Chapter 2. MAE002-Kinematics of particle.pdf

MAE002- Dynamics
Chapter 2 - Kinematics of particles
The motion of the paraglider can be described in terms of its position, velocity, and
acceleration. When landing, the pilot of the paraglider needs to consider the wind
velocity and the relative motion of the glider with respect to the wind. The study of
motion is known as kinematics and is the subject of this chapter.

Objectives
• Describe the relationships between position, velocity,
acceleration, and time.
• Solve kinematics problems.
• Analyze the relative motion of multiple particles.
• Determine the motion of a particle that depends on the
motion of another particle.
• Determine which coordinate system is most appropriate for
solving a curvilinear kinematics problem.
• Calculate the position, velocity, and acceleration of a particle
undergoing curvilinear motion using Cartesian, tangential and
normal, and radial and transverse coordinates.

Contents
1. Introduction
2. Rectilinear Motion
3. Plane Curvilinear Motion
4. Rectangular Coordinates (x-y )
5. Normal and Tangential Coordinates (n-t )
6. Polar Coordinates (r-)
7. Space Curvilinear Motion
8. Relative Motion (Translating Axes)
9. Constrained Motion of Connected Particles
10.Chapter Review

Kinematics: study of the geometry of motion. Relates displacement,
velocity, acceleration, and time without reference to the cause of
motion.
Introduction

Kinematic relationships are used to
help us determine the trajectory of a
golf ball, the orbital speed of a
satellite, and the accelerations during
acrobatic flying.
Introduction

Introduction
 Particle kinetics includes:
• 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
• Rectilinear motion: particle moving
along a straight line
• Position coordinate: defined by
positive or negative distance from a
fixed origin on the line.
• The motion of a particle is known if
the position coordinate for particle is
known for every value of time t.
• May be expressed in the form of a
function, e.g., 3
2
6 t
t
x 

or in the form of a graph x vs. t.

• Instantaneous velocity may be positive
or negative. Magnitude of velocity is
referred to as particle speed.
• Consider particle which occupies position
P at time t and P’ at t+Dt,
t
x
v
t
x
t D
D


D
D


D 0
lim
Average velocity
Instantaneous velocity
• From the definition of a derivative,
dt
dx
t
x
v
t

D
D


D 0
lim
e.g.,
2
3
2
3
12
6
t
t
dt
dx
v
t
t
x





Rectilinear Motion

• Consider particle with velocity v at time
t and v’ at t+Dt,
Instantaneous acceleration
t
v
a
t D
D



D 0
lim
t
dt
dv
a
t
t
v
dt
x
d
dt
dv
t
v
a
t
6
12
3
12
e.g.
lim
2
2
2
0







D
D


D
• From the definition of a derivative,
• Instantaneous acceleration may be:
- positive: increasing positive velocity
or decreasing negative velocity
- negative: decreasing positive velocity
or increasing negative velocity.
Rectilinear Motion

• From our example,
3
2
6 t
t
x 

2
3
12 t
t
dt
dx
v 


t
dt
x
d
dt
dv
a 6
12
2
2




- at t = 2 s, x = 16 m, v = vmax = 12 m/s, a = 0
- at t = 4 s, x = xmax = 32 m, v = 0, a = -12 m/s2
• What are x, v, and a at t = 2 s ?
• 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 ?
Rectilinear Motion

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)
• Can you think of a physical
example of when force is a
function of position?
When force is a function of
velocity?
a spring drag

Acceleration as a function of time, position, or velocity
 
a a t
  
0 0
v t
v
dv a t dt

 
( )
dv
a t
dt

 
vdv a x dx

 
0 0
v x
v x
vdv a x dx

 
 
a a x

and
dx dv
dt a
v dt
 
 
dv
v a v
dx

 
0 0
v t
v
dv
dt
a v

 
 
0 0
x v
x v
vdv
dx
a v

 
 
a a v

( )
dv
a v
dt

If…. Kinematic relationship Integrate

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.
Ball tossed with 10 m/s vertical velocity
from window 20 m above ground.
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
Sample problem 2.1

Sample problem 2.2
 
  t
v
t
v
dt
dv
a
dt
dv
t
t
v
v
81
.
9
81
.
9
s
m
81
.
9
0
0
2
0










  t
t
v 






 2
s
m
81
.
9
s
m
10
 
    2
2
1
0
0
81
.
9
10
81
.
9
10
81
.
9
10
0
t
t
y
t
y
dt
t
dy
t
v
dt
dy
t
t
y
y










  2
2
s
m
905
.
4
s
m
10
m
20 t
t
t
y 














SOLUTION:
• Integrate twice to find v(t) and y(t).

