Kinematics of Particles
Kinematics is the study of motion without reference to the
force which produced the motion. First, we will study the
kinematics of one-dimensional motion, in which case,
force and acceleration are scalars instead of vectors.
F  ma
a 
dv
v
 v
dx
 x
dt
dt
2
a 
d x
dt
2
 x
4-1
Special cases
• There are certain problems where it is possible to solve the
equations of motion without differential equations. These
are listed below:
1) Force and acceleration are only a function of time: F t   m  a t 
2) Force and acceleration are only a function of position:
F x   m  a x 
3) Force and acceleration are only a function of velocity:
F  x   m  a  x 
4-2
(1) Acceleration only a function of time
a t  
Re-arranged as:
When integrated:
dv
dt
dv  a t dt
v2
t2
 dv   a t dt
v1
 v 2  v1
t1
When the starting velocity (v1) is known, the velocity at
any future time t can be written as:
v
t
v1
t1
 dv   a t dt
 v t   v1
v t   v1 
t
 a t dt
t1
4-3
• Likewise, velocity can be written as:
dx
v
dt
Re-arranged as:
dx  v t dt
x2
When integrated:
t2
 dx   v t dt
x1
 x 2  x1
t1
When the starting position (x1) is known, the position at
any future time t can be written as:
t
x  x1   v t dt
t1
4-4
(2) Acceleration only a function of position
ax  
dv
dt
As stated, this equation cannot be integrated with respect to
time, however we can make the useful substitution.
dx
v 
dt 
v
dt
ax  
dx
a  x dx  vdv
dv
dx
v
This equation can be integrated as follows:
x2
 a  x dx 
x1
v2
 vdv 
v1
1
2
v
2
2
 v1
2

x2
v 2   v1  2  a  x dx
2
x1
4-5
Case (2) is useful in solving dynamic problems because it
provides a direct relationship between initial and final velocity if
the intial and final positions are known.
x
v  x    v1  2  a  x dx  f  x 
2
x1
Example 1. Assume the acceleration is due to gravity alone
(i.e. acceleration is constant in space),
The above equation becomes:
v  x    v1  2 g  x
2
Effectively, this equation indicates that if we know the initial velocity of an object
moving under the action of gravity alone, the final velocity can be determined
by the distance it moved.
4-6
Example 2. Assume the object’s acceleration is due
to force applied by a spring.
F   kx  ma
The variable k is known as the spring constant, which is truly a “constant” only
for a linear spring. For non-linear springs, the spring constant may have some
dependence on x (i.e.
). In either case, the acceleration is a function
of position only.
k  k x 
 k 
a x     x
m
1
2

m v v
2
2
2
1

1
2
v2   v  2
2
1

k x x
2
1
2
2

1
2
k
m
mv
x2

xdx   v 
2
1
x1
2
2
 kx
2
2
k
m
  mv
1
2
x
2
2
 x1

2
1
 kx 1

2
2
In other words, the sum of the kinetic energy and the potential energy of a spring
must be constant (i.e. the same at times t1 and t2 and any other time if no other
non-conservative force is present). Thus if we know the length of the spring, then
we can determine the velocity of the ends of the spring at any future time.
4-7
(3) Acceleration only a function of velocity
a v  
dv
dt
This equation can be integrated with respect to time as follows:
t2
 dt
t1
v2

dv
 a v 
v1
v2
t 2  t1 
dv
 a v 
v1
Example 3. Assume the force is due to inertial drag through
a medium such as a projectile moving through air.
In this example, the drag force is often given in the form: F   1 DA  v 2  ma
2
Where D is a proportionality constant known as the drag coefficient, A is the crosssectional area of the object, ρ is the object’s density, and v is the velocity of the
object.
4-8
Given the form of the inertial drag force, the equation can be rearranged and solved for time in terms of the velocity.
a v   
1
2m
DA  v 
2
t2
dv
 dt 
dt
t1
2m
DA 
v1
dv
v
v2
2

2m  1
1


  t 2  t1
DA   v 2 v1 
Interestingly, this equation predicts that provided no other forces
operate on the mass, it will take an infinitely long time for the
projectile to come to a stop.
1
v

