kinematika partikel

1. Kinematics of Particles

2. ContentsProblem 11.5
Sample
Introduction
Rectilinear Motion: Position, Velocity & Acceleration
Graphical Solution of RectilinearDetermination of the Motion of a ParticleMotion Problems
Sample Problem 11.2
Other Graphical Methods
Sample Problem 11.3
Curvilinear Motion: Position, Velocity
Uniform Rectilinear-Motion
& Acceleration
Uniformly Accelerated RectilinearDerivatives of Vector Functions
Motion
Rectangular Components of Velocity
Motion of Several Particles:
and Acceleration
Relative Motion
Motion Relative to a Frame in
Sample Problem 11.4
Translation
Motion of Several Particles:
Tangential and Normal Components
Dependent Motion
Radial and Transverse Components
Sample Problem 11.10
Sample Problem 11.12
11 - 2

3. • Dynamics includes:
Introduction
- Kinematics: study of the geometry of motion. Kinematics is used to
relate displacement, velocity, acceleration, and time without reference to
the cause of motion.
- Kinetics: study of the relations existing between the forces acting on a
body, the mass of the body, and the motion of the body. Kinetics is used
to predict the motion caused by given forces or to determine the forces
required to produce a given motion.
• 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.
11 - 3

4. Rectilinear Motion: Position, Velocity & Acceleration
• Particle moving along a straight line is said
to be in rectilinear motion.
• Position coordinate of a particle is defined
by positive or negative distance of particle
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. Motion of the particle
may be expressed in the form of a function,
e.g.,
x = 6t 2 − t 3
or in the form of a graph x vs. t.
11 - 4

5. Rectilinear Motion: Position, Velocity & Acceleration
• Consider particle which occupies position
P at time t and P’ at t+∆t,
∆x
=
Average velocity
∆t
∆x
= v = lim
Instantaneous velocity
∆t →0 ∆t
• Instantaneous velocity may be positive or
negative. Magnitude of velocity is referred
to as particle speed.
• From the definition of a derivative,
∆x dx
v = lim
=
dt
∆t →0 ∆t
e.g., x = 6t 2 − t 3
dx
v=
= 12t − 3t 2
dt
11 - 5

6. Rectilinear Motion: Position, Velocity & Acceleration
• Consider particle with velocity v at time t and
v’ at t+∆t,
∆v
Instantaneous acceleration = a = lim
∆t →0 ∆t
• Instantaneous acceleration may be:
- positive: increasing positive velocity
or decreasing negative velocity
- negative: decreasing positive velocity
or increasing negative velocity.
• From the definition of a derivative,
∆v dv d 2 x
a = lim
=
= 2
dt dt
∆t →0 ∆t
e.g. v = 12t − 3t 2
dv
a=
= 12 − 6t
dt
11 - 6

7. Rectilinear Motion: Position, Velocity
& Acceleration motion given by
• Consider particle with
x = 6t 2 − t 3
v=
dx
= 12t − 3t 2
dt
dv d 2 x
a=
=
= 12 − 6t
dt dt 2
• at t = 0,
x = 0, v = 0, a = 12 m/s2
• 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
• at t = 6 s,
x = 0, v = -36 m/s, a = 24 m/s2
11 - 7

8. Determination of the Motion of a Particle
• Recall, motion of a particle is known if position is known for all time t.
• Typically, conditions of motion are specified by the type of acceleration
experienced by the particle. Determination of velocity and position requires
two successive integrations.
• Three classes of motion may be defined for:
- 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)
11 - 8

9. Determination of the Motion of a Particle
• Acceleration given as a function of time, a = f(t):
v( t )
t
dv
= a = f (t)
dv = f ( t ) dt
∫ dv = ∫ f ( t ) dt
dt
v
0
dx
= v( t )
dt
x( t )
dx = v( t ) dt
t
v( t ) − v0 = ∫ f ( t ) dt
0
0
t
t
∫ dx = ∫ v( t ) dt
x0
x( t ) − x0 = ∫ v( t ) dt
0
0
• Acceleration given as a function of position, a = f(x):
v=
dx
dx
or dt =
dt
v
v dv = f ( x ) dx
a=
v( x )
dv
dv
or a = v = f ( x )
dt
dx
x
∫ v dv = ∫ f ( x ) dx
v0
x0
1 v( x ) 2
2
2
− 1 v0
2
=
x
∫ f ( x ) dx
x0
11 - 9

