ERT250 DYNAMICS
KINEMATICS OF PARTICLES
MRS SITI KAMARIAH BINTI MD SA’AT
BIOSYSTEMS ENGINEERING
INTRODUCTION
 Kinematics
 Study of the geometry of motion.
 Relates displacement, velocity, acceleration, and time without
reference to the cause of motion.
 Particles
 Not strictly to small particles – possibly as large as cars, rockets
or airplanes.
 The entire bodies will analyze, any rotation to the centre will be
neglected
Kinematics of Particles Cases

Rectilinear motion: position, velocity, and acceleration
of a particle as it moves along a straight line.
 Curvilinear motion: position, velocity, and acceleration
of a particle as it moves along a curved line in two or
three dimensions.
RECTILINEAR MOTION
SUMMARY OF KINEMATIC RELATIONS:
RECTILINEAR MOTION
• Differentiate position to get velocity and acceleration.
v = dx/dt ;
a = dv/dt or a = v d2x/dt2;
a = v dv/dx
• Integrate acceleration for velocity and position.
Position:
Velocity:
v
t
v
s
so
s
t
o


 v dv =  a ds
 ds =  v dt
vo
o
vo
so
v = dv = a dt or
Note that so and vo represent the initial position and
velocity of the particle at t = 0.
Example 11.1
x = 6t 2  t 3
dx
v=
= 12t  3t 2
dt
dv d 2 x
a=
= 2 = 12  6t
dt dt
• What are x, v, and a at t = 2 s ?
- at t = 2 s, x = 16 m, v = vmax = 12 m/s, a = 0
• Note that vmax occurs when a=0, and that the slope of
the velocity curve is zero at this point.
• What are x, v, and a at t = 4 s ?
- at t = 4 s, x = xmax = 32 m, v = 0, a = -12 m/s2
Determination of the Motion of a Particle
• We often determine accelerations from the forces applied
(kinetics will be covered later)
• Generally have three classes of motion
- acceleration given as a function of time, a = f(t)
- acceleration given as a function of position, a = f(x)
- acceleration given as a function of velocity, a = f(v)
Acceleration as a function of time, position, or velocity
If….
a = a t 
a = a  x
a = a v
Kinematic relationship
Integrate
dt =
dx
dv
and a =
v
dt
v dv = a  x  dx
dv
= a (v )
dt
dv
v
= a v
dx
v
t
v0
0
 dv =  a t  dt
dv
= a (t )
dt
v
x
v0
x0
 v dv =  a  x  dx
v
t
dv
v a  v  = 0 dt
0
x
v
v dv
x dx = v a  v 
0
0
Example 11.2 Ball tossed with 10 m/s vertical velocity
from window 20 m above ground.
Determine:
• velocity and elevation above ground at time t,
• highest elevation reached by ball and
corresponding time, and
• time when ball will hit the ground and
corresponding velocity.
SOLUTION:
• Integrate twice to find v(t) and y(t).
• Solve for t when velocity equals zero (time for maximum elevation) and
evaluate corresponding altitude.
• Solve for t when altitude equals zero (time for ground impact) and evaluate
corresponding velocity.
SOLUTION:
• Integrate twice to find v(t) and y(t).
dv
= a = 9.81m s 2
dt
v t 
t
vt   v0 = 9.81t
 dv =   9.81dt
v0
0
vt  = 10
dy
= v = 10  9.81t
dt
y t 
t
 dy =  10  9.81t dt
y0
0
m 
m
  9.81 2  t
s 
s 
y t   y0 = 10t  12 9.81t 2
m
 m 
yt  = 20 m  10 t   4.905 2 t 2
 s 
s 
• Solve for t at which velocity equals zero and evaluate
corresponding altitude.
vt  = 10
m 
m
  9.81 2  t = 0
s 
s 
t = 1.019s
m
 m 
y t  = 20 m  10 t   4.905 2 t 2
 s 
s 
m

