disk has zero initial kinetic energy because it started to rotate from rest
position. However, the final kinetic energy can be solved by the formula, first, determine the
mass moment of inertia of the disk about its center.
1
1
𝐼𝐶 = 𝑚𝑟 2 = (45 𝑘𝑔)(0.25 𝑚)2
2
2
𝑰𝑪 = 𝟏. 𝟒𝟎𝟔 𝒌𝒈 − 𝒎𝟐
1
1
𝑟𝑎𝑑 2
2
𝐾𝐸2 = 𝐼𝐶 𝜔 = (1.406 𝑘𝑔 − 𝑚 ) (7
)
2
2
𝑠
𝑲𝑬𝟐 = 𝟑𝟒. 𝟒𝟓 𝑱
From the principle of work and energy
𝐾𝐸1 + 𝑊𝑀 = 𝐾𝐸2
0 + 27.59𝜃 = 34.45 𝐽
𝜃 = 1.2485 𝑟𝑎𝑑𝑖𝑎𝑛𝑠
𝜽 = 𝟎. 𝟐 𝒓𝒆𝒗𝒐𝒍𝒖𝒕𝒊𝒐𝒏𝒔
The Conservation of Energy
Aside from the principle of the work and energy relationship, the theorem of the
conservation of energy is another useful tool in solving such problems. Situations of rigid
bodies having only applied by conservative forces are much easier to calculate since these
types of forces doesn’t depend on the path of motion, dependent only on the initial and final
positions of the body being observed.
If gravitational and spring forces are applied to the rigid body, their potential
energies will be summated before and after the body’s position. However, non-conservative
forces like friction and other drag-resistant forces may still be present to the motion and
cause changes to the final total amount of the body’s energy. The work of non-conservative
forces is transformed into thermal energy occuring at the contact surface and then
dissipated into the surroundings and may not be recovered.
𝑲𝑬𝟏 + 𝑷𝑬𝟏 = 𝑲𝑬𝟐 + 𝑷𝑬𝟐 + 𝑾𝑵𝑪
𝑊𝑁𝐶 is the work done by non-conservative forces.
The total amount of energy considering the kinetic energy and potential energy
remains constant from the body’s initial position to the final, this principle is stated by the
112
conservation of mechanical energy. This principle of the mechanical energy conservation
also applies to the more complicated systems of rigid bodies such as the pin-connected
bodies and the bodies connected by cords and mesh, also connected with other bodies. The
internal forces at the points of connection will be ignored again from the rigid body’s
analysis.
Example:
1. A 18 𝑘𝑔 wheel with a 500 𝑚𝑚 diameter has a radius of gyration of 𝑘𝑂 = 175 𝑚𝑚.
Attached to it is a 10 𝑘𝑔 block, the block is released from rest making a vertical downward
motion. If the wheel reaches an angular velocity of 7.5 𝑟𝑎𝑑/𝑠 as the block falls, find the total
distance the block fall before the angular velocity is reached. Find the tension in the cord
while the block falls. Ignore the mass of the cord.
Solution:
The mass moment of inertia of the wheel about its center is 𝐼𝐶 = 𝑚𝑘𝑂2 =
(18 𝑘𝑔)(0.175 𝑚)2 = 0.55 𝑘𝑔 − 𝑚2 . The initial kinetic energy is already zero, to determine
the final kinetic energy, we must calculate the velocity of the block, 𝑣𝐵 = 𝑟𝑊 𝜔𝑊 = 0.25𝜔𝑊 .