10. Determination of the Motion of a Particle
• Acceleration given as a function of velocity, a = f(v):
dv
= a = f ( v)
dt
v( t )
∫
v0
dv
= dt
f ( v)
v( t )
∫
v0
t
dv
= ∫ dt
f ( v) 0
dv
=t
f ( v)
dv
v = a = f ( v)
dx
x ( t ) − x0 =
v( t )
∫
v0
v dv
dx =
f ( v)
x( t )
v( t )
x0
v0
∫ dx = ∫
v dv
f ( v)
v dv
f ( v)
11 - 10

11. Sample Problem 11.2
SOLUTION:
• Integrate twice to find v(t) and y(t).
• Solve for t at which velocity equals
zero (time for maximum elevation)
and evaluate corresponding altitude.
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.
• Solve for t at which altitude equals
zero (time for ground impact) and
evaluate corresponding velocity.
11 - 11

12. Sample Problem 11.2
SOLUTION:
• Integrate twice to find v(t) and y(t).
dv
= a = −9.81 m s 2
dt
v( t )
t
v( t ) − v0 = −9.81t
∫ dv = − ∫ 9.81 dt
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 − 1 9.81t 2
2
m
 m 
y ( t ) = 20 m + 10 t −  4.905 2 t 2
 s 
s 
11 - 12

13. Sample Problem 11.2 zero and evaluate
• Solve for t at which velocity equals
corresponding altitude.
v( t ) = 10
m 
m
−  9.81 2  t = 0
s 
s 
t = 1.019 s
• 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
 s 
s 
m

 m
y = 20 m + 10 (1.019 s ) −  4.905 2 (1.019 s ) 2
 s

s 
y = 25.1 m
11 - 13

14. Sample Problem 11.2
•
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 = −1.243 s ( 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
11 - 14

15. Sample Problem 11.3
SOLUTION:
a = − kv
• Integrate a = dv/dt = -kv to find v(t).
• Integrate v(t) = dx/dt to find x(t).
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.
• Integrate a = v dv/dx = -kv to find
v(x).
Determine v(t), x(t), and v(x).
11 - 15

16. Sample Problem 11.3
SOLUTION:
• Integrate a = dv/dt = -kv to find v(t).
v( t )
t
dv
dv
v( t )
a=
= − kv
= − k ∫ dt
ln
= − kt
∫ v
dt
v0
v
0
0
v( t ) = v0 e − kt
• Integrate v(t) = dx/dt to find x(t).
dx
v( t ) =
= v0 e − kt
dt
t
x( t )
t
 1 − kt 
− kt
x ( t ) = v0  − e 
∫ dx = v0 ∫ e dt
 k
0
0
0
x( t ) =
(
v0
1 − e − kt
k
11 - 16
)

17. Sample Problem 11.3
•
Integrate a = v dv/dx = -kv to find v(x).
v
x
v0
dv
a = v = − kv
dx
0
dv = − k dx
∫ dv = −k ∫ dx
v − v0 = − kx
v = v0 − kx
• Alternatively,
(
v0
1 − e − kt
k
)
with
x( t ) =
and
v( t ) = v0 e − kt or e − kt =
then
x( t ) =
v0  v ( t ) 
1 −


k 
v0 

v( t )
v0
v = v0 − kx
11 - 17

18. Uniform Rectilinear Motion
For 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
11 - 18

19. Uniformly Accelerated Rectilinear Motion
For particle in uniformly accelerated rectilinear motion, the acceleration of
the particle is constant.
v
t
v0
dv
= a = constant
dt
0
∫ dv = a ∫ dt
v − v0 = at
v = v0 + at
dx
= v0 + at
dt
x
t
x0
0
∫ dx = ∫ ( v0 + at ) dt
x − x0 = v0 t + 1 at 2
2
x = x0 + v0t + 1 at 2
2
dv
v = a = constant
dx
v
x
v0
x0
∫ v dv = a ∫ dx
1
2
(v 2 − v02 ) = a( x − x0 )
2
v 2 = v0 + 2 a ( x − x0 )
11 - 19

20. 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.
= x B − x A = relative position of B
with respect to A
xB = x A + xB A
xB
A
= v B − v A = relative velocity of B
with respect to A
vB = v A + vB A
vB
A
= a B − a A = relative acceleration of B
with respect to A
aB = a A + aB A
aB
A
11 - 20