• Solve for t when velocity equals zero and
evaluate corresponding altitude.
  0
s
m
81
.
9
s
m
10 2








 t
t
v s
019
.
1

t
• Solve for t when altitude equals zero and evaluate
 
   2
2
2
2
s
019
.
1
s
m
905
.
4
s
019
.
1
s
m
10
m
20
s
m
905
.
4
s
m
10
m
20






























y
t
t
t
y
m
1
.
25

y
Sample problem 2.2

• Solve for t when altitude equals zero and evaluate
  0
s
m
905
.
4
s
m
10
m
20 2
2















 t
t
t
y
 
s
28
.
3
s
meaningles
s
243
.
1



t
t
 
   
s
28
.
3
s
m
81
.
9
s
m
10
s
28
.
3
s
m
81
.
9
s
m
10
2
2
















v
t
t
v
s
m
2
.
22


v
Sample problem 2.2

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).
kv
a 

SOLUTION:
• Integrate a = dv/dt = -kv to find v(t).
• Integrate v(t) = dx/dt to find x(t).
• Integrate a = v dv/dx = -kv to find v(x).
Sample problem 2.2

SOLUTION:
• Integrate a = dv/dt = -kv to find v(t).
 
0 0
0
ln
v t
v
v t
dv dv
a kv k dt kt
dt v v
      
 
  kt
e
v
t
v 
 0
• Integrate v(t) = dx/dt to find x(t).
 
 
0
0 0
0
0 0
1
kt
t
x t
kt kt
dx
v t v e
dt
dx v e dt x t v e
k

 
 
 
  
 
 
 
   
kt
e
k
v
t
x 

 1
0
Sample problem 2.2

• Integrate a = v dv/dx = -kv to find v(x).
kx
v
v
dx
k
dv
dx
k
dv
kv
dx
dv
v
a
x
v
v









 

0
0
0
kx
v
v 
 0
• Alternatively,
   










0
0 1
v
t
v
k
v
t
x
kx
v
v 
 0
   
0
0 or
v
t
v
e
e
v
t
v kt
kt

 

   
kt
e
k
v
t
x 

 1
0
with
and
then
Sample problem 2.2

Uniform rectilinear motion
For a particle in uniform
rectilinear motion, the
acceleration is zero and the
velocity is constant.
vt
x
x
vt
x
x
dt
v
dx
v
dt
dx
t
x
x









0
0
0
0
constant
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.
Careful – these only apply to
uniform rectilinear motion!

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 free-fall
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.
0
0
0
constant
v t
v
dv
a dv a dt v v at
dt
    
 
 
0
2
1
0 0 0 0 2
0
x t
x
dx
v at dx v at dt x x v t at
dt
      
 
 
0 0
2 2
0 0
constant 2
v x
v x
dv
v a vdv a dx v v a x x
dx
     
 
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.


 A
B
A
B x
x
x relative position of B with
respect to A
A
B
A
B x
x
x 



 A
B
A
B v
v
v relative velocity of B with
respect to A
A
B
A
B v
v
v 



 A
B
A
B a
a
a relative acceleration of B
with respect to A
A
B
A
B a
a
a 


Sample Problem 11.4
Ball thrown vertically from 12 m level
in elevator shaft with initial velocity of
18 m/s. At same instant, open-platform
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.
SOLUTION:
• Ball: uniformly accelerated rectilinear
motion.
• Elevator: 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:
• Ball: uniformly accelerated rectilinear motion.
2
2
2
2
1
0
0
2
0
s
m
905
.
4
s
m
18
m
12
s
m
81
.
9
s
m
18
t
t
at
t
v
y
y
t
at
v
v
B
B




























• Elevator: uniform rectilinear motion.
t
t
v
y
y
v
E
E
E











s
m
2
m
5
s
m
2
0
Sample Problem 11.4

• Write equation for relative position of ball
with respect to elevator and solve for zero
relative position, i.e., impact.
• Substitute impact time into equations for position of elevator and
relative velocity of ball with respect to elevator.
 
