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2/21/2017
1
Rotational Kinematics and
Dynamics
Uniform Circular MotionUniform Circular Motion
Uniform Circular Motion
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Uniform Circular MotionUniform Circular Motion
• An object that moves at uniform speed in a circle of constant radius is said to be in uniform circular motion.
• Question: Why is uniform circular motion accelerated motion?
• Answer: Although the speed is constant, the velocity is not constant since an object in uniform circular motion is continually changing direction.
Centrifugal ForceCentrifugal Force
• Question: What is centrifugal force?
• Answer: That’s easy. Centrifugal force is the force that flings an object in circular motion outward. Right?
• Wrong! Centrifugal force is a myth!
• There is no outward directed force in circular motion. To explain why this is the case, let’s review Newton’s 1st Law.
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NewtonNewton’’s 1s 1stst Law and carsLaw and cars
•When a car accelerates forward suddenly, you as a passenger feel as if you are flung backward.
• You are in fact NOT flung backward. Your body’s inertia resists acceleration and wants to remain at rest as the car accelerates forward.
•When a car brakes suddenly, you as a passenger feel as if you are flung forward.
• You are NOT flung forward. Your body’s inertia resists acceleration and wants to remain at constant velocity as the car decelerates.
When a car turnsWhen a car turns
• You feel as if you are flung to the outside. You call this apparent, but nonexistent, force “centrifugal force”.
• You are NOT flung to the outside. Your inertia resists the inward acceleration and your body simply wants to keep moving in straight line motion!
• As with all other types of acceleration, your body feels as if it is being flung in the opposite direction of the actual acceleration. The force on your body, and the resulting acceleration, actually point inward.
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Centripetal AccelerationCentripetal Acceleration
• Centripetal (or center-seeking) acceleration points toward the center of the circle and keeps an object moving in circular motion.
• This type of acceleration is at right angles to the velocity.
• This type of acceleration doesn’t speed up an object, or slow it down, it just turns the object.
Centripetal AccelerationCentripetal Acceleration
• ac = v2/r – ac: centripetal acceleration
in m/s2
– v: tangential speed in m/s
– r: radius in meters
v ac
Centripetal acceleration always points toward center of circle!
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Centripetal ForceCentripetal Force
• A force responsible for centripetal acceleration is referred to as a centripetal force.
• Centripetal force is simply mass times centripetal acceleration.
• Fc = m ac
• Fc = m v2 / r – Fc: centripetal force in N
– v: tangential speed in m/s
– r: radius in meters
Fc
Always toward
center of circle!
Any force can be centripetalAny force can be centripetal
• The name “centripetal” can be applied to any force in situations when that force is causing an object to move in a circle.
• You can identify the real force or combination of forces which are causing the centripetal acceleration.
• Any kind of force can act as a centripetal force.
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Highway Curves, Banked and Unbanked
When a car goes around a curve, there must be a
net force towards the center of the circle of which
the curve is an arc. If the road is flat, that force is
supplied by friction.
If the frictional force is
insufficient, the car will tend
to move more nearly in a
straight line, as the skid
marks show.
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As long as the tires do not slip, the friction is static. If the tires do
start to slip, the friction is kinetic, which is bad in two ways:
1. The kinetic frictional force is smaller than the static.
2. The static frictional force can point towards the center of the
circle, but the kinetic frictional force opposes the direction of
motion, making it very difficult to regain control of the car and
continue around the curve.
Highway Curves, Banked and Unbanked
To keep a car on the road in the curve
the static frictional force must be equal
to or greater than the centripetal force.
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Sample problemSample problem • A 1200-kg car rounds a corner of radius r = 45 m. If the coefficient of
static friction between tires and the road is 0.93 and the coefficient of kinetic friction between tires and the road is 0.75, what is the maximum velocity the car can have without skidding?
fs = Fcentripetal
μFN = mv2/r
(.93)(1200kg)(10m/s2) = (1200kg)v2/45m
v2 = 418.5
v = 20.5 m/s
Sample problemSample problem You whirl a 2.0 kg stone in a horizontal circle about your head. The rope attached to the stone is 1.5 m long.
a) What is the tension in the rope? (The rope makes a 10o angle with the horizontal).
b) How fast is the stone moving?
a) FT = 19.6N/sin10 = 113N
b)FC = 113cos10 = 111N
111N = (2.0kg)v2/1.48
82.14 = v2
v = 9.06m/s
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To keep the car on the
track in a vertical
loop, the centripetal
acceleration must
equal the acceleration
due to gravity.