21. Sample Problem 11.4
SOLUTION:
• Substitute initial position and velocity
and constant acceleration of ball into
general equations for uniformly
accelerated rectilinear motion.
• Substitute initial position and constant
velocity of elevator into equation for
uniform rectilinear motion.
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.
• 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.
11 - 21

22. Sample Problem 11.4
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 + v0 t + 1 at 2 = 12 m + 18 t −  4.905 2 t 2
2
 s 
s 
• Substitute initial position and constant velocity of
elevator into equation for uniform rectilinear
motion. m
vE = 2
s
 m
y E = y0 + v E t = 5 m +  2 t
 s
11 - 22

23. Sample Problem 11.4
• Write equation for relative position of ball with respect to
elevator and solve for zero relative position, i.e., impact.
yB
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
vB
E
= (18 − 9.81t ) − 2
= 16 − 9.81( 3.65)
vB
E
= −19.81
m
s
11 - 23

24. Motion of Several Particles:
Dependent Motion on position of one
• Position of a particle may depend
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 x B + 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
11 - 24

25. 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.
• Pulley D has uniform rectilinear motion.
Pulley D is attached to a collar which
Calculate change of position at time t.
is pulled down at 3 in./s. At t = 0,
collar A starts moving down from K • Block B motion is dependent on motions
of collar A and pulley D. Write motion
with constant acceleration and zero
relationship and solve for change of block
initial velocity. Knowing that
B position at time t.
velocity of collar A is 12 in./s as it
passes L, determine the change in
• Differentiate motion relation twice to
elevation, velocity, and acceleration
develop equations for velocity and
of block B when block A is at L.
acceleration of block B.
11 - 25

26. 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
v 2 = ( v A ) 0 + 2a A [ x A − ( x A ) 0 ]
A
2
 in. 
12  = 2a A ( 8 in.)
 s 
aA = 9
in.
s2
v A = ( v A ) 0 + a At
12
in.
in.
=9 2t
s
s
t = 1.333 s
11 - 26

27. Sample Problem rectilinear motion. Calculate
11.5
• Pulley D has uniform
change of position at time t.
xD = ( xD ) 0 + vDt
 in. 
x D − ( x D ) 0 =  3 (1.333 s ) = 4 in.
 s 
• 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 x D + x B = ( x A ) 0 + 2( x D ) 0 + ( x B ) 0
[ x A − ( x A ) 0 ] + 2[ xD − ( x D ) 0 ] + [ x B − ( x B ) 0 ] = 0
( 8 in.) + 2( 4 in.) + [ x B − ( x B ) 0 ] = 0
x B − ( x B ) 0 = −16 in.
11 - 27

28. Sample Problem 11.5 to develop
• Differentiate motion relation twice
equations for velocity and acceleration of block B.
x A + 2 x D + x B = constant
v A + 2v D + v B = 0
 in.   in. 
12  + 2 3  + v B = 0
 s   s 
v B = 18
in.
s
a A + 2a D + a B = 0
 in. 
 9 2  + vB = 0
 s 
a B = −9
in.
s2
11 - 28

29. Graphical Solution of RectilinearMotion Problems
• 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.
11 - 29

30. Graphical Solution of RectilinearMotion Problems
• 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.
11 - 30

31. Other Graphical Methods
• Moment-area method to determine particle position
at time t directly from the a-t curve:
x1 − x0 = area under v − t curve
= v0t1 +
v1
∫ ( t1 − t ) dv
v0
using dv = a dt ,
x1 − x0 = v0t1 +
v1
∫ ( t1 − t ) a dt
v0
v1
∫ ( t1 − t ) a dt = first moment of area under a-t curve
v0
with respect to t = t1 line.
x1 = x0 + v0t1 + ( area under a-t curve)( t1 − t )
t = abscissa of centroid C
11 - 31

32. Other Graphical Methods
• Method to determine particle acceleration
from v-x curve:
dv
a=v
dx
= AB tan θ
= BC = subnormal to v-x curve
11 - 32