1
v1

DA  t
2m
You might have experienced this effect when riding in a boat, and
it is this effect which underlies the constant movement of solar
bodies.
4-9
When acceleration is a function of position, time and
velocity, the equations of motion become much more
difficult to solve. a  a  x , v , t   a  x , x , t 
Example 4. Assume the objects acceleration is due to a
spring, and inertial drag through a medium.
F   kx  
1
2
a  x , x   
k
m
x
1
2m
2
DA  v  ma  x , x 
2
DA  x  x
x 
1
2m
DA  x 
2
k
x0
m
The solution to this second order differential equation, which can
only be solved numerically on a computer. It cannot be solved using
the methods introduced previously in cases (1-3).
4 - 10
Problem 1
A projectile enters a resisting medium at x = 0 with an
initial velocity vo = 900 ft/s and travels 4 in. before
coming to rest. Assuming that the velocity of the
projectile is defined by the relation v = vo - kx, where
v is expressed in ft/s and x is in feet, determine (a) the
initial acceleration of the projectile, (b) the time
required for the projectile to penetrate 3.9 in. into the
resisting medium.
4 - 11
Problem 1
A projectile enters a resisting medium at x = 0 with an initial
velocity vo = 900 ft/s and travels 4 in. before coming to rest.
Assuming that the velocity of the projectile is defined by the
relation v = vo - kx, where v is expressed in ft/s and x is in feet,
determine (a) the initial acceleration of the projectile, (b) the
time required for the projectile to penetrate 3.9 in. into the
resisting medium.
1. Determine a(x) for a given v(x). Substitute v(x) in the formula
a = v dv
dx
differentiate and obtain a(x).
2. Determine t(x) for a given v(x). Substitute v(x) in the formula
dx
v = dt
rewrite as
dt = dx
v
Integrate and obtain t(x).
4 - 12
Problem 1 Solution
v = vo - k x
0 = 900 ft/s - k ( 4 ft)
12
k = 2700 s-1
Determine k:
Determine a(x) for a given v(x).
a = v dv
dx
a = ( vo - k x )(- k )
(a) at x = 0
a = - k vo
a = - ( 2700 s-1 )( 900 ft/s )
a = - 2.43 x 106 ft/s2
4 - 13
Problem 1 Solution
Determine t(x) for a given v(x).
v=
vo - k x =
dx
dt
dx
dt
vo - k x
-1
dx
du
dt =
=
k v u
0 ( vo - k x )
0
o
vo - k x
-1
t = ln (
)
vo
k
t
At x =
t=
x
3.9
12
-1
2700
ln [ 1 -
2700 (3.9)
]
(900) 12
t = 1.366 x 10-3 s
4 - 14
Problem 2
A
C
B
In the position shown, collar B moves
to the left with a velocity of 150 mm/s.
Determine (a) the velocity of collar A,
(b) the velocity of portion C of the
cable, (c) the relative velocity of portion
C of the cable with respect to collar B.
4 - 15
Problem 2
In the position shown, collar B moves
to the left with a velocity of 150 mm/s.
Determine (a) the velocity of collar A,
C
(b) the velocity of portion C of the
cable, (c) the relative velocity of portion
B
C of the cable with respect to collar B.
1. Dependent motion of two or more particles:
A
1a. Draw a sketch of the system: Select a coordinate system,
indicating clearly a positive sense for each of the coordinate axes.
The displacements, velocities, and accelerations have positive
values in the direction of the coordinate axes.
1b. Write the equation describing the constraint: When particles
are connected with a cable, its length which remains constant is
4 - 16
the constraint.
Problem 2
A
C
B
In the position shown, collar B moves
to the left with a velocity of 150 mm/s.
Determine (a) the velocity of collar A,
(b) the velocity of portion C of the
cable, (c) the relative velocity of portion
C of the cable with respect to collar B.
1c. Differentiate the equation describing the constraint: This
gives the the corresponding relations among velocities and
accelerations of the various particles.
4 - 17
Problem 2 Solution
Draw a sketch of the system.
xA
Write the equation describing
the constraint.
A
C
B
xB
(a)
xC
The total length of the cable:
constant = 2xA + xB + (xB - xA)
vB = 150 mm/s
Differentiate the equation describing the constraint.
0 = vA + 2vB
vB = 150 mm/s
vA = - 2 vB
vA = -300 mm/s
vA = 300 mm/s
4 - 18
Problem 2 Solution
xA
A
(b) The length of the cable from
the right end to an arbitrary
point on portion C of the cable:
C
B
xB
xC
vB = 150 mm/s
(c) Relative velocity of
portion C:
vC = vB + vC/B
constant = 2xA + xC
0 = 2vA + vC
vC = - 2vA
vC = 600 mm/s
600 = 150 + vC/B
vC/B = 450 mm/s
4 - 19