On a frictionless banked curve, the centripetal force is
the horizontal component of the normal force. The
vertical component of the normal force balances the
car’s weight.
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r
vmFF Nc
2
sin
cos
cos
mgF
mgF
N
N
tancF mg
The turns at the Daytona International Speedway have a maximum
radius of 316 m and are steeply banked at 31 degrees. Suppose these
turns were frictionless. At what speed would the cars have to travel
around them to stay on the track?
2
2
2
2
tan
tan
(9.8) tan 31316
1861
43 /
mvmg
r
vg
r
v
v
v m s
tancF mg
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AssignmentAssignment
pg. 155 #1,7,15pg. 155 #1,7,15
pg. 156pg. 156--158 158
#5,9,11,13,21,25,28,42,44,48#5,9,11,13,21,25,28,42,44,48
Rotational Kinematics and
Dynamics
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In the simplest kind of rotation,
points on a rigid object move on
circular paths around an axis of
rotation.
The angle through which the
object rotates is called the
angular displacement.
o
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rs
r
s
Radius
length Arcradians)(in
For a full revolution:
360rad 2 rad 22
r
r
Example 1 Adjacent Synchronous Satellites
Synchronous satellites are put into
an orbit whose radius is 4.23×107m.
If the angular separation of the two
satellites is 2.00 degrees, find the
arc length that separates them.
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8.1 Rotational Motion and Angular Displacement
rad 0349.0deg360
rad 2deg00.2
miles) (920 m1048.1
rad 0349.0m1023.4
6
7
rs
r
s
Radius
length Arcradians)(in
DEFINITION OF AVERAGE ANGULAR VELOCITY
ttt o
o
SI Unit of Angular Velocity: radian per second (rad/s)
timeElapsed
ntdisplacemeAngular locity angular ve Average
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Example 3 Gymnast on a High Bar
A gymnast on a high bar swings through two revolutions in a
time of 1.90 s.
Find the average angular velocity of the gymnast.
rad 6.12rev 1
rad 2rev 00.2
srad63.6s 90.1
rad 6.12
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Changing angular velocity means that an angular acceleration
is occurring.
DEFINITION OF AVERAGE ANGULAR ACCELERATION
ttt o
o
timeElapsed
locityangular vein Change on acceleratiangular Average
SI Unit of Angular acceleration: radian per second squared (rad/s2)
Example 4 A Jet Revving Its Engines
As seen from the front of the
engine, the fan blades are
rotating with an angular
speed of -110 rad/s. As the
plane takes off, the angular
velocity of the blades reaches
-330 rad/s in a time of 14 s.
Find the angular acceleration, assuming it to
be constant.
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2srad16s 14
srad110srad330
atvv o
tvvx o 21
axvv o 222
2
21 attvx o
Five kinematic variables:
1. displacement, x
2. acceleration (constant), a
3. final velocity (at time t), v
4. initial velocity, vo
5. elapsed time, t
Recall the equations of kinematics for constant acceleration.
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to
to 21
222 o
2
21 tto
The equations of rotational kinematics for constant
angular acceleration:
ANGULAR DISPLACEMENT
ANGULAR VELOCITY
ANGULAR ACCELERATION
TIME
8.3 The Equations of Rotational Kinematics
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8.3 The Equations of Rotational Kinematics
Example 5 Blending with a Blender
The blades are whirling with an
angular velocity of +375 rad/s when
the “puree” button is pushed in.
When the “blend” button is pushed,
the blades accelerate and reach a
greater angular velocity after the
blades have rotated through an
angular displacement of +44.0 rad.
The angular acceleration has a
constant value of +1740 rad/s2.
Find the final angular velocity of the blades.
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θ α ω ωo t
+44.0 rad +1740 rad/s2 ? +375 rad/s
222 o
srad542rad0.44srad17402srad375
2
22
2
o
velocityl tangentiaTv
speed l tangentiaTv
rad/s)in ( rvT
)rad/sin ( 2raT
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21
Example 6 A Helicopter Blade
A helicopter blade has an angular speed of 6.50 rev/s and an
angular acceleration of 1.30 rev/s2.
For point 1 on the blade, find
the magnitude of (a) the
tangential speed and (b) the
tangential acceleration.