1
1
1
1
𝐾𝐸2 = 𝐼𝐶 𝜔𝑊 2 + 𝑚𝐵 𝑣𝐵 2 = (0.55 𝑘𝑔 − 𝑚2 )𝜔𝑊 2 + (10 𝑘𝑔)(0.25𝜔𝑊 )2
2
2
2
2
𝐾𝐸2 = 0.588𝜔𝑊 2
Since, the wheel reaches the angular velocity of 7.5 𝑟𝑎𝑑/𝑠, therefore
𝐾𝐸2 = 0.588(7.5 𝑟𝑎𝑑/𝑠)2
𝑲𝑬𝟐 = 𝟑𝟑. 𝟎𝟕𝟓 𝑱
To determine the total distance that the block falls, we use the conservation of
energy. We already know the initial and final kinetic energies, next we find the potential
energies of the block. Referring to the free-body diagram, the points set from 𝑦1 to 𝑦2 with
113
𝑦1 set as zero, therefore, the initial potential energy is zero. For the final potential energy,
𝑃𝐸2 = 𝑚𝑔𝑦2 . The value of 𝑦2 will be the value of the total distance needed.
By conservation of energy
𝐾𝐸1 + 𝑃𝐸1 = 𝐾𝐸2 + 𝑃𝐸2
0 + 0 = 33.075 𝐽 + 𝑚𝑔𝑦2
Since, the displacement of points goes downwards, 𝑦2 is negative.
0 + 0 = 33.075 𝐽 + (10 𝑘𝑔) (9.81
𝑚
) (−𝑦2 )
𝑠2
𝒚𝟐 = 𝟎. 𝟑𝟑𝟕 𝒎
The total distance travelled by the block is 0.337 𝑚.
From the principle of work and energy, the force of tension 𝐹𝑇 can be determined.
𝑟𝑎𝑑
The velocity of the block is 𝑣𝐵 = 𝑟𝑊 𝜔𝑊 = (0.25 𝑚) (7.5 ) = 1.875 𝑚/𝑠.
𝑠
𝐾𝐸1 + 𝑊 = 𝐾𝐸2
0 + (10 𝑘𝑔) (9.81
𝑚
1
𝑚 2
(0.337
(10
)
𝑚)
−
𝐹
(0.337
𝑚)
=
𝑘𝑔)
(1.875
)
𝑇
𝑠2
2
𝑠
𝑭𝑻 = 𝟏𝟗. 𝟖𝟔 𝑵
The force of tension in the cord is 19.86 𝑁.
References
Hibbler, R. C., 2016 – Engineering Mechanics: Dynamics Fourteenth Edition, Pearson
Prentice Hall Pearson Education, Inc.
114
Assessing Learning
Activity V
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
1. A one meter long rod with a mass of 12.5 𝑘𝑔 is pin-connected at a wall. The other end of
the rod is attached to a spring of 𝑘 = 40 𝑁/𝑚 stiffness. When the spring is unstretched, the
angle the rod makes to the wall is 25°. If the rod is being rotated, the spring remains its
horizontal position due to the roller support. The rod is subjected to a couple Moment of
𝑀 = 20 𝑁 − 𝑚. Find the angular velocity of the rod at the angle 𝜃 = 75°.
115
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
2. A half-meter rod with a mass of 20 𝑘𝑔 is pushed by a force of 295 𝑁. Before the force acts
at the right end of the rod, the angle it makes to the horizontal is 𝜃 = 0°. Find the angular
velocity of the rod at the angle of 𝜃 = 45°.
116
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
3. The linkage below consists of two rods and a bar all connected by pin. Each rod weighs
5 𝑘𝑔 and has length of 1.3 𝑚, 𝐴𝐵 and 𝐶𝐷, and the bar 𝐵𝐷 weighs 15 𝑘𝑔 and a length of 2 𝑚.
At an angle of 𝜃 = 0°, the left rod is rotating with an angular velocity of 3 𝑟𝑎𝑑/𝑠. Find the
angular velocity of the left rod if the angle they make is 𝜃 = 55°, assuming the right rod is
subjected to a couple moment of 𝑀 = 30 𝑁 − 𝑚.
