Post on 19-Jan-2023
VECTOR MECHANICS FOR ENGINEERS: DYNAMICSDYNAMICS
Tenth Tenth Edition in Edition in SI UnitsSI Units
Ferdinand P. BeerFerdinand P. BeerE. Russell Johnston, Jr.E. Russell Johnston, Jr.Phillip J. CornwellPhillip J. CornwellLecture Notes:Lecture Notes:Brian P. SelfBrian P. SelfCalifornia Polytechnic State UniversityCalifornia Polytechnic State University
Sanjeev SanghiIndian Institute of Technology, Delhi
CHAPTER
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11 Kinematics of
Particles
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Contents
11 - 2
IntroductionRectilinear Motion: Position, Velocity & Acceleration
Determination of the Motion of a Particle
Sample Problem 11.2Sample Problem 11.3Uniform Rectilinear-Motion
Uniformly Accelerated Rectilinear-Motion
Motion of Several Particles: Relative Motion
Sample Problem 11.4Motion of Several Particles: Dependent Motion
Sample Problem 11.5Graphical Solution of Rectilinear-Motion Problems
Other Graphical MethodsCurvilinear Motion: Position, Velocity & Acceleration
Derivatives of Vector Functions
Rectangular Components of Velocity and Acceleration
Motion Relative to a Frame in Translation
Tangential and Normal Components
Radial and Transverse Components
Sample Problem 11.10Sample Problem 11.12
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units
11 - 3
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.
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Introduction
11 - 4
•Dynamics includes:
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.
Kinematics: study of the geometry of motion. Relates displacement, velocity, acceleration, and time without reference to the cause of motion.Fthrust
Flift
Fdrag
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Introduction
11 - 5
•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.
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units
Rectilinear Motion: Position, Velocity & Acceleration
11 - 6
•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.,
326 ttx or in the form of a graph x vs. t.
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units
Rectilinear Motion: Position, Velocity & Acceleration
11 - 7
• 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,
txv
tx
t
0lim
Average velocity
Instantaneous velocity
• From the definition of a derivative,
dtdx
txv
t
0lim
e.g.,
2
32
312
6
ttdtdxv
ttx
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units
Rectilinear Motion: Position, Velocity & Acceleration
11 - 8
• Consider particle with velocity v at time t and v’ at t+t, Instantaneous
acceleration tva
t
0lim
tdtdva
ttvdt
xddtdv
tva
t
612
312e.g.
lim
222
0
• From the definition of a derivative,
• Instantaneous acceleration may be:- positive: increasing positive velocity
or decreasing negative velocity- negative: decreasing
positive velocityor increasing negative velocity.
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Concept Quiz
11 - 9
What is true about the kinematics of a particle?
a) The velocity of a particle is always positive
b) The velocity of a particle is equal to the slope of the position-time graph
c) If the position of a particle is zero, then the velocity must zero
d) If the velocity of a particle is zero, then its acceleration must be zero
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units
Rectilinear Motion: Position, Velocity & Acceleration
11 - 10
• From our example,326 ttx
2312 ttdtdxv
tdt
xddtdva 6122
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 ?
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units
Determination of the Motion of a Particle
11 - 11
•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
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units
Acceleration as a function of time, position, or velocity
11 - 12
a a t 0 0
v t
v
dv a t dt ()dv a tdt
vdv a x dx
0 0
v x
v x
vdv a x dx a a x and dx dvdt a
v dt
dvv a vdx
0 0
v t
v
dv dta v
0 0
x v
x v
vdvdxa v
a a v
( )dv a vdt
If…. Kinematic relationship Integrate
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Sample Problem 11.2
11 - 13
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 corresponding velocity.
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Sample Problem 11.2
11 - 14
tvtvdtdv
adtdv
ttv
v81.981.9
sm81.9
00
2
0
ttv
2s
m81.9sm10
2
210
081.91081.910
81.910
0
ttytydttdy
tvdtdy
tty
y
22sm905.4s
m10m20 ttty
SOLUTION:• Integrate twice to find v(t) and y(t).
