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![Page 1: 1 ENGR 214 Engineering Mechanics II: Dynamics Summer 2011 Dr. Mustafa Arafa American University in Cairo Mechanical Engineering Department mharafa@aucegypt.edu.](https://reader031.fdocuments.us/reader031/viewer/2022012305/56649c8a5503460f9494405f/html5/thumbnails/1.jpg)
1
ENGR 214Engineering Mechanics II:
Dynamics
Summer 2011
Dr. Mustafa ArafaAmerican University in Cairo
Mechanical Engineering [email protected]
![Page 2: 1 ENGR 214 Engineering Mechanics II: Dynamics Summer 2011 Dr. Mustafa Arafa American University in Cairo Mechanical Engineering Department mharafa@aucegypt.edu.](https://reader031.fdocuments.us/reader031/viewer/2022012305/56649c8a5503460f9494405f/html5/thumbnails/2.jpg)
2
Course Information• Textbook: Vector Mechanics for Engineers: Dynamics, F.P. Beer and E.R.
Johnston, McGraw-Hill, 8th edition, 2007. • Grading: attendance 6%; homework & quizzes 15%; mid-term exams 20%, 16%,
13%; final exam 30%; bonus project 5%.• Lecture notes: will be posted on my website. Additional material will also be
covered on the board. Please print out the notes beforehand & bring them to class.• Bonus project:
• Apply the course material to the analysis of a real system• Use Working Model to model and solve some problems• Search the internet for related software, download & test the software on
simple problems• Develop your own software to solve a class of problems• Develop some teaching aids for part of the course
• My advice: practice makes perfect!
![Page 3: 1 ENGR 214 Engineering Mechanics II: Dynamics Summer 2011 Dr. Mustafa Arafa American University in Cairo Mechanical Engineering Department mharafa@aucegypt.edu.](https://reader031.fdocuments.us/reader031/viewer/2022012305/56649c8a5503460f9494405f/html5/thumbnails/3.jpg)
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KINEMATICS
KINETICSNEWTON’S LAW
KINETICSENERGY & MOMENTUM
PARTICLE SYSTEM OF PARTICLES
RIGID BODIES
Chapter 11
Chapter 12
Chapter 13
Chapter 14
Chapter 15
Chapter 16
Chapter 17
Course Outline
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All figures taken from Vector Mechanics for Engineers: Dynamics, Beer and Johnston, 2004
ENGR 214Chapter 11
Kinematics of Particles
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Introduction• Dynamics includes:
- Kinematics: study of the motion (displacement, velocity, acceleration, & time) without reference to the cause of motion (i.e. regardless of forces).
- Kinetics: study of the forces acting on a body, and the resulting motion caused by the given forces.
• 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.
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Rectilinear Motion: Position, Velocity & Acceleration
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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 (+ or -) 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.,
326 ttx
or in the form of a graph x vs. t.
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Rectilinear Motion: Position, Velocity & Acceleration
• 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+t,
t
xv
t
x
t
0lim
Average velocity
Instantaneous velocity
• From the definition of a derivative,
dt
dx
t
xv
t
0lim
e.g.,
2
32
312
6
ttdt
dxv
ttx
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Rectilinear Motion: Position, Velocity & Acceleration
• Consider particle with velocity v at time t and v’ at t+t,
Instantaneous accelerationt
va
t
0lim
tdt
dva
ttv
dt
xd
dt
dv
t
va
t
612
312e.g.
lim
2
2
2
0
• From the definition of a derivative,
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Rectilinear Motion: Position, Velocity & Acceleration
• Consider particle with motion given by326 ttx
2312 ttdt
dxv
tdt
xd
dt
dva 612
2
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
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11
Determining the Motion of a Particle
• Recall, motion is defined if position x is known for all time t.
• If the acceleration is given, we can determine velocity and position by 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)
dxv
dt
dva
dt
2
2
d xa
dt
dv dv dx dva v
dt dx dt dx
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Determining the Motion of a Particle
• Acceleration given as a function of time, a = f(t):
0
0
0 0
( ) ( ) ( ) ( )v t t
v
dva f t dv f t dt dv f t dt v v f t dt
dt
0
0
0 0
x t t
x
dxv dx vdt dx vdt x x vdt
dt
0 0 0
2 20
1 1( ) ( ) ( ) ( )
2 2
v x x
v x x
dva f x v vdv f x dx vdv f x dx v v f x dx
dx
•Acceleration given as a function of position, a = f(x):
0 0
x t
x
dx dx dxv dt dt
dt v v
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13
Determining the Motion of a Particle
0 00
( )( ) ( ) ( )
v t v
v v
dv dv dv dva f v dt dt t
dt f v f v f v
0 0 0
0( )( ) ( ) ( )
x v v
x v v
dv vdv vdv vdva f v v dx dx x x
dx f v f v f v
• Acceleration given as a function of velocity, a = f(v):
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Summary
Procedure:1. Establish a coordinate system & specify an origin2. Remember: x,v,a,t are related by:
3. When integrating, either use limits (if known) or add a constant of integration
dxv
dt
dva
dt
2
2
d xa
dt
dv dv dx dva v
dt dx dt dx
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Sample Problem 11.2
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.
