Kinetics particles

1. ENGINEERING DYNAMICS
Dr. THABIT SALIM NASSOR
Kinematics of Particles

2. Kinematics of Particles
Contents
• Rectilinear motion; Derivation of motion
equations for velocity and acceleration.
• Curvilinear motion in a single plane
• Rectangular coordinates for motion of particle
• Polar coordinates for motion of particle
• Space curvilinear motion in 3D (Velocity and
acceleration)

3. Kinematics of Particles
ME101 - Division III Kaustubh Dasgupta 3
• Motion
– Constrained :: confined to a specific path
– Unconstrained :: not confined to a specific path
• Choice of coordinates
– Position of P at any time t
• rectangular (i.e., Cartesian) coordinates x, y, z
• cylindrical coordinates r, θ, z
• spherical coordinates R, θ, Φ
– Path variables
• Measurements along the tangent t and normal n to
the curve

4. Kinematics of Particles
• Choice of coordinates
4

5. Kinematics of Particles
Rectilinear Motion
• Motion along a straight line
t t+Δt
5

6. •Motion along a straight line
Position at any instance of time t
:: specified by its distance s measured from some convenient reference point O
fixed on the line
:: (disp. is negative if the particle moves in the negative s-direction).
Velocity of the particle: Acceleration of the particle
Kinematics of Particles :: Rectilinear Motion
t t+Δt
+ve or –ve depending
on +ve or –ve displacement
6
Both are
vector quantities
+ve or –ve depending on whether
velocity is increasing or decreasing

7. Kinematics of Particles
Rectilinear Motion:
Graphical Interpretations
Using s-t curve, v-t & a-t curves can be plotted.
Area under v-t curve during time dt ; vdt == ds
• Net disp from t1 to t2 = corresponding area under
v-t curve à
or s2 - s1 = (area under v-t curve)
Area under a-t curve during time dt ; adt == dv
• Net change in velocity from t1 to t2 = corresponding
area under a-t curve à
or v2 - v1 = (area under a-t curve)
ò
ò
2
2
s t
s1 t1
ds = vdt
ò
7
ò
2
2
v t
v1 t1
dv = adt

8. Kinematics of Particles
Rectilinear Motion:
Graphical Interpretations
Two additional graphical relations:
Area under a-s curve during disp ds; ads == vdv
• Net area under a-s curve betn position
coordinates s1 and s2 à
or ½ (v2
2 – v1
2) = (area under a-s curve)
Slope of v-s curve at any point A = dv/ds
• Construct a normal AB to the curve at A. From
similar triangles:
à
• Vel and posn coordinate axes should have the same
numerical scales so that the accln read on the x-axis in
meters will represent the actual accln in m/s2
ò
ò
2 s
s1
v
v1
vdv = 2
ads
v ds
CB
=
dv CB = v
dv
= a (acceleration)
ds
8

9. Kinematics of Particles :: Rectilinear Motion
Analytical Integration to find the position coordinate
Acceleration may be specified as a function of time, velocity, or position
coordinate, or as a combined function of these.
(a) Constant Acceleration
At the beginning of the interval à t = 0, s = s0, v = v0
For a time interval t: integrating the following two equations
Substituting in the following equation and integrating will give the position
coordinate:
Equations applicable for Constant Acceleration and for time interval 0 to t
v =
ds
dt
a = dv
dt
vdv = ads
9

10. Kinematics of Particles :: Rectilinear Motion
Analytical Integration to find the position coordinate
(b) Acceleration given as a function of time, a = f(t)
At the beginning of the interval à t = 0, s = s0, v = v0
For a time interval t: integrating the following equation
à
Substituting in the following equation and integrating will give the position
coordinate:
Alternatively, following second order differential equation may be solved to
get the position coordinate:
v =
ds
dt
a =
dv
dt
f (t) =
dv
dt
10

11. Kinematics of Particles :: Rectilinear Motion
Analytical Integration to find the position coordinate
(c) Acceleration given as a function of velocity, a = f(v)
At the beginning of the interval à t = 0, s = s0, v = v0
For a time interval t: Substituting a and integrating the following equation
à
Solve for v as a function of t and integrate the following equation to get the
position coordinate:
Alternatively, substitute a = f(v) in the following equation and integrate to
get the position coordinate :
v = ds
dt
a = dv
dt
f (v) =
dv
dt
vdv = ads
11

12. Kinematics of Particles: Rectilinear Motion
Analytical Integration to find the position coordinate
(d) Acceleration given as a function of displacement, a = f(s)
At the beginning of the interval à t = 0, s = s0, v = v0
For a time interval t: substituting a and integrating the following equation
Solve for v as a function of s : v = g(s), substitute in the following equation
and integrate to get the position coordinate:
It gives t as a function of s. Rearrange to obtain s as a function of t to get
the position coordinate.
In all these cases, if integration is difficult, graphical, analytical, or
computer methods can be utilized.
v = ds
dt
vdv = ads
12