 m
y = 20 m  10 1.019 s    4.905 2 1.019 s 2
 s

s 
y = 25.1m
• Solve for t at which altitude equals zero and
evaluate corresponding velocity.
m
 m 
yt  = 20 m  10 t   4.905 2 t 2 = 0
 s 
s 
t = 1.243s meaningless 
t = 3.28 s
vt  = 10
m 
m
  9.81 2  t
s 
s 
v3.28 s  = 10
m 
m
  9.81 2  3.28 s 
s 
s 
v = 22.2
m
s
Problem 11.3
SOLUTION:
• Integrate a = dv/dt = -kv to find v(t).
a = kv
• Integrate v(t) = dx/dt to find x(t).
• Integrate a = v dv/dx = -kv to find v(x).
Brake mechanism used to reduce gun recoil
consists of piston attached to barrel moving
in fixed cylinder filled with oil. As barrel
recoils with initial velocity v0, piston moves
and oil is forced through orifices in piston,
causing piston and cylinder to decelerate at
rate proportional to their velocity.
Determine v(t), x(t), and v(x).
SOLUTION:
• Integrate a = dv/dt = -kv to find v(t).
v
dv
a=
= kv
dt
t
dv
v v = k 0 dt
0
ln
v t 
= kt
v0
vt  = v0 e kt
• Integrate v(t) = dx/dt to find x(t).
v t  =
dx
= v0 e  kt
dt
x
t
0
0
 kt
dx
=
v
e
0

 dt
t
 1

x  t  = v0   e  kt 
 k
0

v
xt  = 0 1  e  kt
k

• Integrate a = v dv/dx = -kv to find v(x).
dv
a = v = kv
dx
dv = k dx
v
x
v0
0
 dv = k  dx
v  v0 = kx
v = v0  kx
• Alternatively,
with
and
then

v
xt  = 0 1  e  kt
k

vt 
vt  = v0 e kt or e kt =
v0
v  vt  

xt  = 0 1 
k 
v0 
v = v0  kx
Uniform Rectilinear Motion
 During free-fall, a parachutist
reaches terminal velocity when
her weight equals the drag force.
If motion is in a straight line,
this is uniform rectilinear
motion.
For a particle in uniform
rectilinear motion, the
acceleration is zero and the
velocity is constant.
dx
= v = constant
dt
x
t
x0
0
 dx = v  dt
x  x0 = vt
x = x0  vt
Uniformly Accelerated Rectilinear Motion
If forces applied to a body
are constant (and in a
constant direction), then
you have uniformly
accelerated rectilinear
motion.
Another example is freefall when drag is negligible
UNIFORMLY ACCELERATED RECTILINEAR MOTION
For a particle in uniformly accelerated rectilinear motion, the acceleration of
the particle is constant. You may recognize these constant acceleration
equations from your physics courses.
dv
= a = constant
dt
dx
= v0  at
dt
x
v
t
 dv = a  dt
v0
0
t
 dx =   v0  at  dt
x0
dv
v = a = constant
dx
v = v0  at
x = x0  v0t  12 at 2
0
v
x
v0
x0
 v dv = a  dx
v 2 = v02  2a  x  x0 
Careful – these only apply to uniformly accelerated rectilinear motion!
Motion of Several Particles
We may be interested in the motion of several different
particles, whose motion may be independent or linked
together.
Motion of Several Particles: Relative Motion
• For particles moving along the same line, time should be recorded from
the same starting instant and displacements should be measured from
the same origin in the same direction.
xB A = xB  x A = relative position of B with
xB = x A  xB A
respect to A
vB A = vB  v A = relative velocity of B with
vB = v A  vB A
respect to A
a B A = a B  a A = relative acceleration of B with
aB = a A  aB A
respect to A
Problem 11.4
SOLUTION:
• Substitute initial position and velocity and
constant acceleration of ball into general
equations for uniformly accelerated
rectilinear motion.
Ball thrown vertically from 12 m level
in elevator shaft with initial velocity of
18 m/s. At same instant, openplatform elevator passes 5 m level
moving upward at 2 m/s.
Determine (a) when and where ball hits
elevator and (b) relative velocity of ball
and elevator at contact.
• Substitute initial position and constant
velocity of elevator into equation for
uniform rectilinear motion.
• Write equation for relative position of ball
with respect to elevator and solve for zero
relative position, i.e., impact.
• Substitute impact time into equation for
position of elevator and relative velocity
of ball with respect to elevator.
SOLUTION:
• Substitute initial position and velocity and constant acceleration
of ball into general equations for uniformly accelerated rectilinear
motion.
v B = v0  at = 18
m 
m
  9.81 2 t
s 
s 
m
 m 
y B = y0  v0t  12 at 2 = 12 m  18 t   4.905 2 t 2
 s 
s 
• Substitute initial position and constant velocity of elevator into
equation for uniform rectilinear motion.
vE = 2
m
s
 m
y E = y0  v E t = 5 m   2 t
 s
• Write equation for relative position of ball with respect to elevator and
solve for zero relative position, i.e., impact.