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Sample Problem 11.2
tvtvdtdv
adt
dv
ttv
v
81.981.9
sm81.9
00
2
0
ttv
2s
m81.9
s
m10
2
21
00
81.91081.910
81.910
0
ttytydttdy
tvdt
dy
tty
y
22s
m905.4
s
m10m20 ttty
SOLUTION:• Integrate twice to find v(t) and y(t).
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17
Sample Problem 11.2
• Solve for t at which velocity equals zero and evaluate corresponding altitude.
0s
m81.9
s
m10
2
ttv
s019.1t
22
22
s019.1s
m905.4s019.1
s
m10m20
s
m905.4
s
m10m20
y
ttty
m1.25y
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18
Sample Problem 11.2
• Solve for t at which altitude equals zero and evaluate corresponding velocity.
0s
m905.4
s
m10m20 2
2
ttty
s28.3
smeaningles s243.1
t
t
s28.3s
m81.9
s
m10s28.3
s
m81.9
s
m10
2
2
v
ttv
s
m2.22v
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What if the ball is tossed downwards with the same speed? (The audience is thinking …)
vo= - 10 m/s
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20
Uniform Rectilinear Motion
Uniform rectilinear motion acceleration = 0 velocity = constant
vtxx
vtxx
dtvdx
vdt
dx
tx
x
0
0
00
constant
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21
Uniformly Accelerated Rectilinear MotionUniformly accelerated motion acceleration = constant
atvv
atvvdtadvadt
dv tv
v
0
000
constant
221
00
221
000
000
attvxx
attvxxdtatvdxatvdt
dx tx
x
020
2
020
221
2
constant00
xxavv
xxavvdxadvvadx
dvv
x
x
v
v
Also:
Application: free fall
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22
Motion of Several Particles: Relative Motion
• For particles moving along the same line, displacements should be measured from the same origin in the same direction.
ABAB xxx relative position of B with respect to A
ABAB xxx
ABAB vvv relative velocity of B with respect to A
ABAB vvv
ABAB aaa relative acceleration of B with respect to A
ABAB aaa
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23
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.
Sample Problem 11.4
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Sample Problem 11.4 SOLUTION:
• Ball: uniformly accelerated motion (given initial position and velocity).
• Elevator: constant velocity (given initial position and velocity)
• 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.
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25
Sample Problem 11.4
SOLUTION:
• Ball: uniformly accelerated rectilinear motion.
22
221
00
20
s
m905.4
s
m18m12
s
m81.9
s
m18
ttattvyy
tatvv
B
B
• Elevator: uniform rectilinear motion.
ttvyy
v
EE
E
s
m2m5
s
m2
0
![Page 26: 1 ENGR 214 Engineering Mechanics II: Dynamics Summer 2011 Dr. Mustafa Arafa American University in Cairo Mechanical Engineering Department mharafa@aucegypt.edu.](https://reader031.fdocuments.us/reader031/viewer/2022012305/56649c8a5503460f9494405f/html5/thumbnails/26.jpg)
26
Sample Problem 11.4
• Relative position of ball with respect to elevator:
025905.41812 2 ttty EB
s65.3
smeaningles s39.0
t
t
• Substitute impact time into equations for position of elevator and relative velocity of ball with respect to elevator. 65.325Ey
m3.12Ey
65.381.916
281.918
tv EB
s
m81.19EBv
![Page 27: 1 ENGR 214 Engineering Mechanics II: Dynamics Summer 2011 Dr. Mustafa Arafa American University in Cairo Mechanical Engineering Department mharafa@aucegypt.edu.](https://reader031.fdocuments.us/reader031/viewer/2022012305/56649c8a5503460f9494405f/html5/thumbnails/27.jpg)
27
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.
BA xx 2 constant (one degree of freedom)
• Positions of three blocks are dependent.
CBA xxx 22 constant (two degrees of freedom)
• For linearly related positions, similar relations hold between velocities and accelerations.