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Sample Problem 11.2
11 - 15
• Solve for t when velocity equals zero and evaluate corresponding altitude. 0
sm81.9s
m10 2
ttv
s019.1t
• Solve for t when altitude equals zero and evaluate corresponding velocity.
22
22
s019.1sm905.4s019.1s
m10m20
sm905.4s
m10m20
y
ttty
m1.25y
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Sample Problem 11.2
11 - 16
• Solve for t when altitude equals zero and evaluate corresponding velocity.
0sm905.4s
m10m20 22
ttty
s28.3
smeaningles s243.1
tt
s28.3sm81.9s
m10s28.3
sm81.9s
m10
2
2
v
ttv
sm2.22v
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Sample Problem 11.3
11 - 17
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).
kva
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).
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Sample Problem 11.3
11 - 18
SOLUTION:• Integrate a = dv/dt = -kv to find v(t).
0 00
lnv t
v
v tdv dva kv k dt ktdt v v
ktevtv 0
• Integrate v(t) = dx/dt to find x(t).
0
0 000 0
1
kt
tx tkt kt
dxv t v edt
dx v e dt x t v ek
ktekvtx 10
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Sample Problem 11.3
11 - 19
• Integrate a = v dv/dx = -kv to find v(x).
kxvv
dxkdvdxkdvkvdxdvva
xv
v
0
00
kxvv 0
• Alternatively,
00 1
vtv
kvtx
kxvv 0
0
0 or v
tveevtv ktkt
ktekvtx 10wit
handthen
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Group Problem Solving
11 - 20
A bowling ball is dropped from a boat so that it strikes the surface of a lake with a speed of 5 m/s. Assuming the ball experiences a downward acceleration of a =10 - 0.01v2 when in the water, determine the velocity of the ball when it strikes the bottom of the lake.
Which integral should you choose?
0 0
v t
v
dv a t dt 0 0
v x
v x
vdv a x dx
0 0
v t
v
dv dta v
0 0
x v
x v
vdvdxa v
(a)
(b)
(c)
(d)
+y
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Concept Question
11 - 21
When will the bowling ball start slowing down?
A bowling ball is dropped from a boat so that it strikes the surface of a lake with a speed of 5 m/s. Assuming the ball experiences a downward acceleration of a =10 - 0.01v2 when in the water, determine the velocity of the ball when it strikes the bottom of the lake. The velocity would have to be high enough for the 0.01 v2 term to be bigger than 10
+y
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Group Problem Solving
11 - 22
SOLUTION:• Determine the proper kinematic relationship to apply (is acceleration a function of time, velocity, or position?• Determine the total distance the car travels in one-half lap• Integrate to determine the velocity after one-half lap
The car starts from rest and accelerates according to the relationship
23 0.001a v It travels around a circular track that has a radius of 200 meters. Calculate the velocity of the car after it has travelled halfway around the track. What is the car’s maximum possible speed?
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Group Problem Solving
11 - 23
200 m
dvv a vdx
0 0
x v
x v
vdvdxa v
Choose the proper kinematic relationship
Given: 23 0.001a v vo = 0, r = 200 m
Find: v after ½ lap
Maximum speed
Acceleration is a function of velocity, and we also can determine distance. Time is not involved in the problem, so we choose:
Determine total distance travelled3.14(200) 628.32 mx r
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Group Problem Solving
11 - 24
0 0
x v
x v
vdvdxa v
Determine the full integral, including limits
628.32
20 0 3 0.001
v vdx dvv
Evaluate the interval and solve for v2
01628.32 ln 3 0.0010.002
vv
2628.32( 0.002) ln 3 0.001 ln 3 0.001(0)v
2ln 3 0.001 1.2566 1.0986= 0.15802v
2 0.158023 0.001v e Take the exponential of each side
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Group Problem Solving
11 - 25
2 0.158023 0.001v e Solve for v
0.158022 3 2146.20.001
ev
46.3268 m/sv
How do you determine the maximum speed the car can reach?23 0.001a v Velocity is a maximum
when acceleration is zeroThis occurs when 20.001 3v
30.001maxv max 54.772 m/sv
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Uniform Rectilinear Motion
11 - 26
For a particle in uniform rectilinear motion, the acceleration is zero and the velocity is constant.
vtxxvtxx
dtvdx
vdtdx
tx
x
00
00
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!