33. Curvilinear Motion: Position,
• Particle Acceleration
Velocity &moving along a curve other than a straight line
is in curvilinear motion.
• 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 particle which occupies position P


r′
defined by at time t and P’ defined by
at t +
r


∆t, 
∆r dr
v = lim
=
dt
∆t →0 ∆t
= instantaneous velocity (vector)
∆s ds
=
dt
∆t →0 ∆t
v = lim
= instantaneous speed (scalar)
11 - 33

34. Curvilinear Motion: Position,

• Consider velocity v of particle at time t and velocity
Velocity v& Acceleration
′
at t + ∆t,


∆v dv

a = lim
=
dt
∆t →0 ∆t
= instantaneous acceleration (vector)
• In general, acceleration vector is not tangent to
particle path and velocity vector.
11 - 34

35. 
Derivatives P( u ) be a vector function of scalar variable u,
of Vector Functions
• Let




dP
∆P
P( u + ∆u ) − P ( u )
= lim
= lim
du ∆u →0 ∆u ∆u →0
∆u
• Derivative of vector sum,

 

d ( P + Q ) dP dQ
=
+
du
du du
• Derivative of product of scalar and vector functions,



d ( f P ) df
dP
=
P+ f
du
du
du
• Derivative of scalar product and vector product,

 

d ( P • Q ) dP   dQ
=
•Q + P•
du
du
du

 

  dQ
d ( P × Q ) dP
=
×Q + P×
du
du
du
11 - 35

36. Rectangular Components of
• & Acceleration
VelocityWhen 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 = xi + y j + zk



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 = i +  j + k
x y z
dt
dt
dt



= ax i + a y j + az k
11 - 36

37. Rectangular Components of
• Rectangular components particularly
Velocity &component accelerations can beeffective
Acceleration integrated
when
independently, e.g., motion of a projectile,
a x =  = 0
x
a y =  = − g
y
a z =  = 0
z
with initial conditions,
( vx ) 0 , v y , ( vz ) 0 = 0
x0 = y 0 = z 0 = 0
( )0
Integrating twice yields
vx = ( vx ) 0
x = ( vx ) 0 t
( ) 0 − gt
y = ( v y ) y − 1 gt 2
2
0
vy = vy
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.
11 - 37

38. Motion Relative to a Frame in
• Designate one frame as
of
Translation the fixed frame thereference.
All other frames not rigidly attached to
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 rA and rB .

rB A joining A and B defines the position of
• Vector
B with respect to the moving frame Ax’y’z’ and

 
rB = rA + rB A
• Differentiating twice,




vB = v A + vB A vB



a B = a A + aB
A

aB
A
A
= velocity of B relative to 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.
11 - 38

39. Tangential and Normal
Componentsparticle is tangent to path of
• Velocity vector of
particle. In general, acceleration vector is not.
Wish to express acceleration vector in terms of
tangential and normal components.


• et and et′ are tangential unit vectors for the
particle path at P and P’. When drawn with
  
respect to the same origin, ∆et = et′ − et and
∆θ is the angle between them.
∆et = 2 sin ( ∆θ 2)

∆et
sin ( ∆θ 2 ) 

lim
= lim
en = en
∆θ →0 ∆θ
∆θ →0 ∆θ 2

det

en =
dθ
11 - 39

40. Tangential and Normal


• With the velocity vector expressed as v = ve
Components may be written as
the particle acceleration
t



de dv 
de dθ ds
 dv dv 
a=
= et + v
= et + v
dt dt
dt dt
dθ ds dt
but 
det 
ds
= en
ρ dθ = ds
=v
dθ
dt
After substituting,
dv
v2
 dv  v 2 
a = et + en
at =
an =
dt
ρ
dt
ρ
• Tangential component of acceleration reflects
change of speed and normal component reflects
change of direction.
• Tangential component may be positive or
negative. Normal component always points
toward center of path curvature.
11 - 40

41. Tangential and Normal
Components and normal acceleration
• Relations for tangential
also apply for particle moving along space curve.
 dv  v 2 
a = et + en
dt
ρ
dv
at =
dt
v2
an =
ρ
• Plane containing tangential and normal unit
vectors is called the osculating plane.
• Normal to the osculating plane is found from

 
eb = et × en

en = principal normal

eb = binormal
• Acceleration has no component along binormal.
11 - 41

42. Radial and Transverse Components
• When particle position is given in polar coordinates,
it is convenient to express velocity and acceleration
with components parallel and perpendicular to OP.