y B E = 12  18t  4.905t 2  5  2t  = 0
t = 0.39 s meaningless 
t = 3.65 s
• Substitute impact time into equations for position of elevator and relative
velocity of ball with respect to elevator.
y E = 5  23.65
y E = 12.3 m
v B E = 18  9.81t   2
= 16  9.813.65
vB
E
= 19.81
m
s
Motion of Several Particles: Dependent Motion
• Position of a particle may depend on position of one or more other
particles.
• Position of block B depends on position of block A. Since rope is of
constant length, it follows that sum of lengths of segments must be
constant.
x A  2 x B = constant (one degree of freedom)
• Positions of three blocks are dependent.
2 x A  2 xB  xC = constant (two degrees of freedom)
• For linearly related positions, similar relations hold between
velocities and accelerations.
dx
dx A
dx
 2 B  C = 0 or 2v A  2v B  vC = 0
dt
dt
dt
dv
dv
dv
2 A  2 B  C = 0 or 2a A  2a B  aC = 0
dt
dt
dt
2
Problem 11.5
SOLUTION:
• Define origin at upper horizontal surface with
positive displacement downward.
• Collar A has uniformly accelerated rectilinear
motion. Solve for acceleration and time t to
reach L.
Pulley D is attached to a collar
• Pulley D has uniform rectilinear motion.
which is pulled down at 75 mm/s.
Calculate change of position at time t.
At t = 0, collar A starts moving
• Block B motion is dependent on motions of
down from K with constant
collar A and pulley D. Write motion
acceleration and zero initial velocity. relationship and solve for change of block B
Knowing that velocity of collar A is
position at time t.
300 mm/s as it passes L, determine
• Differentiate motion relation twice to develop
the change in elevation, velocity, and equations for velocity and acceleration of block
acceleration of block B when block
B.
A is at L.
SOLUTION:
• Define origin at upper horizontal surface with positive
displacement downward.
• Collar A has uniformly accelerated rectilinear motion. Solve for
acceleration and time t to reach L.
2
v A2 =  v A 0  2a A  x A   x A 0 
2
mm 

 300
 = 2a A  200 mm 
s 

mm
a A = 225 2
s
v A = (v A )0 + a At
300
mm
mm
= 225 2 t
s
s
t = 1.333 s
• Pulley D has uniform rectilinear motion. Calculate change of
position at time t.
x D = ( x D ) 0 + vD t
æ mm ö
xD - ( xD ) 0 = ç 75
÷ (1.333s) = 100 mm
è
s ø
• Block B motion is dependent on motions of collar A and pulley
D. Write motion relationship and solve for change of block B
position at time t.
Total length of cable remains constant,
x A  2 xD  x B =  x A  0  2  x D  0   x B  0
 x A   x A 0   2  xD   xD 0    xB   xB 0  = 0



 