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Uniformly Accelerated Rectilinear Motion
11 - 27
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
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Uniformly Accelerated Rectilinear Motion
11 - 28
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
00
constantv t
v
dv a dv a dt v v atdt
0
210 0 0 0 2
0
x t
x
dx v at dx v at dt x x v t atdt
0 0
2 20 0constant 2
v x
v x
dvv a vdv a dx v v a x xdx
Careful – these only apply to uniformly accelerated rectilinear motion!
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Motion of Several Particles
11 - 29
We may be interested in the motion of several different particles, whose motion may be independent or linked together.
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units
Motion of Several Particles: Relative Motion
11 - 30
• 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. 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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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Sample Problem 11.4
11 - 31
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:• 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.• 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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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Sample Problem 11.4
11 - 32
SOLUTION:• Substitute initial position and velocity and constant acceleration of ball into general equations for uniformly accelerated rectilinear motion.
22
22100
20
sm905.4s
m18m12
sm81.9s
m18
ttattvyy
tatvv
B
B
• Substitute initial position and constant velocity of elevator into equation for uniform rectilinear motion.
ttvyy
v
EE
E
sm2m5
sm2
0
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Sample Problem 11.4
11 - 33
• Write equation for relative position of ball with respect to elevator and solve for zero relative position, i.e., impact. 025905.41812 2 ttty EB
s65.3
smeaningles s39.0
tt
• 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
sm81.19EBv
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units
Motion of Several Particles: Dependent Motion
11 - 34
• 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
dvdt
dvdt
dv
vvvdt
dxdt
dxdt
dx
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Sample Problem 11.5
11 - 35
Pulley D is attached to a collar which is pulled down at 75 mm/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 300 mm/s as it passes L, determine the change in elevation, velocity, and acceleration of block B when block 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.
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Sample Problem 11.5
11 - 36
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.
( )02
mm mm300 225 1.333 ss s
= +
= =
A A Av v a t
t t
220 02
2
2
mm mm300 2 200mm 225s s
A A A A A
A A
v v a x x
a a
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Sample Problem 11.5
11 - 37
• 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. Total length of cable remains constant,
0 0 0
0 0 0
0
2 22 0
200mm 2 100mm 0
A D B A D B
A A D D B B
B B
x x x x x x
x x x x x x
x x
( )0 400mmB Bx x- = -
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Sample Problem 11.5
11 - 38
• Differentiate motion relation twice to develop equations for velocity and acceleration of block B.
2 constant2 0mm mm300 2 75 0s s
A D B
A D B
B
x x xv v v
v
mm mm450 450s sBv = - =
2
2 0mm225 2(0) 0s
A D B
B
a a a
a
2 2m m mm225 225s sBa = - =
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Group Problem Solving
11 - 39
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
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Group Problem Solving
11 - 40
Given: vA= 6 m/s left Find: vBxA
yB
This length is constant no matter how the blocks move
Sketch your system and choose coordinates
Differentiate the constraint equation to get velocity
constnts3 aA Bx y L Define your constraint equation(s)
6 m/s + 3 0Bv
2 m/sB v
Note that as xA gets bigger, yB gets smaller.
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units
Graphical Solution of Rectilinear-Motion Problems
11 - 41
Engineers often collect position, velocity, and acceleration data. Graphical solutions are often useful in analyzing these data.
0
20
40
60
80
100
120
140
160
180
47.76 47.77 47.78 47.79 47.8 47.81Tim e (s)
Acceleration (g)
Acceleration data from a head impact during a round of boxing.
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units
Graphical Solution of Rectilinear-Motion Problems
11 - 42
•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.
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units
Graphical Solution of Rectilinear-Motion Problems
11 - 43
• 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.
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Other Graphical Methods
11 - 44
• Moment-area method to determine particle position at time t directly from the a-t curve:
1
0110
01 curve under areav
vdvtttv
tvxx
using dv = a dt ,
1
011001
v
vdtatttvxx
1
01
v
vdtatt first moment of area
under a-t curve with respect to t = t1 line.
