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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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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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Assignment pg. 277-279

#1,5,9,12,18,19,25

Rotational Dynamics

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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.

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kgM

sradmMN

I

3.33

)/120)()15(.2

1(45 22

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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

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Assignment pg. 280-282

#31,32,45,48,49,57,61