 200 mm   2 100 mm    xB   xB 0  = 0
xB - (xB )0 = - 400mm
• Differentiate motion relation twice to develop equations for
velocity and acceleration of block B.
x A  2 xD  xB = constant
v A  2vD  vB = 0
mm 

 mm 
300

2


 75
  vB = 0
s 
s 


vB = - 450
mm
mm
= 450
s
s
a A  2aD  aB = 0
mm 

225
 2(0)  aB = 0

2 
s 

aB = - 225
mm
mm
=
225
s2
s2
CURVILINEAR MOTION
Curvilinear Motion: Position, Velocity &
Acceleration
The softball and the car both undergo curvilinear motion.
A particle moving along a curve other than a straight line is
in curvilinear motion.
Curvilinear Motion: Position, Velocity &
Acceleration
• The position vector of a particle at time t is defined by a vector between origin O of a fixed
reference frame and the position occupied by particle.

• Consider a particle which occupies position P defined by r at time t and P’ defined
by r  at t + Dt,
Curvilinear Motion: Position, Velocity &
Acceleration
Instantaneous velocity (vector)
Dr dr
v = lim
=
Dt  0 D t
dt
Instantaneous speed (scalar)
Ds ds
v = lim
=
Dt  0 D t
dt
Curvilinear Motion: Position, Velocity &
Acceleration

• Consider velocity v of a particle at time t and velocity v at t + Dt,
Dv dv
=
= instantaneous acceleration (vector)
Dt  0 Dt
dt
a = lim
• In general, the acceleration vector is not tangent to the
particle path and velocity vector.
Rectangular Components of Velocity & Acceleration
• When position vector of particle P is given by its rectangular
components,




r = xi  y j  zk
• Velocity vector,



 dx  dy  dz 
v = i  j  k = xi  y j  zk
dt
dt
dt



= vx i  v y j  vz k
• Acceleration vector,



 d 2 x d 2 y  d 2 z 
a = 2 i  2 j  2 k = xi  y j  zk
dt
dt
dt



= ax i  a y j  az k
Rectangular Components of Velocity & Acceleration
• Rectangular components particularly effective when
component accelerations can be integrated independently, e.g.,
motion of a projectile,
a x = x = 0
a y = y =  g
with initial conditions,
x0 = y 0 = z 0 = 0
a z = z = 0
v x 0 , v y 0 , v z 0 = 0
Integrating twice yields
v x = v x 0
x = v x 0 t
v y = v y   gt
0
y = v y  y  12 gt 2
0
vz = 0
z=0
• Motion in horizontal direction is uniform.
• Motion in vertical direction is uniformly accelerated.
• Motion of projectile could be replaced by two independent
rectilinear motions.
PROJECTILE MOTION
KINEMATIC EQUATIONS: HORIZONTAL MOTION
Since ax = 0, the velocity in the horizontal direction remains constant
(vx = vox) and the position in the x direction can be determined by:
x = xo + (vox) t
Why is ax equal to zero (assuming movement through the air)?
KINEMATIC EQUATIONS: VERTICAL MOTION
Since the positive y-axis is directed upward, ay = – g. Application of
the constant acceleration equations yields:
vy = voy – g t
y = yo + (voy) t – ½ g t2
vy2 = voy2 – 2 g (y – yo)
For any given problem, only two of these three equations can be used. Why?
Problem 11.7
SOLUTION:
• Consider the vertical and horizontal motion
separately (they are independent)
• Apply equations of motion in y-direction
• Apply equations of motion in x-direction
• Determine time t for projectile to hit the ground,
A projectile is fired from the edge of a
use this to find the horizontal distance
150-m cliff with an initial velocity of
180 m/s at an angle of 30°with the
horizontal. Neglecting air resistance, • Maximum elevation occurs when v =0
y
find (a) the horizontal distance from the
gun to the point where the projectile
strikes the ground, (b) the greatest
elevation above the ground reached by
the projectile.
SOLUTION:
Given:
(v)o =180 m/s
(a)y = - 9.81 m/s2
(y)o =150 m
(a)x = 0 m/s2
Vertical motion – uniformly accelerated:
Horizontal motion – uniformly accelerated:
Choose positive x to the right as shown
SOLUTION:
Horizontal distance
Projectile strikes the ground at:
Substitute into equation (1) above
Solving for t, we take the positive root
Substitute t into equation (4)
Maximum elevation occurs when vy=0
Maximum elevation above the ground =
EXERCISE
Given: Projectile is fired with vA=150 m/s at point A.
Find: The horizontal distance it travels (R) and the time in the air.
Solution:
1) Place the coordinate system at point A.
Then, write the equation for horizontal motion.
+ xB = xA + vAx tAB
where xB = R, xA = 0, vAx = 150 (4/5) m/s
Range, R will be R = 120 tAB
2) Now write a vertical motion equation. Use the distance equation.
+ yB = yA + vAy tAB – 0.5 g tAB2
where yB = – 150, yA = 0, and vAy = 150(3/5) m/s
We get the following equation: –150 = 90 tAB + 0.5 (– 9.81)tAB2
Solving for tAB first, tAB = 19.89 s.
Then, R = 120 tAB = 120 (19.89) = 2387 m
Motion Relative to a Frame in Translation
A soccer player must consider
the relative motion of the ball
and her teammates when
making a pass.
It is critical for a pilot to
know the relative motion
of his aircraft with respect
to the aircraft carrier to
make a safe landing.
Motion Relative to a Frame in Translation
• Designate one frame as the fixed frame of reference. All other frames
not rigidly attached to the fixed reference frame are moving frames
of reference.
• Position vectors for particles A and B with respect to the fixed
frame of reference Oxyz are rA and rB .