Cttta-ttvxx
centroid of abscissacurve under area 11001
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Other Graphical Methods
11 - 45
• Method to determine particle acceleration from v-x curve:
BCAB
dxdvva
tansubnormal to v-x curve
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units
Curvilinear Motion: Position, Velocity & Acceleration
11 - 46
The softball and the car both undergo curvilinear motion.
•A particle moving along a curve other than a straight line is in curvilinear motion.
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units
Curvilinear Motion: Position, Velocity & Acceleration
11 - 47
• 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 + t,
rr
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units
Curvilinear Motion: Position, Velocity & Acceleration
11 - 48
0lim
t
s dsvt dt
Instantaneous velocity (vector)
Instantaneous speed (scalar)
0lim
t
r drvt dt
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units
Curvilinear Motion: Position, Velocity & Acceleration
11 - 49
0lim
t
v dvat dt
instantaneous acceleration (vector)
• Consider velocity of a particle at time t and velocity at t + t,
v v
• In general, the acceleration vector is not tangent to the particle path and velocity vector.
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Derivatives of Vector Functions
11 - 50
uP
• Let be a vector function of scalar variable u,
uuPuuP
uP
duPd
uu
00limlim
• Derivative of vector sum,
duQd
duPd
duQPd
duPdfP
dudf
duPfd
• Derivative of product of scalar and vector functions,
• Derivative of scalar product and vector product,
duQdPQ
duPd
duQPd
duQdPQ
duPd
duQPd
Delete or put in “bonus” slides
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units
Rectangular Components of Velocity & Acceleration
11 - 51
• When position vector of particle P is given by its rectangular components,kzjyixr
• Velocity vector,
kvjviv
kzjyixkdtdzj
dtdyi
dtdxv
zyx
• Acceleration vector,
kajaia
kzjyixkdt
zdjdt
ydidt
xda
zyx
22
22
22
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units
Rectangular Components of Velocity & Acceleration
11 - 52
• Rectangular components particularly effective when component accelerations can be integrated independently, e.g., motion of a projectile,
00 zagyaxa zyx
with initial conditions, 0,,0 000000 zyx vvvzyx
Integrating twice yields
00
221
0000
zgtyvytvxvgtvvvv
yx
zyyxx
• Motion in horizontal direction is uniform.• Motion in vertical direction is uniformly accelerated.
• Motion of projectile could be replaced by two independent rectilinear motions.
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Tenth Editionin SI Units Sample Problem 11.7
11 - 53
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
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Tenth Editionin SI Units Sample Problem 11.7
11 - 54
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 – uniformly accelerated: Choose positive x to the right as shown
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Tenth Editionin SI Units Sample Problem 11.7
11 - 55
SOLUTION:Horizontal distanceProjectile 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 =
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Tenth Editionin SI Units Concept Quiz
11 - 56
If you fire a projectile from 150 meters above the ground (see Ex Problem 11.7), what launch angle will give you the greatest horizontal distance x?a)A launch angle of 45b)A launch angle less than 45c)A launch angle greater than
45d)It depends on the launch
velocity
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Tenth Editionin SI Units Group Problem Solving
11 - 57
A baseball pitching machine “throws” baseballs with a horizontal velocity v0. If you want the height h to be 1 m, 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 1 m.• Calculate v0=0
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Tenth Editionin SI Units Group Problem Solving
11 - 58
Analyze the motion in the y-direction
Given: x= 12 m, yo = 1.5 m, yf= 1 m. Find: vo
20
1(0) 2fy y t gt
2 210.5 m (9.81 m/s )2 t
213.5 5 2gt
0.3193 st
Analyze the motion in the x-direction
0 00 ( )xx v t v t
012 m ( )(0.3193 s)v
0 37.6 m/s 135.3 km/hv
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Tenth Editionin SI Units
Motion Relative to a Frame in Translation
11 - 59
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.
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Tenth Editionin SI Units
Motion Relative to a Frame in Translation
11 - 60
• 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
. and BA rr
• Vector joining A and B defines the position of B with respect to the moving frame Ax’y’z’ and
ABr
ABAB rrr
• Differentiating twice,ABv velocity of B relative to A.