65
.
3
2
5

E
y m
3
.
12

E
y
 
 
65
.
3
81
.
9
16
2
81
.
9
18




 t
v E
B
s
m
81
.
19


E
B
v
Sample Problem 11.4
2
/ (12 18 4.905 ) (5 2 ) 0
B E
y t t t
     
0.39 (meaningless)
3.65
t s
t s
 


Sample Problem 11.5
Car A is travelling at a constant 90 mi/h when she passes a parked
police officer B, who gives chase when the car passes her. The officer
accelerates at a constant rate until she reaches the speed of 105
mi/h. Thereafter, her speed remains constant. The police officer
catches the car 3 mi from her starting point. Determine the initial
acceleration of the police officer.
?

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.

 B
A x
x 2 constant (one degree of freedom)
• Positions of three blocks are dependent.


 C
B
A x
x
x 2
2 constant (two degrees of freedom)
• For linearly related positions, similar relations hold
between velocities and accelerations.
0
2
2
or
0
2
2
0
2
2
or
0
2
2












C
B
A
C
B
A
C
B
A
C
B
A
a
a
a
dt
dv
dt
dv
dt
dv
v
v
v
dt
dx
dt
dx
dt
dx

Pulley D is attached to a collar
which is pulled down at 3 in./s. At
t = 0, collar A starts moving down
from K with constant acceleration
and zero initial velocity. Knowing
that velocity of collar A is 12 in./s
as it passes L, determine the
change in elevation, velocity, and
acceleration of block B when
collar 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.
• Pulley D has uniform rectilinear motion.
Calculate change of position at time t.
• 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.
• Differentiate motion relation twice to
develop equations for velocity and
acceleration of block B.
Sample problem 11.5

Sample 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.
   
 
  2
2
0
2
0
2
s
in.
9
in.
8
2
s
in.
12
2











A
A
A
A
A
A
A
a
a
x
x
a
v
v
 
s
333
.
1
s
in.
9
s
in.
12 2
0




t
t
t
a
v
v A
A
A

• Pulley D has uniform rectilinear motion.
Calculate change of position at time t.
 
    in.
4
s
333
.
1
s
in.
3
0
0











D
D
D
D
D
x
x
t
v
x
x
• 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,
     
 
   
   
 
     
  0
in.
4
2
in.
8
0
2
2
2
0
0
0
0
0
0
0















B
B
B
B
D
D
A
A
B
D
A
B
D
A
x
x
x
x
x
x
x
x
x
x
x
x
x
x
  in.
16
0 

 B
B x
x
Sample problem 11.5

• Differentiate motion relation twice to
develop equations for velocity and
acceleration of block B.
0
s
in.
3
2
s
in.
12
0
2
constant
2





















B
B
D
A
B
D
A
v
v
v
v
x
x
x
s
in.
18

B
v
0
s
in.
9
0
2
2











B
B
D
A
v
a
a
a
2
s
in.
9


B
a
Sample problem 11.5

Group problem solving
Slider block A moves to the left with a
constant velocity of 6 m/s. Determine
the velocity of block B.
SOLUTION STEPS:
• Sketch your system and choose
coordinate system
• Write out constraint equation
• Differentiate the constraint equation to
get velocity

Given: vA= 6 m/s left Find: vB
xA
yB
This length is constant no matter how the
blocks move
• Sketch your system and choose
coordinates
• Differentiate the constraint equation
to get velocity
const nts
3 a
A B
x y L
  
• Define your constraint equation(s)
6 m/s + 3 0
B
v  2 m/s
B  
v
Note that as xA gets bigger, yB gets smaller.
Group problem solving

Graphical Solution of Rectilinear-Motion Problems
Engineers often collect position, velocity, and acceleration data.
Graphical solutions are often useful in analyzing these data.
Data Fideltity / Highest Recorded Punch
0
20
40
60
80
100
120
140
160
180
47.76 47.77 47.78 47.79 47.8 47.81
Time (s)
Acceleration
(g)
Acceleration data from a
head impact during a round
of boxing.

• Given the x-t curve, the v-t curve is equal to the x-t curve slope.
• Given the v-t curve, the a-t curve is equal to the v-t curve slope.

• Given the a-t curve, the change in velocity between t1 and t2 is
equal to the area under the a-t curve between t1 and t2.
• Given the v-t curve, the change in position between t1 and t2 is equal
to the area under the v-t curve between t1 and t2.

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.