• Vector rB A joining A and B defines the position of B with
respect to the moving frame Ax’y’z’ and

 
rB = rA  rB A
• Differentiating twice,



vB = v A  vB A



a B = a A  aB A

vB A = velocity of B relative to A.

a B A = acceleration of B relative to A.
• Absolute motion of B can be obtained by combining motion of A
with relative motion of B with respect to moving reference frame
attached to A.
Problem 11.9
SOLUTION:
• Define inertial axes for the system
• Determine the position, speed, and acceleration of
car A at t = 5 s
• Determine the position, speed, and acceleration of
car B at t = 5 s
• Using vectors (Eqs 11.31, 11.33, and 11.34) or a
graphical approach, determine the relative position,
velocity, and acceleration
Automobile A is traveling east at the constant
speed of 36 km/h. As automobile A crosses the
intersection shown, automobile B starts from
rest 35 m north of the intersection and moves
south with a constant acceleration of 1.2 m/s2.
Determine the position, velocity, and
acceleration of B relative to A 5 s after A crosses
the intersection.
Problem 11.9
• Define axes along the road
SOLUTION:
Given:
vA=36 km/h, aA= 0, (xA)0 = 0
(vB)0= 0, aB= - 1.2 m/s2, (yA)0 = 35 m
Determine motion of Automobile A:
We have uniform motion for A so:
At t = 5 s
SOLUTION:
Determine motion of Automobile B:
We have uniform acceleration for B so:
At t = 5 s
11 - 48
SOLUTION:
We can solve the problems geometrically, and apply the arctangent relationship:
Or we can solve the problems using vectors to obtain equivalent results:
rB = rA  rB/ A
20 j = 50i  rB/ A
rB/ A = 20 j  50i (m)
aB = a A  aB/ A
v B = v A  v B/ A
6 j = 10i  v B/ A
v B/ A = 6 j  10i (m/s)
1.2 j = 0i  aB/ A
aB/ A = 1.2 j (m/s 2 )
Physically, a rider in car A would “see” car B travelling south and west.
TANGENT AND NORMAL
COMPONENTS
Tangential and Normal Components
If we have an idea of the path of a vehicle, it is often convenient to
analyze the motion using tangential and normal components
(sometimes called path coordinates).
NORMAL AND TANGENTIAL COMPONENTS
When a particle moves along a curved path, it is sometimes convenient to
describe its motion using coordinates other than Cartesian. When the path of
motion is known, normal (n) and tangential (t) coordinates are often used.
In the n-t coordinate system, the
origin is located on the particle (the
origin moves with the particle).
The t-axis is tangent to the path (curve) at the instant considered, positive in the
direction of the particle’s motion.
The n-axis is perpendicular to the t-axis with the positive direction toward the
center of curvature of the curve.
The positive n and t directions are defined by the unit
vectors un and ut, respectively.
The center of curvature, O’, always lies on the concave
side of the curve.
The radius of curvature, r, is defined as the
perpendicular distance from the curve to the center of
curvature at that point.
The position of the particle at any instant is defined
by the distance, s, along the curve from a fixed reference point.
VELOCITY IN THE n-t COORDINATE SYSTEM
The velocity vector is always tangent to the