ABAB vvv
ABa acceleration of B relative to A.
ABAB aaa
• 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.
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Tenth Editionin SI Units Sample Problem 11.9
11 - 61
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
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Tenth Editionin SI Units Sample Problem 11.9
11 - 62
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 mDetermine motion of
Automobile A:
We have uniform motion for A so:
At t = 5 s
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Tenth Editionin SI Units Sample Problem 11.9
11 - 63
SOLUTION:
Determine motion of Automobile B:We have uniform acceleration for B so:
At t = 5 s
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Tenth Editionin SI Units Sample Problem 11.9
11 - 64
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/Ar r r B A B/Av v v B A B/Aa a a
20 5020 50 (m)
B/A
B/A
j i rr j i
6 106 10 (m/s)
B/A
B/A
j i vv j i 2
1.2 01.2 (m/s )
B/A
B/A
j i aa j
Physically, a rider in car A would “see” car B travelling south and west.
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Tenth Editionin SI Units Concept Quiz
11 - 65
If you are sitting in train B looking out the window, it which direction does it appear that train A is moving?
a)
b)
c)
d)25o
25o
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Tangential and Normal Components
11 - 66
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).
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Tangential and Normal Components
11 - 67
• 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
2dv vdt
t na e e
v tv e
• The acceleration can have components in both the en and et directions
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Tenth Editionin SI Units Tangential and Normal Components
11 - 68
• To derive the acceleration vector in tangential and normal components, define the motion of a particle as shown in the figure.
• are tangential unit vectors for the particle path at P and P’. When drawn with respect to the same origin, and
is the angle between them.
tt ee and
ttt eee
dede
eee
e
tn
nnt
t
22sinlimlim
2sin2
00
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Tenth Editionin SI Units Tangential and Normal Components
11 - 69
tevv • With the velocity vector expressed asthe particle acceleration may be written as dt
dsdsd
dedve
dtdv
dtedve
dtdv
dtvda tt
butv
dtdsdsde
ded
nt
After substituting,
22 vadtdvaeve
dtdva ntnt
• The tangential component of acceleration reflects change of speed and the normal component reflects change of direction.• The tangential component may be positive or negative. Normal component always points toward center of path curvature.
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Tenth Editionin SI Units Tangential and Normal Components
11 - 70
22 vadtdvaeve
dtdva ntnt
• Relations for tangential and normal acceleration also apply for particle moving along a space curve.
• The plane containing tangential and normal unit vectors is called the osculating plane.
ntb eee
• The normal to the osculating plane is found from
binormale
normalprincipal e
b
n
• Acceleration has no component along the binormal.
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Tenth Editionin SI Units Sample Problem 11.10
11 - 71
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.
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
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Tenth Editionin SI Units
21.041m/s=a 53.1a =
2
20.833m /stan0.625m /s
n
t
aa
a = =
2 22(25m/s) 0.833m /s750mn
var
= = =
220m/s 25m /saverage 0.625m /s8st tva at
D -= = = = -D
km 1000m 1 h90km/h = 90 25m/sh 1km 3600s72km/h = 20m/s
Sample Problem 11.8
11 - 72
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 25 m/s
2 2 2 2( 0.625) (0.833)n ta a a
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Tangential and Normal Components
11 - 73
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
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Tenth Editionin SI Units Group Problem Solving
11 - 74
SOLUTION:
• Define your coordinate system• Calculate the tangential velocity and tangential acceleration
• Determine overall acceleration magnitude
• Calculate the normal acceleration
The tangential acceleration of the centrifuge cab is given bywhere t is in seconds and at is in m/s2. If the centrifuge starts from fest, determine the total acceleration magnitude of the cab after 10 seconds.
20.5 (m/s )ta t
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Tenth Editionin SI Units Group Problem Solving
11 - 75
In the side view, the tangential direction points into the “page”
Define your coordinate system
8 met
en
en
Top View
Determine the tangential velocity0.5ta t
2 200
0.5 0.25 0.25t ttv t dt t t
20.25 10 25 m/stv
Determine the normal acceleration 2 2
225 78.125 m/s8t
nvar
Determine the total acceleration magnitude
22 2 278.125 + (0.5)(10) mag n ta a a 278.285 m/smaga
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Tenth Editionin SI Units Group Problem Solving
11 - 76
a)The accelerations would remain the same
b)The an would increase and the at would decrease
c)The an and at would both increased)The an would decrease and the at
would increase
Notice that the normal acceleration is much higher than the tangential acceleration. What would happen if, for a given tangential velocity and acceleration, the arm radius was doubled?