• 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 at time t
and P’ defined by at t + Dt,
r

r


0
lim
t
s ds
v
t dt
D 
D
 
D
Instantaneous velocity (vector) Instantaneous speed (scalar)
0
lim
t
r dr
v
t dt
D 
D
 
D

0
lim
t
v dv
a
t dt
D 
D
  
D
instantaneous acceleration (vector)
• Consider velocity of a particle at time t and velocity at t + Dt,
v

v


• 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,
k
z
j
y
i
x
r







• Velocity vector,
k
v
j
v
i
v
k
z
j
y
i
x
k
dt
dz
j
dt
dy
i
dt
dx
v
z
y
x






















• Acceleration vector,
k
a
j
a
i
a
k
z
j
y
i
x
k
dt
z
d
j
dt
y
d
i
dt
x
d
a
z
y
x
























 2
2
2
2
2
2

• Rectangular components particularly effective
when component accelerations can be integrated
independently, e.g., motion of a projectile,
0
0 





 z
a
g
y
a
x
a z
y
x 





with initial conditions,
      0
,
,
0 0
0
0
0
0
0 


 z
y
x v
v
v
z
y
x
Integrating twice yields
   
    0
0
2
2
1
0
0
0
0








z
gt
y
v
y
t
v
x
v
gt
v
v
v
v
y
x
z
y
y
x
x
• Motion in horizontal direction is uniform.
• Motion in vertical direction is uniformly
accelerated.
• Motion of projectile could be replaced by two
independent rectilinear motions.
Rectangular Components of Velocity & Acceleration

Sample Problem 11.7
A projectile is fired from the
edge of a 150-m cliff with an
initial velocity of 180 m/s at an
angle of 30°with the
horizontal. Neglecting air
resistance, 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:
• 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, use this to find the
horizontal distance
• Maximum elevation occurs when vy=0

SOLUTION:
Given: (v)o =180 m/s (y)o =150 m
(a)y = - 9.81 m/s2 (a)x = 0 m/s2
Vertical motion – uniformly accelerated:
Horizontal motion – uniform rectilinear:
Choose positive x to the right as shown
Sample Problem 11.7

SOLUTION:
Horizontal distance
Projectile strikes the ground at:
Solving for t, we take the positive root
Maximum elevation occurs when vy=0
Substitute into equation (1) above
Substitute t into equation (4)
Maximum elevation above the ground =
Sample Problem 11.7

Group Problem Solving
A baseball pitching machine “throws” baseballs with a horizontal
velocity v0. If you want the height h to be 42 in., determine the value of
v0.
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 fall to 42 inches
• Calculate v0=0

Group Problem Solving
• Analyze the motion in the
y-direction
Given: x= 40 ft, yo = 5 ft, yf= 42 in. Find: vo
2
0
1
(0)
2
f
y y t gt
  
2 2
1
1.5 ft (32.2 ft/s )
2
t
  
2
1
3.5 5
2
gt
 
0.305234 s
t 
• Analyze the motion in the
x-direction
0 0
0 ( )
x
x v t v t
  
0
40 ft ( )(0.305234 s)
v

0 131.047 ft/s 89.4 mi/h
v  

Motion Relative to a Frame in Translation
A soccer player must consider
the relative motion of the ball
and his 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.

• 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.
• Vector joining A and B defines the position of B with respect to the
moving frame Ax’y’z’ and
A
B
r

A
B
A
B r
r
r





• Differentiating twice,

A
B
v

velocity of B relative to A.
A
B
A
B v
v
v






A
B
a

acceleration of B relative to A.
A
B
A
B a
a
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.
• Position vectors for particles A and B with
respect to the fixed frame of reference Oxyz are
and
A B
r r

Sample Problem 11.9
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.
SOLUTION:
• Define inertial axes for the system
• Determine the position, speed, and
acceleration of car A 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
• Determine the position, speed, and
acceleration of car B at t = 5 s

SOLUTION:
• Define axes along the road
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
Sample Problem 11.9

SOLUTION:
Determine motion of Automobile B:
We have uniform acceleration for B so:
At t = 5 s
Sample Problem 11.9

SOLUTION:
We can solve the problems geometrically, and apply the arctangent relationship:
Or we can solve the problems using vectors to obtain equivalent results:
 
B A B/A
r r r  
B A B/A
v v v  
B A B/A
a a a
20 50
20 50 (m)
 
 
B/A
B/A
j i r
r j i
6 10
6 10 (m/s)
  
  
B/A
B/A
j i v
v j i
2
1.2 0
1.2 (m/s )
  
 
B/A
B/A
j i a
a j
Physically, a driver in car A would “see” car B travelling south and west.
Sample Problem 11.9

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).
Non-rectangular components: radial distance and angular
displacement