117
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
4. Two gears support a heavy cylinder that weighs 65 𝑘𝑔. The smaller gear weighs 12 𝑘𝑔
and a radius of gyration of 125 𝑚𝑚 about its center of mass. Attached to the bigger gear is a
drum that holds the cylinder, the total mass of the bigger gear and the drum is 35 𝑘𝑔, the
radius of gyration of both the bigger gear and the drum about their center of mass is
150 𝑚𝑚. If the cylinder falls a distance of 2.5 𝑚, find the velocity the cylinder reached. The
diameters of the smaller gear, bigger gear and the drum are 150 𝑚𝑚, 200 𝑚𝑚 and 100 𝑚𝑚
respectively.
118
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
5. The steel linkage below is made of two bars connected by pin, each bar weighs 15 𝑘𝑔 and
has a length of 2.8 𝑚. At the right end of the bar attached a one meter diameter disk with a
mass of 5 𝑘𝑔 allowing the whole linkage to change positions. The bars are initially at the
angle of 𝜃 = 65° then suddenly released. Find the angular velocities of the bars at the angle
𝜃 = 32°. Assume the disk rolls without slipping.
119
UNIT VI. PLANE MOTION OF KINETICS OF RIGID BODIES: IMPULSE
AND MOMENTUM
Overview
The impulse-momentum method is particularly convenient in situations when
forces act for very small-time intervals during which the forces may vary, as in an impact or
sudden blow. The method is also useful for solving in which a system gains or loses mass.
Still other cases will involve a combination of work-energy and impulse-momentum
methods, as in satellite motion.
Learning Objectives:
At the end of this Unit, I am able to:
1. Differentiate and relate Impulse and Momentum.
2. Calculate Linear and Angular Momentum.
3. Define the Conservation of Momentum.
4. Differentiate the different types of collisions.
Topics
6.1 Impulsive Motion
6.2 Linear Impulse-Momentum
6.3 Conservation of Momentum
6.4 Elastic Impact
6.5 Angular Impulse and Momentum
120
Pre-Test
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
1. A man kicks the 150-g ball such that it leaves the ground at an angle of 60° and strikes the
ground at the same elevation a distance of 12 m away. Determine the impulse of his foot on
the ball at A. Neglect the impulse caused by the ball’s weight while it’s being kicked.
121
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
2. The uniform beam has a weight of 5000 lb. Determine the average tension in each of the
two cables AB and AC if the beam is given an upward speed of 8 ft/s in 1.5 s starting from
rest. Neglect the mass of the cables.
122
Lesson Proper
PLANE MOTION OF KINETICS OF RIGID BODIES: IMPULSE AND MOMENTUM
Impulsive Motion
According to Mark D. Ardema author of Analytical Dynamics an impulsive force is a
force that tends to infinity at an isolated instant, say 𝑡𝑗 , such that its time integral remains
bounded
𝑡
𝑙𝑖𝑚 ∫ 𝐹 𝑟 (𝑥 𝑠 , 𝑥̇ 𝑠 , 𝜏) 𝑑𝑟 = 𝑃𝑟
𝑡→𝑘𝑗 𝑡
𝑗
Where:
𝑃𝑟 = impulse of force
𝐹 𝑟 = force
𝑡𝑗 = time
𝑥 𝑠 = denotes the position vector of particle s in an inertial frame
𝐹 𝑟 = the component of the impulsive force on particle r
As we have seen, the momentum of a particle changes if a net force acts on the
particle. Knowing the change in momentum caused by a force is useful in solving some
types of problems. To begin building a better understanding of this important concept, let
us assume that a single force F acts on a particle and that this force may vary with time.
According to Newton’s second law,
𝑑𝑝 = 𝐹𝑑𝑡
If the momentum of the particle changes from pi at time ti to pf at time t f ,
integrating Equation gives
𝑡𝑓
𝛥𝑃 = 𝑃𝑓 − 𝑃𝑖 = ∫ 𝐹 𝑑𝑡
𝑡𝑖
Impulse is a vector defined by
𝑡𝑓
𝐼 = ∫ 𝐹 𝑑𝑡 = 𝛥𝑃
𝑡𝑖
The impulse of the force F acting on a particle equals the change in the momentum
of the particle caused by that force. Also known as the impulse–momentum theorem is
equivalent to Newton’s second law.