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Tenth Editionin SI Units Radial and Transverse Components
11 - 77
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.
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Tenth Editionin SI Units Radial and Transverse Components
11 - 78
• 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 • The particle velocity vector is
• The particle acceleration vector is
rerr
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Radial and Transverse Components
11 - 79
• We can derive the velocity and acceleration relationships by recognizing that the unit vectors change direction.
rr e
dede
ded
dtde
dtd
ded
dted rr
dtde
dtd
ded
dted
r
erer
edtdre
dtdr
dtedre
dtdrer
dtdv
r
rr
rr
• The particle velocity vector is
• Similarly, the particle acceleration vector is
errerrdted
dtdre
dtdre
dtd
dtdr
dted
dtdre
dtrd
edtdre
dtdr
dtda
r
rr
r
2222
22
rerr
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Tenth Editionin SI Units Concept Quiz
11 - 80
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.
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Tenth Editionin SI Units Radial and Transverse Components
11 - 81
• When particle position is given in cylindrical coordinates, it is convenient to express the velocity and acceleration vectors using the unit vectors
. and ,, keeR
• Position vector,kzeRr R
• Velocity vector,kzeReR
dtrdv R
• Acceleration vector,
kzeRReRRdtvda R
22
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Tenth Editionin SI Units Sample Problem 11.12
11 - 82
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.
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Tenth Editionin SI Units Sample Problem 11.12
11 - 83
SOLUTION:• Evaluate time t for = 30o.
s 869.1rad524.0300.152
tt
• Evaluate radial and angular positions, and first and second derivatives at time t.
2
2
sm24.0sm449.024.0m 481.012.09.0
rtr
tr
2
2
srad30.0srad561.030.0
rad524.015.0
tt
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Tenth Editionin SI Units Sample Problem 11.12
11 - 84
• Calculate velocity and acceleration.
rr
r
vvvvv
rvsrv
122 tan
sm270.0srad561.0m481.0m449.0
0.31sm524.0 v
rr
r
aaaaa
rra
rra
122
2
2
2
22
2
tan
sm359.0srad561.0sm449.02srad3.0m481.0
2sm391.0
srad561.0m481.0sm240.0
6.42sm531.0 a
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Vector Mechanics for Engineers: Vector Mechanics for Engineers: DynamicsDynamics
Tenth Editionin SI Units Sample Problem 11.12
11 - 85
• Evaluate acceleration with respect to arm.Motion of collar with respect to arm is rectilinear and defined by coordinate r. 2sm240.0ra OAB
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Tenth Editionin SI Units Group Problem Solving
11 - 86
SOLUTION:
• Define your coordinate system
• Calculate the angular velocity after three revolutions
• Determine overall acceleration magnitude
• Calculate the radial and transverse accelerations
The angular acceleration of the centrifuge arm varies according to
where is measured in radians. If the centrifuge starts from rest, determine the acceleration magnitude after the gondola has travelled two full rotations.
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Tenth Editionin SI Units Group Problem Solving
11 - 87
In the side view, the transverse direction points into the “page”
Define your coordinate system
8 me
er
er
Top View
Determine the angular velocity
Evaluate the integral
Acceleration is a function of position, so use:
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Tenth Editionin SI Units Group Problem Solving
11 - 88
er
Determine the angular velocity
Determine the angular acceleration
Find the radial and transverse accelerations
22 2 2( 63.166)+ 5.0265 mag ra a a 2 63.365 m/smaga
Magnitude:
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Tenth Editionin SI Units Group Problem Solving
11 - 89
You could now have additional acceleration terms. This might give you more control over how quickly the acceleration of the gondola changes (this is known as the G-onset rate).
What would happen if you designed the centrifuge so that the arm could extend from 6 to 10 meters?
r