13. Kinematics of Particles: Rectilinear Motion
13
Example
Position coordinate of a particle confined to move along a straight line is given by
s = 2t3 – 24t + 6, where s is measured in meters from a convenient origin and t is
in seconds. Determine: (a) time reqd for the particle to reach a velocity of 72 m/s
from its initial condition at t = 0, (b) acceleration of the particle when v = 30 m/s,
and (c) net disp of the particle during the interval from t = 1 s to t = 4 s.
Solution
Differentiating s = 2t3 – 24t + 6 à v = 6t2 – 24 m/s
à a = 12t m/s2
(a) v = 72 m/s à t = ± 4 s
(- 4 s happened before initiation of motion à no physical interest.)
à t = 4 s
(b) v = 30 m/s à t = 3 sec à a = 36 m/s2
(c) t = 1 s to 4 s. Using s = 2t3 – 24t + 6
Δs = s4 – s1 = [2(43) – 24(4) +6] – [2(13) – 24(1) + 6]
Δs = 54 m

14. Working example-1
The position coordinate of a particle which is confined to move along a straight line is given
by s = 2t3-24t+6, where s is measured in meters from a convenient origin and t is in seconds.
Determine;
(a) the time required for the particle to reach a velocity of 72 m/s from its initial condition at
t=0.
(b) the acceleration of the particle when 𝑣=30 m/s.
(c) the net displacement of the particle during the interval from t =1 s to t=4 s.

15. Working example-1
The position coordinate of a particle which is confined to move along a straight
line is given by s = 2t3-24t+6, where s is measured in meters from a convenient
origin and t is in seconds. Determine;
(a) the time required for the particle to reach a velocity of 72 m/s from its initial
condition at t=0.
Given;
s = 2t3+24t+6
to=0 , 𝑣𝑓 = 72m/s
RTC;
Time tf when 𝑣𝑓=72m/s
Solution (a)
From; eq.1
𝑣 =
)*
)+
= 𝑆̇,
𝑣𝑓 =
)(2t3+24t+6)
)+
𝑣𝑓 = 6t2+24
Therefore; t = 4
2
3 𝑠

16. Working example-1
The position coordinate of a particle which is confined to move along a straight
line is given by s = 2t3-24t+6, where s is measured in meters from a convenient
origin and t is in seconds. Determine;
(b) the acceleration of the particle when 𝑣=30 m/s.
Given;
s = 2t3+24t+6
to=0 , 𝑣𝑓 = 30m/s
RTC; 𝒂 at 𝑣𝑓 Solution (b)
From; eq.2
𝑎 =
)9
)+
= 𝑣̇,
𝑎 =
)(6t2+24)
)+
𝒂 = 12t
from; 𝑣𝑓 = 6t2+24
t = 3
2
3
𝑠 , 3s
Therefore
a=12x3=36 m/s2

17. Working example-1
The position coordinate of a particle which is confined to move along a straight line
is given by s = 2t3-24t+6, where s is measured in meters from a convenient origin
and t is in seconds. Determine;
(c) the net displacement of the particle during the interval from t =1 s to t=4 s.
Given;
s = 2t 3+24t+6
RTC; ∆𝑆 between t=1s to t=4s
Solution;
Net displacement S= S (4)-S(1)
Where; S(1)=2(1)3-24(1)+6 = -16m
S(2)= 2(4)3-24(4)+6 = 38m
Therefore;
S= 54m

18. Working example-2
A particle moves along the x-axis with an initial velocity vx=50 ft/sec at the origin
when t=0. For the first 4 seconds it has no acceleration, and thereafter it is acted
on by a retarding force which gives it a constant acceleration ax= -10 ft/sec2.
Calculate the velocity and the x-coordinate of the particle for the conditions of t=8
sec and t=12 sec and find the maximum positive x-coordinate reached by the
particle.
At t=8 sec and t=12 sec, the velocities along x axis are;

19. Working example-2
A particle moves along the x-axis with an initial velocity vx=50 ft/sec at the origin
when t=0. For the first 4 seconds it has no acceleration, and thereafter it is acted
on by a retarding force which gives it a constant acceleration ax=-10 ft/sec2.
Calculate the velocity and the x-coordinate of the particle for the conditions of t=8
sec and t=12 sec and find the maximum positive x-coordinate reached by the
particle.
Solution continue;
The distance travelled by the particle (x-coordinate) after 4 sec, is obtained by
adding the distance covered in 4 sec and that after change of acceleration; hence
from eq. 1;

20. Working example-2
A particle moves along the x-axis with an initial velocity vx=50 ft/sec at the origin when
t=0. For the first 4 seconds it has no acceleration, and thereafter it is acted on by a
retarding force which gives it a constant acceleration ax=-10 ft/sec2. Calculate the
velocity and the x-coordinate of the particle for the conditions of t=8 sec and t=12 sec
and find the maximum positive x-coordinate reached by the particle.
Solution continue;
It was noted the x coordinate when t=8 sec was greater than that of t=12sec. That is
because the motion is in negative during t=12 sec. Hence the max x-coordinate is
reached when the 𝑣x = 0. (t=9sec). Therefore;

21. Working example-3
Solution;
Since the acceleration is given in term of displacement,
then from from eqn. 3.

22. Working example-3
From eg.1, substitute the value of 𝑣;