• The tangential direction (et) is tangent to the path of the particle.
This velocity vector of a particle is in this direction
x
y
et
en
• The normal direction (en) is perpendicular to et and points towards
the inside of the curve.
v= vt et
 = the instantaneous radius of curvature
2
t n
dv v
a e e
dt 
 
t
v ve

• The acceleration can have components in both the en and et directions


2
2
v
a
dt
dv
a
e
v
e
dt
dv
a n
t
n
t 







A motorist is traveling on a curved
section of highway of radius 2500 ft
at the speed of 60 mi/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 45
mi/h, determine the acceleration of the
automobile immediately after the
brakes have been applied.
SOLUTION:
• Define your coordinate system
• Calculate the tangential velocity
and tangential acceleration
• Determine overall acceleration
magnitude after the brakes have
been applied
• Calculate the normal acceleration

SOLUTION:
• Define your coordinate system
et
en
• Determine velocity and acceleration in the
tangential direction
• The deceleration constant, therefore
• Immediately after the brakes are applied, the
speed is still 88 ft/s
2 2 2 2
2.75 3.10
n t
a a a
   

In 2001, a race scheduled at the Texas Motor Speedway was
cancelled because the normal accelerations were too high and
caused some drivers to experience excessive g-loads (similar to
fighter pilots) and possibly pass out. What are some things that
could be done to solve this problem?
Some possibilities:
• Reduce the allowed speed
• Increase the turn radius
(difficult and costly)
• Have the racers wear g-suits

Radial and Transverse Components
By knowing the distance to the aircraft
and the angle of the radar, air traffic
controllers can track aircraft.
Fire truck ladders can rotate as well as
extend; the motion of the end of the
ladder can be analyzed using radial and
transverse components.

• The position of a particle P is expressed as a distance r from the
origin O to P – this defines the radial direction er. The transverse
direction e is perpendicular to er
r
v r e r e

 
• The particle velocity vector is
• The particle acceleration vector is
   
2
2
r
a r r e r r e
  
   
r
e
r
r



Radial and Transverse Components
Polar coordinates (r,)

If you are travelling in a perfect circle, what is always true about
radial/transverse coordinates and normal/tangential coordinates?
a) The er direction is identical to the en direction.
b) The e direction is perpendicular to the en direction.
c) The e direction is parallel to the er direction.
Concept Quiz

Rotation of the arm about O is
defined by  = 0.15t2 where  is in
radians and t in seconds. Collar B
slides along the arm such that r =
0.9 - 0.12t2 where r is in meters.
After the arm has rotated through
30o, determine (a) the total velocity
of the collar, (b) the total
acceleration of the collar, and (c)
the relative acceleration of the
collar with respect to the arm.
SOLUTION:
• Evaluate time t for  = 30o.
• Evaluate radial and angular
positions, and first and second
derivatives at time t.
• Calculate velocity and acceleration
in cylindrical coordinates.
• Evaluate acceleration with respect
to arm.
Sample Problem 11.12

SOLUTION:
• Evaluate time t for  = 30o.
s
869
.
1
rad
524
.
0
30
0.15 2





t
t

• Evaluate radial and angular positions, and
first and second derivatives at time t.
2
2
s
m
24
.
0
s
m
449
.
0
24
.
0
m
481
.
0
12
.
0
9
.
0









r
t
r
t
r



2
2
s
rad
30
.
0
s
rad
561
.
0
30
.
0
rad
524
.
0
15
.
0










 t
t

• Calculate velocity and acceleration.
  
r
r
r
v
v
v
v
v
r
v
s
r
v





1
2
2
tan
s
m
270
.
0
s
rad
561
.
0
m
481
.
0
m
449
.
0














 0
.
31
s
m
524
.
0 
v
  
     
r
r
r
a
a
a
a
a
r
r
a
r
r
a







1
2
2
2
2
2
2
2
2
tan
s
m
359
.
0
s
rad
561
.
0
s
m
449
.
0
2
s
rad
3
.
0
m
481
.
0
2
s
m
391
.
0
s
rad
561
.
0
m
481
.
0
s
m
240
.
0



























 6
.
42
s
m
531
.
0 
a

• Evaluate acceleration with respect to
arm.
Motion of collar with respect to arm
is rectilinear and defined by
coordinate r.
2
s
m
240
.
0


 r
a OA
B 


Sample Problem
During a parasailing ride, the boat is traveling at a constant 30 km/hr
with a 200-m long tow line. At the instant shown, the angle between
the line and the water is 30° and is increasing at a constant rate of 2°/s.
Determine the velocity and acceleration of the parasailer at this instant.

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