8.4 Angular Variables and Tangential Variables
srad 8.40rev 1
rad 2
s
rev 50.6
sm122srad8.40m 3.00 rvT
22 sm5.24srad17.8m 3.00 raT
2
2srad 17.8
rev 1
rad 2
s
rev 30.1
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rad/s)in ( 2
2
ra
r
va
c
Tc
Example 7 A Discus Thrower
Starting from rest, the thrower
accelerates the discus to a final
angular speed of +15.0 rad/s in
a time of 0.270 s before releasing it.
During the acceleration, the discus
moves in a circular arc of radius
0.810 m.
Find the magnitude of the total
acceleration.
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2
22
sm182
srad0.15m 810.0
rac
2sm0.45
s 0.270
srad0.15m 810.0
t
ω-ωrra o
T
22222 sm187sm0.45sm182 cT aaa
Example 8 An Accelerating Car
Starting from rest, the car accelerates
for 20.0 s with a constant linear
acceleration of 0.800 m/s2. The
radius of the tires is 0.330 m.
What is the angle through which
each wheel has rotated?
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2
21 tto
θ α ω ωo t ? -2.42 rad/s2 0 rad/s 20.0 s
22
srad42.2m 0.330
sm800.0
r
a
rad 484s 0.20srad42.2222
21
Assignment pg. 241-244
#4,12,22,30,34,40,44,51,58
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Rotational Dynamics
In pure translational motion, all points on an
object travel on parallel paths.
The most general motion is a combination of
translation and rotation.
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According to Newton’s second law, a net force
causes an object to have an acceleration.
What causes an object to have an angular
acceleration?
TORQUE
The amount of torque depends on where and in
what direction the force is applied, as well as the
location of the axis of rotation.
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The amount of torque depends on where and in
what direction the force is applied, as well as the
location of the axis of rotation.
9.1 The Action of Forces and Torques on Rigid Objects
DEFINITION OF TORQUE
Magnitude of Torque = (Magnitude of the force) x (Lever arm)
sinrFDirection: The torque is positive when the force tends to produce
a counterclockwise rotation about the axis.
SI Unit of Torque: newton x meter (N·m)
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Example 2 The Achilles Tendon
The tendon exerts a force of magnitude
790 N. Determine the torque (magnitude
and direction) of this force about the
ankle joint.
790 N
F
o2
2
m)cos55106.3(
m106.355cos
mN 15
55cosm106.3N 720 2
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If a rigid body is in equilibrium, neither its linear motion
nor its rotational motion changes.
0 xF 0yF 0
0 yx aa 0
EQUILIBRIUM OF A RIGID BODY
A rigid body is in equilibrium if it has zero translational
acceleration and zero angular acceleration. In equilibrium,
the sum of the externally applied forces is zero, and the
sum of the externally applied torques is zero.
0 0yF0 xF
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Reasoning Strategy 1. Select the object to which the equations for equilibrium are to be applied.
2. Draw a free-body diagram that shows all of the external forces acting on the
object.
3. Choose a convenient set of x, y axes and resolve all forces into components
that lie along these axes.
4. Apply the equations that specify the balance of forces at equilibrium. (Set the
net force in the x and y directions equal to zero.)
5. Select a convenient axis of rotation. Set the sum of the torques about this
axis equal to zero.
6. Solve the equations for the desired unknown quantities.
Example 3 A Diving Board
A woman whose weight is 530 N is
poised at the right end of a diving board
with length 3.90 m. The board has
negligible weight and is supported by
a fulcrum 1.40 m away from the left
end.
Find the forces that the bolt and the
fulcrum exert on the board.
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022 WWF
N 1480
m 1.40
m 90.3N 5302 F
2
2
WWF
021 WFFFy
0N 530N 14801 F
N 9501 F
Example 5 Bodybuilding
The arm is horizontal and weighs 31.0 N. The deltoid muscle can
supply 1840 N of force. What is the weight of the heaviest dumbbell
he can hold?
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0 Mddaa MWW
0.13sinm 150.0M
N 1.86
m 620.0
0.13sinm 150.0N 1840m 280.0N 0.31
d
Maad
MWW
DEFINITION OF CENTER OF GRAVITY
The center of gravity of a rigid body is the point at
which its weight can be considered to act when the
torque due to the weight is being calculated.
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When an object has a symmetrical shape and its weight is
distributed uniformly, the center of gravity lies at its
geometrical center.