022or022
022or022
CBACBA
CBACBA
aaadt
dv
dt
dv
dt
dv
vvvdt
dx
dt
dx
dt
dx
![Page 28: 1 ENGR 214 Engineering Mechanics II: Dynamics Summer 2011 Dr. Mustafa Arafa American University in Cairo Mechanical Engineering Department mharafa@aucegypt.edu.](https://reader031.fdocuments.us/reader031/viewer/2022012305/56649c8a5503460f9494405f/html5/thumbnails/28.jpg)
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Applications
![Page 29: 1 ENGR 214 Engineering Mechanics II: Dynamics Summer 2011 Dr. Mustafa Arafa American University in Cairo Mechanical Engineering Department mharafa@aucegypt.edu.](https://reader031.fdocuments.us/reader031/viewer/2022012305/56649c8a5503460f9494405f/html5/thumbnails/29.jpg)
29
Sample Problem 11.5
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 block A is at L.
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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
020
2
s
in.9in.82
s
in.12
2
AA
AAAAA
aa
xxavv
s 333.1s
in.9
s
in.12
2
0
tt
tavv AAA
![Page 31: 1 ENGR 214 Engineering Mechanics II: Dynamics Summer 2011 Dr. Mustafa Arafa American University in Cairo Mechanical Engineering Department mharafa@aucegypt.edu.](https://reader031.fdocuments.us/reader031/viewer/2022012305/56649c8a5503460f9494405f/html5/thumbnails/31.jpg)
31
Sample Problem 11.5
• Pulley D has uniform rectilinear motion. Calculate change of position at time t.
in. 4s333.1s
in.30
0
DD
DDD
xx
tvxx
• 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,
0in.42in.8
02
22
0
000
000
BB
BBDDAA
BDABDA
xx
xxxxxx
xxxxxx
in.160 BB xx
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32
Sample Problem 11.5
• Differentiate motion relation twice to develop equations for velocity and acceleration of block B.
0s
in.32
s
in.12
02
constant2
B
BDA
BDA
v
vvv
xxx
in.18
sBv
2
2 0
in.9 0
s
A D B
B
a a a
a
2s
in.9Ba
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Curvilinear Motion
http://news.yahoo.com/photos/ss/441/im:/070123/ids_photos_wl/r2207709100.jpg
A particle moving along a curve other than a straight line is said to be in curvilinear motion.
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34
Curvilinear Motion: Position, Velocity & Acceleration
• 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 defined by at time t and P’ defined by at t + t,
r
r
dt
ds
t
sv
dt
rd
t
rv
t
t
0
0
lim
lim
instantaneous velocity (vector)
instantaneous speed (scalar)
Velocity is tangent to path
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35
Curvilinear Motion: Position, Velocity & Acceleration
dt
vd
t
va
t
0
lim
instantaneous acceleration (vector)
• Consider velocity of particle at time t and velocity at t + t,
v
v
• In general, acceleration vector is not tangent to particle path and velocity vector.
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36
Rectangular Components of Velocity & Acceleration
• Position vector of particle P given by its rectangular components:
kzjyixr
• Velocity vector,
kvjviv
kzjyixkdt
dzj
dt
dyi
dt
dxv
zyx
• Acceleration vector,
kajaia
kzjyixkdt
zdj
dt
ydi
dt
xda
zyx
2
2
2
2
2
2
![Page 37: 1 ENGR 214 Engineering Mechanics II: Dynamics Summer 2011 Dr. Mustafa Arafa American University in Cairo Mechanical Engineering Department mharafa@aucegypt.edu.](https://reader031.fdocuments.us/reader031/viewer/2022012305/56649c8a5503460f9494405f/html5/thumbnails/37.jpg)
37
Rectangular Components of Velocity & Acceleration
0 0 0 0 00 x yx y z v v given
• Rectangular components are useful when acceleration components can be integrated independently, ex: motion of a projectile.
00 zagyaxa zyx
with initial conditions,
Therefore:
• Motion in horizontal direction is uniform.
• Motion in vertical direction is uniformly accelerated.
• Motion of projectile could be replaced by two independent rectilinear motions.
0 0
2
0 0
1
2
x x y y
x y
v v v v gt
x v t y v t gt
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38
ExampleA 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. Find (a) the range, and (b) maximum height.
x
y
0v v at 21
0 0 2x x v t at
2 20 02v v a x x
Remember:
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39
ExampleCar A is traveling at a constant speed of 36 km/h. As A crosses intersection, B starts from rest 35 m north of intersection and moves with a constant acceleration of 1.2 m/s2. Determine the speed, velocity and acceleration of B relative to A 5 seconds after A crosses intersection.
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Tangential and Normal Components• Velocity vector of particle is tangent to path
of particle. In general, acceleration vector is not. Wish to express acceleration vector in terms of tangential and normal components.