path of motion (t-direction).
The magnitude is determined by taking the time derivative of the path function, s(t).
v = v ut where v = s = ds/dt
.
Here v defines the magnitude of the velocity (speed) and
ut defines the direction of the velocity vector.
ACCELERATION IN THE n-t COORDINATE SYSTEM
Acceleration is the time rate of change of velocity:
.
.
a = dv/dt = d(vut)/dt = vut + vut
.
Here v represents the change in the magnitude of
velocity and ut represents the rate of change in.
the direction of ut.
After mathematical manipulation, the acceleration
vector can be expressed as:
.
a = v ut + (v2/r) un = at ut + an un.
ACCELERATION IN THE n-t COORDINATE SYSTEM
(continued)
So, there are two components to the acceleration vector:
a = at ut + an un
• The tangential component is tangent to the curve and in the direction of increasing or
decreasing velocity.
at = v . or at ds = v dv
• The normal or centripetal component is always directed toward the center of curvature
of the curve. an = v2/r
• The magnitude of the acceleration vector is
a = [(at)2 + (an)2]0.5
SPECIAL CASES OF MOTION
There are some special cases of motion to consider.
1) The particle moves along a straight line.
.
r
 => an = v2/r = 0 => a = at = v
The tangential component represents the time rate of change in the
magnitude of the velocity.
2) The particle
moves along a curve at constant speed.
.
at = v = 0 => a = an = v2/r
The normal component represents the time rate of change in the
direction of the velocity.
3) The tangential component of acceleration is constant, at = (at)c.
In this case,
s = so + vo t + (1/2) (at)c t2
v = vo + (at)c t
v2 = (vo)2 + 2 (at)c (s – so)
As before, so and vo are the initial position and velocity of the particle at t = 0. How are these
equations related to projectile motion equations? Why?
4)
The particle moves along a path expressed as y = f(x).
The radius of curvature, r, at any point on the path can be calculated from
[ 1  (dy/dx)2
r = ________________
d2y/dx
2
]3/2
Problem 11.10
SOLUTION:
• Define your coordinate system
• Calculate the tangential velocity and tangential
acceleration
• Calculate the normal acceleration
A motorist is traveling on a curved section of
highway of radius 750 m at the speed of 90
km/h. The motorist suddenly applies the
brakes, causing the automobile to slow down
at a constant rate. Knowing that after 8 s the
speed has been reduced to 72 km/h,
determine the acceleration of the automobile
immediately after the brakes have been
applied.
• Determine overall acceleration magnitude after the
brakes have been applied
SOLUTION:
• Define your coordinate system
• Determine velocity and acceleration in the
tangential direction
et
en
 km   1000 m  1 h 
90 km/h =  90

 = 25 m/s

h
1
km
3600
s




72 km/h = 20 m/s
• The deceleration constant, therefore
at = average at =
Dv 20 m/s - 25 m/s
=
= - 0.625 m/s2
Dt
8s
• Immediately after the brakes are applied, the
speed is still 25 m/s
v 2 (25 m/s)2
an =
=
= 0.833 m/s2
r
750 m
an 0.833 m/s2
a = an2  at2 = (0.625)2  (0.833)2 tan a = a =
0.625 m/s2
t
a = 1.041 m/s2
a = 53.1
THE END OF KINEMATICS OF
PARTICLES
ASSIGNMENT 1
SUBMIT NEXT WEEK