21
2211
WW
xWxWxcg
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Example 6 The Center of Gravity of an Arm
The horizontal arm is composed
of three parts: the upper arm (17 N),
the lower arm (11 N), and the hand
(4.2 N).
Find the center of gravity of the
arm relative to the shoulder joint.
21
2211
WW
xWxWxcg
m 28.0
N 2.4N 11N 17
m 61.0N 2.4m 38.0N 11m 13.0N 17
cg
cg
x
x
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A uniform beam is supported by a stout piece of line as shown.
The beam weighs 175 N. The cable makes an angle of 75.0°
as shown. Find (a) the tension in the cable and (b) the force
exerted on the end of the beam by the wall.
A uniform beam is supported by a stout piece of line as shown.
The beam weighs 175 N. The cable makes an angle of 75.0°
as shown. Find (a) the tension in the cable and (b) the force
exerted on the end of the beam by the wall.
NF
NNNcompy
NNcompxb
NT
mNmTa
o
o
o
6.905.874.23
5.8775sin)6.90(175:
4.2375cos)6.90(:)
6.90
)2)(175(75sin)00.4()
22
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TT maF
raT
rFT
2mr
Moment of Inertia, I
ROTATIONAL ANALOG OF NEWTON’S SECOND LAW FOR
A RIGID BODY ROTATING ABOUT A FIXED AXIS
onaccelerati
Angular
inertia
ofMoment torqueexternalNet
I
2mrIRequirement: Angular acceleration
must be expressed in radians/s2.
2/21/2017
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2
2
18000
)/1.0(1800
mkgI
sradINm
I
A certain merry-go-round is accelerated uniformly from rest and attains
an angular speed of 1.5 rad/s in the first 15 seconds. If the net applied
torque is 1800 N m, what is the moment of inertia of the merry-go-
round?
sec7.4
0.22
)/571.0(2
1028.6
2
1
/571.0
)175(100
2
22
2
2
2
t
t
tsradrad
tt
srad
kgmNm
I
o
A 100-N m torque acts on a wheel with a moment of inertia
175 kg × m2. If the wheel starts from rest, how long will it
take the wheel to make one revolution?
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2
2
2
385.0
)/0.10(85.3
/0.1000.3
/0.30
85.311.1)0.15)(330.0(
mkgI
sradINm
I
srads
srad
t
NmNmNmRF frT
Example 12 Hoisting a Crate
The combined moment of inertia of the dual pulley is 50.0 kg·m2. The
crate weighs 4420 N. A tension of 2150 N is maintained in the cable
attached to the motor. Find the angular acceleration of the dual pulley.
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ITT 2211 yy mamgTF
2
equal
ymamgT 22ya
2ya
ImamgT y 211
ImmgT 2211
2
22
2
2
2
211
srad3.6m 200.0kg 451mkg 46.0
m 200.0sm80.9kg 451m 600.0N 2150
mI
mgT
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DEFINITION OF ROTATIONAL WORK
The rotational work done by a constant torque in turning
an object through an angle is:
RW
Requirement: The angle must
be expressed in radians.
SI Unit of Rotational Work: joule (J)
2
21 IKER
DEFINITION OF ROTATIONAL KINETIC ENERGY
The rotational kinetic energy of a rigid rotating object is:
Requirement: The angular speed must
be expressed in rad/s.
SI Unit of Rotational Kinetic Energy: joule (J)
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DEFINITION OF ANGULAR MOMENTUM
The angular momentum L of a body rotating about a fixed axis
is the product of the body’s moment of inertia and its angular
velocity with respect to that axis:
IL
Requirement: The angular speed must
be expressed in rad/s.
SI Unit of Angular Momentum: kg·m2/s
Example 15 A Satellite in an Elliptical Orbit
An artificial satellite is placed in an
elliptical orbit about the earth. Its point
of closest approach is 8.37x106m
from the center of the earth, and
its point of greatest distance is
25.1x106m from the center of
the earth.
The speed of the satellite at the
perigee is 8450 m/s. Find the speed
at the apogee.
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9.6 Angular Momentum
IL
angular momentum conservation
PPAA II
rvmrI 2
P
PP
A
AA
r
vmr
r
vmr 22
9.6 Angular Momentum
PPAA vrvr
sm2820
m 1025.1
sm8450m 1037.86
6
A
PPA
r
vrv
P
PP
A
AA
r
vmr
r
vmr 22