Because the force imparting an impulse can generally vary in time, it is convenient
to define a time-averaged force
1 𝑡𝑓
𝐹̅ = ∫ 𝐹 𝑑𝑡
𝛥𝑡 𝑡𝑖
123
Where 𝛥𝑡 = 𝑡𝑓 − 𝑡𝑖 . Therefore,
𝐼 = 𝐹̅ 𝛥𝑡
if the force acting on the particle is constant, 𝐹̅ = 𝐹
𝐼 = 𝐹𝛥𝑡
Linear Impulse-Momentum
The motion of the mass center of any body, rigid or nonrigid, is governed by F=ma,
where F=force, m=mass, and a=acceleration. Replacing acceleration “a” by its equivalent
dv/dt, we obtain
𝐹=𝑚
𝑑𝑣
𝑑𝑡
𝐹 𝑑𝑡 = 𝑚 𝑑𝑣
which is the differential form of the linear impulse-momentum equation for all systems
which neither gain nor lose mass. Deriving the equation, we obtain
F Δt = m Δv
F (tf – to) = m (vf – vo)
Where tf: final time
to: initial time
vf: final velocity
vo: initial velocity
Momentum
Momentum (P) is the product mass (m) and velocity (v), thus
P = mv
Relating to the above linear impulse-momentum equation
F Δt = m Δv
therefore, Impulse is equal to the change in momentum
I = ΔP
124
*Take Note: Solving the units simultaneously we have, N-s = kg-m/s
Examples:
1. A constant force of 50 N is applied to a 20 kg block for 10 seconds. (a) What is the impulse
acting on the block? (b) What is the change in the momentum of block? (c) What is the final
velocity of the block if it was originally at rest? (d) What is the final velocity of the block if it
was originally moving at 15 m/s?
Solution:
Solving for the impulse acting on the black
I = F Δt
I = 50 N (10 s)
I = 500 N-s (a)
since I = ΔP therefore ΔP = 500 kg-m/s (b)
*Take Note: N-s = kg m/s
Solving for the final velocity when the block is at rest
ΔP = m (vf – vo)
500 kg-m/s = 20 kg (vf – 0)
25 m/s = vf – 0
vf = 25 m/s (c)
Solving for the final velocity when the block is moving at a speed of 15 m/s
ΔP = m (vf – vo)
500 kg m/s = 20 kg (vf – 15 m/s)
25 m/s = vf – 15 m/s
vf = 40 m/s (d)
125
2. A 0.20 kg ball was struck by a baseball bat from rest up to a speed of 35 m/s. The ball was
in contact with the bat for 0.02 seconds. (a) What is the change in the momentum of the ball?
(b) What was the impulse exerted on the ball? (c) Calculate the average force exerted on the
ball by the bat.
Solution:
Solving for the change in momentum of the ball
ΔP = m (vf – vo)
ΔP = 0.20 kg (35 m/s – 0 m/s)
ΔP = 7 kg-m/s (a)
since I = ΔP therefore I = 7 N-s (b)
Solving for the average force exerted by the bat
I = F Δt
7 N-s = F (0.02 s)
F = 350 N (c)
3. A 0.25 kg tennis ball moves east at a speed of 50 m/s and strikes a wall. The ball bounces
back at a speed of 50 m/s. The contact time between the wall and the ball was 0.015 seconds.
What average force was exerted by the wall on the ball?