• are tangential unit vectors for the particle path at P and P’. When drawn with respect to the same origin,
tt ee and
t t tde e e
t t te e de
tde dFrom geometry:
t nde d e
tn
dee
d
d
tde
![Page 41: 1 ENGR 214 Engineering Mechanics II: Dynamics Summer 2011 Dr. Mustafa Arafa American University in Cairo Mechanical Engineering Department mharafa@aucegypt.edu.](https://reader031.fdocuments.us/reader031/viewer/2022012305/56649c8a5503460f9494405f/html5/thumbnails/41.jpg)
41
2 2
t n t n
dv v dv va e e a a
dt dt
Tangential and Normal Components
tevv• With the velocity vector expressed as
the particle acceleration may be written as
t tt t
de dedv dv dv d dsa e v e v
dt dt dt dt d ds dt
but
vdt
dsdsde
d
edn
t
After substituting,
• 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.
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rr re
Radial and Transverse Components• If particle position is given in polar coordinates,
we can express velocity and acceleration with components parallel and perpendicular to OP.
rr e
d
ede
d
ed
dt
de
dt
d
d
ed
dt
ed rr
dt
de
dt
d
d
ed
dt
edr
rr r
d dr dev re e r
dt dt dt
• Particle velocity vector:
• Similarly, particle acceleration:
r
rr
r r
da re r e
dtde de
re r r e r e rdt dt
d dre re r e r e r e
dt dt
rerr
• Particle position vector:
r r
dr dv e r e re r e
dt dt
2 2ra r r e r r e
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Sample Problem 11.10
A motorist is traveling on curved section of highway at 60 mph. The motorist applies brakes causing a constant deceleration.
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.
![Page 44: 1 ENGR 214 Engineering Mechanics II: Dynamics Summer 2011 Dr. Mustafa Arafa American University in Cairo Mechanical Engineering Department mharafa@aucegypt.edu.](https://reader031.fdocuments.us/reader031/viewer/2022012305/56649c8a5503460f9494405f/html5/thumbnails/44.jpg)
44
Sample Problem 11.10
ft/s66mph45
ft/s88mph60
SOLUTION:
• Calculate tangential and normal components of acceleration.
2
22
2
s
ft10.3
ft2500
sft88
s
ft75.2
s 8
sft8866
v
a
t
va
n
t
• Determine acceleration magnitude and direction with respect to tangent to curve.
2222 10.375.2 nt aaa 2s
ft14.4a
75.2
10.3tantan 11
t
n
a
a 4.48
![Page 45: 1 ENGR 214 Engineering Mechanics II: Dynamics Summer 2011 Dr. Mustafa Arafa American University in Cairo Mechanical Engineering Department mharafa@aucegypt.edu.](https://reader031.fdocuments.us/reader031/viewer/2022012305/56649c8a5503460f9494405f/html5/thumbnails/45.jpg)
45
Sample Problem 11.11
Determine the minimum radius of curvature of the trajectory described by the projectile.
2
n
va
Recall:
2
n
v
a
Minimum , occurs for small v and large an
2155.9
24809.81
m
v is min and an is max
a
an
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46
Sample Problem 11.12
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.
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47
Sample Problem 11.12
SOLUTION:
• Evaluate time t for = 30o.
s 869.1rad524.030
0.15 2
t
t
• Evaluate radial and angular positions, and first and second derivatives at time t.
2
2
sm24.0
sm449.024.0
m 481.012.09.0
r
tr
tr
2
2
srad30.0
srad561.030.0
rad524.015.0
t
t
![Page 48: 1 ENGR 214 Engineering Mechanics II: Dynamics Summer 2011 Dr. Mustafa Arafa American University in Cairo Mechanical Engineering Department mharafa@aucegypt.edu.](https://reader031.fdocuments.us/reader031/viewer/2022012305/56649c8a5503460f9494405f/html5/thumbnails/48.jpg)
48
Sample Problem 11.12
• Calculate velocity and acceleration.
rr
r
v
vvvv
rv
srv
122 tan
sm270.0srad561.0m481.0
m449.0
0.31sm524.0 v
rr
r
a
aaaa
rra
rra
122
2
2
2
22
2
tan
sm359.0
srad561.0sm449.02srad3.0m481.0
2
sm391.0
srad561.0m481.0sm240.0
6.42sm531.0 a
![Page 49: 1 ENGR 214 Engineering Mechanics II: Dynamics Summer 2011 Dr. Mustafa Arafa American University in Cairo Mechanical Engineering Department mharafa@aucegypt.edu.](https://reader031.fdocuments.us/reader031/viewer/2022012305/56649c8a5503460f9494405f/html5/thumbnails/49.jpg)
49
Sample Problem 11.12
• Evaluate acceleration with respect to arm.
Motion of collar with respect to arm is rectilinear and defined by coordinate r.
2sm240.0ra OAB