Dynamics of Rigid
Bodies
CET 0223
PAMANTASAN NG LUNGSOD NG MAYNILA
(University of the City of Manila)
COLLEGE OF ENGINEERING AND TECHNOLOGY
CIVIL ENGINEERING DEPARTMENT
ENGR. GIEZEL MARIE M. ARCEO
KINEMATICS OF PARTICLES
Translation : Rectilinear Kinematics
2. Erratic Motion
is a type of motion that occurs whenever a given motion
parameter (such as s, v, or a) is not continuous. It can also occur when
the function has jumps or sharp bends.
When a particle has erratic or changing motion then its position,
velocity, and acceleration cannot be described by a single continuous
mathematical function along the entire path.
Erratic Motion
A B C
Erratic Motion
Recall : Graph for function with no variable : straight line
Graph for polynomial of 1st degree, y = x Graph for 2nd degree polynomials, y = 𝑥 2
Erratic Motion
𝑑𝑠
V = dS = V dt
𝑑𝑡
𝑑𝑉
a = dV = a dt
𝑑𝑡
Erratic Motion
Equations Equations
𝑡𝑑 𝑉 = 𝑆𝑑
𝑆1 𝑡1
s – t Graph s (Given) 𝑆0 𝑑𝑆 =𝑡0 𝑉 𝑑𝑡
𝑆2 𝑡2
𝑆1 𝑑𝑆 =𝑡1 𝑉 𝑑𝑡
Take
Integral
𝑑𝑠 𝑡𝑑 𝑎 = 𝑉𝑑
𝑉1 𝑡1
v – t Graph 𝑑𝑡 𝑉0 𝑑𝑉 =𝑡0 𝑎 𝑑𝑡
𝑉2 𝑡2
𝑉1 𝑑𝑉 =𝑡1 𝑎 𝑑𝑡
a – t Graph 𝑑𝑉 a (Given)
Take 𝑑𝑡
Derivative
Erratic Motion
S2
S1
Take
Derivative
S0
t0
t1 t2
Erratic Motion
Time, t s – t graph v – t graph a – t graph
Equation (s) Equation (V) Equation (a)
0 to t1
t1 to t2
Time, t s – t graph v – t graph a – t graph
Values (s) Values (V) Values (a)
At time t1
At time t2
Erratic Motion
Problem 1:
A bicycle moves along a straight road such that its position is described by the
graph shown in Fig. below. Construct the v-t and a-t graphs for 0 ≤ t ≤ 30 s.
Given :
s-t graph
t = from 0 to 10s to 30s
Required :
v-t graph
a-t graph
Erratic Motion
Take Derivative
Problem 1:
Time, t s – t graph v – t graph a – t graph
Equation (s) Equation (V) Equation (a)
0 to 10s 𝑡2
10 to 30s 20t - 100
Time, t s – t graph v – t graph a – t graph
Values (s) Values (V) Values (a)
At 10s 100 ft
At 30s 500 ft
Erratic Motion
Problem 1:
Solution : Solve for Equations
Take Derivative
s – t graph
Equation (s)
𝑑𝑠 𝑑𝑉
𝑡2 V= = 𝑡 2 = 2t a= = 2t = 2
𝑑𝑡 𝑑𝑡
𝑑𝑠 𝑑𝑉
20t - 100 V= = 20𝑡 − 100 = 20 a= = 𝟐𝟎 = 0
𝑑𝑡 𝑑𝑡
Erratic Motion
Problem 1:
Solution : Solve for VALUES
Time, t v – t graph a – t graph
Values (V) Values (a)
At 10s 2t = 2(10) = 20 2
At 30s 20 0
Erratic Motion
Take Derivative
Problem 1:
Time, t s – t graph v – t graph a – t graph
Equation (s) Equation (V) Equation (a)
0 to 10s 𝑡2 2t 2
10 to 30s 20t - 100 20 0
Time, t s – t graph v – t graph a – t graph
Values (s) Values (V) Values (a)
At 10s 100 ft 20 2
At 30s 500 ft 20 0
Erratic Motion
Time, t s – t graph v – t graph a – t graph
Equation (s) Equation (V) Equation (a)
0 to 10s 𝑡2 2t 2
Problem 1: 10 to 30s 20t - 100 20 0
Solution : DRAW GRAPH Time, t s – t graph
Values (s)
v – t graph
Values (V)
a – t graph
Values (a)
At 10s 100 ft 20 2
At 30s 500 ft 20 0
v – t graph a – t graph
Erratic Motion
Problem 2 :
The car in Fig. below starts from rest and travels along a straight track such that
it accelerates at 10 m/s2 for 10 s, and then decelerates at 2 m/s2. Draw the v-
t and s-t graphs and determine the time t2 needed to stop the car. How far has
the car traveled?