Solution:
I = ΔP
F Δt = m (vf – vo)
F (0.015 s) = 0.25 kg [-50 m/s – (+50 m/s)]
F (0.015 s) = 0.25 kg (-100 m/s)
F = -1666.67 N
126
Conservation of Momentum
Momentum is conserved by all collisions as well as in all explosions. In the
conservation of momentum; the final total momentum is equal to the initial total
momentum. The essential effect of collision is to redistribute the total momentum of the
colliding objects. All collisions conserve momentum, but not all of them conserve kinetic
energy as well. Collision falls into three categories:
Elastic Collision
Inelastic Collision
Completely Inelastic Collision
Elastic Collision is a collision which conserves kinetic energy.
Inelastic Collision is a collision which does not conserve kinetic energy. Some kinetic
energy is converted into heat energy, sound energy, etc.
Completely Inelastic Collision is the collision in which the objects stick together
afterward. In such collisions the kinetic energy loss is maximum.
Pbefore impact = Pafter impact
m1 v1 + m2 v2 = m1 v1´+ m2 v2´
m1 (v1 - v1´) = m2 (v2´- v2)
127
Example:
1. A 1550 kg tank, initially at rest, fires a 60 kg shell horizontally from its cannon with a
speed of 525 m/s. What is the maximum possible recoil velocity of the tank?
Solution:
Solving the recoil velocity of the tank (vc´). Taking the initial velocity of both tank
and shell be equal to zero.
Pbefore firing = Pafter firing
mc vc + ms vs = mc vc´+ ms vs´
1550 kg (0 m/s) + 60 kg (0 m/s) = 1550 kg (vc´) + 60 kg (525 m/s)
1550 kg (vc´) = -31,500 kg-m/s
vc´ = -20.32 m/s “negative sign means opposite direction of the shell which is the recoil”.
2. A 588.6 kN car moving at 1 km/hr instantaneously collides a stationary 392.4 kN car. If the
collision is perfectly inelastic, what is the velocity of the cars after collision?
Solution:
Since the collision is perfectly inelastic, there are two possibilities: the moving car will
either stop or will be coupled with the stationary car and move in one common direction and
velocity. Considering that the two cars did not stop so it is obvious that the two moved in one
common direction.
128
Pbefore impact = Pafter impact
m1 v1 + m2 v2 = (m1 + m2) v
𝑊1
𝑊2
(𝑚1 + 𝑚2 )
𝑣1 +
𝑣2 =
𝑣
𝑔
𝑔
𝑔
588.6 𝑘𝑁 𝑘𝑚
392.4 𝑘𝑁
𝑘𝑚
(588.6 + 392.4)
(1
)+
(0
) =
𝑣
𝑔
ℎ𝑟
𝑔
ℎ𝑟
𝑔
v = 0.60 km/hr
Elastic Impact
Coefficient of Restitution (e) is the ratio between the relative speeds of two
colliding objects after and before they collide.
𝛥𝑣𝑎𝑓𝑡𝑒𝑟 𝑖𝑚𝑝𝑎𝑐𝑡
𝑣2 ´ − 𝑣1 ´
𝑒=
=
𝛥𝑣𝑏𝑒𝑓𝑜𝑟𝑒 𝑖𝑚𝑝𝑎𝑐𝑡
𝑣1 − 𝑣2
when:
e = 1.0
e=0
0<e<1
- perfectly elastic collision
-for completely (perfectly inelastic collision)
- for inelastic collision
Special Cases (e):
a. Dropped and Rebounds
ℎ2
𝑒=√
ℎ1
where: h2 & h1 are the elevation
129
b. Thrown at an angle β, and rebounds at an angle θ
e = cot β (tan θ)
Examples:
1. A ball strikes the floor vertically and rebounds with a coefficient of restitution of 0.80. If
the mass of the ball is 0.11 slug and the velocity is 5 ft/s just before it hits, compute the
momentum of the ball as it leaves the floor.
Solution:
First, solve the velocity of the ball after it hits the floor.