r = re r

der 
= eθ
dθ
• The particle velocity vector is

der dr 
dr 
dθ 
 d 
v = ( r er ) = e r + r
= er + r
eθ
dt
dt
dt
dt
dt


= r er + rθ eθ


deθ

= − er
dθ


der der dθ  dθ
=
= eθ
dt
dθ dt
dt


deθ deθ dθ
 dθ
=
= − er
dt
dθ dt
dt
• Similarly, the particle acceleration vector is
dθ  
 d  dr 
a =  er + r
eθ 
dt  dt
dt 


d 2 r  dr der dr dθ 
d 2θ 
dθ deθ
= 2 er +
+
eθ + r 2 eθ + r
dt dt dt dt
dt dt
dt
dt



=  − rθ 2 er + ( rθ + 2rθ ) eθ
r

(
)
11 - 42

43. Radial and Transverse position is given in cylindrical
Components
• When particle
coordinates, it is convenient to express the
velocity and acceleration vectors using the unit

 
vectors eR , eθ , and k .
• Position vector,



r = R e R +z k
• Velocity vector,


 dr  

 eθ + z k
v=
= R e R + Rθ

dt
• Acceleration vector,


 dv

2 



a=
= R − Rθ e R + ( Rθ + 2 Rθ ) eθ +  k
z
dt
(
)
11 - 43

44. Sample Problem 11.10
SOLUTION:
• Calculate tangential and normal
components of acceleration.
• Determine acceleration magnitude and
direction with respect to tangent to
curve.
A motorist is traveling on curved
section of highway at 60 mph. The
motorist applies brakes causing a
constant deceleration rate.
Knowing that after 8 s the speed has
been reduced to 45 mph, determine
the acceleration of the automobile
immediately after the brakes are
applied.
11 - 44

45. SampleSOLUTION:
Problem 11.10
• Calculate tangential and normal components of
acceleration.
∆v ( 66 − 88) ft s
ft
at =
=
= −2.75 2
∆t
8s
s
v 2 ( 88 ft s ) 2
ft
an =
=
= 3.10 2
ρ
2500 ft
s
60 mph = 88 ft/s
45 mph = 66 ft/s
• Determine acceleration magnitude and direction
with respect to tangent to curve.
ft
2
2
2
2
a = 4.14 2
a = at + a n = ( − 2.75) + 3.10
s
α = tan
−1 a n
at
= tan
−1 3.10
2.75
α = 48.4°
11 - 45

46. Sample Problem 11.12
SOLUTION:
• Evaluate time t for θ = 30o.
• Evaluate radial and angular positions,
and first and second derivatives at
time t.
Rotation of the arm about O is defined
by θ = 0.15t2 where θ 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.
• Calculate velocity and acceleration in
cylindrical coordinates.
• Evaluate acceleration with respect to
arm.
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.
11 - 46

47. Sample Problem 11.12
SOLUTION:
• Evaluate time t for θ = 30o.
θ = 0.15 t 2
= 30° = 0.524 rad
t = 1.869 s
• Evaluate radial and angular positions, and first
and second derivatives at time t.
r = 0.9 − 0.12 t 2 = 0.481 m
r = −0.24 t = −0.449 m s

r
 = −0.24 m s 2
θ = 0.15 t 2 = 0.524 rad
θ = 0.30 t = 0.561 rad s

θ = 0.30 rad s 2
11 - 47

48. Sample Problem acceleration.
11.12
• Calculate velocity and

vr = r = −0.449 m s
vθ = rθ = ( 0.481m )( 0.561rad s ) = 0.270 m s
v
β = tan −1 θ
vr
2
2
v = vr + vθ
v = 0.524 m s
β = 31.0°
ar =  − rθ 2
r
= −0.240 m s 2 − ( 0.481m ) ( 0.561rad s ) 2
= −0.391m s 2


aθ = rθ + 2rθ
(
)
= ( 0.481m ) 0.3 rad s 2 + 2( − 0.449 m s )( 0.561rad s )
= −0.359 m s 2
2
2
a = ar + aθ
a
γ = tan −1 θ
ar
a = 0.531m s
γ = 42.6°
11 - 48

49. Sample •Problem 11.12 to arm.
Evaluate acceleration with respect
Motion of collar with respect to arm is rectilinear
and defined by coordinate r.
a B OA = r = −0.240 m s 2

11 - 49