Given :
a-t graph
a = 10 m/s2 to -2 m/s2
Required :
v-t graph
t0 s-t graph
t1 t2
Value of t2
Value of s2
Erratic Motion
Take the Integral
Problem 2 :
Time, t s – t graph v – t graph a – t graph
Equation (s) Equation (V) Equation (a)
0 to 10s 10
10s to t2 -2
Time, t s – t graph v – t graph a – t graph
Values (s) Values (V) Values (a)
At 10s 10
At t2 -2
Erratic Motion
Problem 2: Solution : Solve for Equations
When a = - 2 When a = 10
Limits Value: Limits Value:
V2 = ? t2 = t V1 = ? t1 = t Take the Integral
V1 = 100 t1 = 10 V0 = 0 t0 = 0
𝑉2 𝑡2 𝑉1 𝑡1 a – t graph
𝑉1 𝑑𝑉 =𝑡1 𝑎 𝑑𝑡 𝑉0 𝑑𝑉 =𝑡0 𝑎 𝑑𝑡
Equation (a)
𝑉2 𝑡 𝑉1 𝑡
100 𝑑𝑉 =10 −2 𝑑𝑡 0 𝑑𝑉 =0 10 𝑑𝑡
𝑉2 𝑡
100 𝑑𝑉 = −2 10 𝑑𝑡 𝑉1
0 𝑑𝑉 =10 0 𝑑𝑡
𝑡
10
V = -2t V = 10t
Apply the limits : Apply the limits :
𝑡
V 100
𝑉2
= -2t 10 V 𝑉1= 10t
𝑡
0 0
V2 – 100 = -2(t) – (-2)(10) V1 – 0 = 10(t) – 10(0) -2
V2 = -2 (t) + 120 m/s V1 = 10t m/s
Solve for the values, t1=10, V1 = 10(10) = 100 m/s
Erratic Motion
Problem 2: Solution : Solve for Equations
When v = -2 (t) + 120 When v = 10t
Limits Value: Limits Value:
S1 = ? t1 = t
S2 = ? t2 = t
S0 = 0 t0 = 0
Take the Integral
S1 = 500 t1 = 10
𝑆2 𝑡2 𝑆1 𝑡1 v – t graph
𝑆1 𝑑𝑆 =𝑡1 𝑉 𝑑𝑡 𝑆0 𝑑𝑆 =𝑡0 𝑉 𝑑𝑡 Equation (V)
𝑆1 𝑡
𝑆2
500 𝑑𝑆
𝑡
=( 𝟎𝟏−2 (t) + 120)𝑑𝑡 0 𝑑𝑆 =0 10𝑡𝑑𝑡
𝑆1 𝑡 10t
S = (−(𝑡)2 +120(t) 0 𝑑𝑆 =100 𝑡𝑑𝑡
Apply the limits : 𝑡2
𝑆2 𝑡 S = 10 = 5𝑡 2
S = (−(𝑡)2 +120(t) 𝟐
500 10 Apply the limits :
𝑡
S2-500= (−(𝑡)2 +120(t) – (-(10)2+(120)(10)) S 𝑆1= 5𝑡 2 -2 (t) + 120
0 0
S2 = −(𝒕)𝟐 +120(t) - 600 S1 – 0 = 5((𝑡)2 ) – 5((0)2 )
S1 = 5𝑡 2
Solve for the values, t1=10, S1 = 5(10)2= 500 m
Erratic Motion
Problem 2:
Solution : Solve for VALUES
At t2, the car already stops , therefore V2 = 0
Using equation,
V2 = -2 (t) + 120
0 = -2 (t) + 120
t2 = 60 s
Solve for S2
Using equation
S2 = −(𝑡)2 +120(t) - 600
S2 = −(60)2 +120(60) - 600
S2 = 3000m
Erratic Motion
Take the Integral
Problem 2 :
Time, t s – t graph v – t graph a – t graph
Equation (s) Equation (V) Equation (a)
0 to 10s 5𝑡 2 10t 10
10s to t2 −(𝒕)𝟐 +120(t) - 600 -2t + 120 -2
Time, t s – t graph v – t graph a – t graph
Values (s) Values (V) Values (a)
At 10s 500 100 10
At t2 (60s) 3000 0 -2
Erratic Motion Time, t s – t graph
Equation (s)
v – t graph
Equation (V)
a – t graph
Equation (a)
Problem 2 : 0 to 10s 5𝑡 2 10t 10
10s to t2 −(𝑡)2 +120(t) – 600 -2 (t) + 120 -2
Solution : GRAPH Time, t s – t graph v – t graph a – t graph
Values (s) Values (V) Values (a)
At 10s 500 100 10
At t2(60s) 3000 0 -2
v – t graph s – t graph