𝑒=
𝑣2
𝑣1
0.80 =
𝑣2
5 𝑓𝑡/𝑠
v2 = 4 ft/s
Solving for the momentum of the ball after it hits the floor
P = mv
P = 0.11 slug (4 ft/s)
P = 0.44 slug-ft/s
130
2. A ball is thrown at an angle of 35° from the horizontal towards a smooth floor. At what
angle will it rebound if the coefficient of restitution between the ball and the floor is 0.75?
Solution:
e = cot β (tan θ)
𝑒=
1
(𝑡𝑎𝑛 𝜃)
𝑡𝑎𝑛 𝛽
𝜃 = 𝑡𝑎𝑛−1[𝑒 (𝑡𝑎𝑛 𝜃)]
𝜃 = 𝑡𝑎𝑛−1[0.75 (𝑡𝑎𝑛 35)]
θ = 27.71°
Angular Impulse and Momentum
Angular Impulse (J) is the product of the Linear Impulse (I) and the moment arm r.
J = I r sin θ
J = F Δt r sin θ
Angular Momentum (L) is the product of Linear Momentum and the moment arms.
L = P r sin θ
L = mV r sin θ
In relation with Angular Motion, we can also use the following equations:
L = m (r ω) r
L = mr2ω
L = Io ω
where: Io = rotational inertia
ω = angular velocity
131
*Take Note: Rotational Inertia may vary depending on the mass moment of inertia of different
shape of the object. Below is the mass moment of inertia of common shapes.
Solid Sphere: I = 2/5 mr2
Thin-walled Hollow Sphere: I = 2/3 mr2
Solid Cylinder: I = ½ mr2
Hollow Cylinder: I = ½ m (R2 - r2)
Right Circular Cone: I = 3/10 mr2
Examples:
1. A 10 kg disk of radius 3 meters spinning at 15 rad/s. Calculate the angular momentum of
the disk.
Solution:
L = Io ω
Io. D
Solving for the angular momentum, first, we have to solve for the rotational inertia
Io = ½ mr2 = ½ (10 kg) (3 m)2 = 45 kg-m2
L = Io ω
L = 45 kg-m2 (15 rad/s)
L = 675 kg-m2 rad/s
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References
Ardema, M. D. (2004). Analytical Dynamics: Theory and Applications (1st ed.). Springer.
Borisov, A. V., & Mamaev, I. S. (2019). Rigid Body Dynamics. Higher Education Press and
Walter de Gruyter.
Capote, R. S., & Mandawe, J. A. (2007). Mathematics & Basic Engineering Sciences. JAM
Publisher.
HIBBELER, R. C. (2016). ENGINEERING MECHANICS DYNAMICS (Fourteenth Edition ed.).
Singer, F. L. (1975). ENGINEERING MECHANICS STATICS AND DYNAMICS (3rd Edition ed.).
HARPER & ROW PUBLISHER.
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Assessing Learning
Activity VI
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
1. A sky rocket of mass 3 x 104 kg, starting from rest, is acted on by a net force of 5 x 105 N
for 20 seconds. What is the final velocity of the sky rocket?
134
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
2. A truck full of cement has a mass of 45,000 kg. It travels north at a speed of 20 m/s. (a)
Calculate the truck’s momentum. (b) How fast must an 800 kg car to have the same
momentum?
135
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
3. A tennis ball approaches a player’s racket horizontally at 14 m/s. After it is struck, its
velocity is horizontal in the opposite direction with magnitude 25 m/s. The ball has a mass
of 0.07 kg, and it is in contact with the racket for 0.02 seconds. What is the average force
acted on the ball?
136
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
4. A 1000 kg car traveling east at a velocity of 15 m/s has a head-on collision with a 4000 kg
truck traveling west at a velocity of 10 m/s. If the vehicles are lock together after the
collision, what is their final velocity?
137
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
5. A gun is shot into a 500 N block which is hanging from a rope of 1.8 m long. The weight of
the bullet is equal to 5 N with a muzzle velocity of 320 m/s. How high will the block swing
after it was hit by the bullet?
138