Physics 101Lecture 10Rotation

Assist. Prof. Dr. Ali ÖVGÜNEMU Physics Department

www.aovgun.com

December 21, 2018

Rotational Motionq Angular Position and Radiansq Angular Velocityq Angular Accelerationq Rigid Object under Constant

Angular Acceleration q Angular and Translational

Quantitiesq Rotational Kinetic Energyq Moments of Inertia

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Angle and Radianq What is the circumference S ?

q q can be defined as the arc length s along a circle divided by the radius r:

q q is a pure number, but commonly is given the artificial unit, radian (“rad”)

rqs

rq =

rs )2( p=rs

=p2

s

q Whenever using rotational equations, you must use angles expressed in radians

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Conversionsq Comparing degrees and radians

q Converting from degrees to radians

q Converting from radians to degrees

!180)( =radp

3601 57.32

radp°

= = °

( ) ( )180

rad degreespq q=°

!360)(2 =radp

)(180)(deg radrees qp

q!

=

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Rigid ObjectqA rigid object is one that is nondeformable

n The relative locations of all particles making up the object remain constant

n All real objects are deformable to some extent, but the rigid object model is very useful in many situations where the deformation is negligible

qThis simplification allows analysis of the motion of an extended object

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Angular Positionq Axis of rotation is the center of the

discq Choose a fixed reference lineq Point P is at a fixed distance r from

the originq As the particle moves, the only

coordinate that changes is qq As the particle moves through q, it

moves though an arc length s.q The angle q, measured in radians,

is called the angular position.

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Angular Displacementq The angular displacement

is defined as the angle the object rotates through during some time interval

q SI unit: radian (rad)q This is the angle that the

reference line of length rsweeps out

f iq q qD = -

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Average and Instantaneous Angular Speed

q The average angular speed, ωavg, of a rotating rigid object is the ratio of the angular displacement to the time interval

q The instantaneous angular speed is defined as the limit of the average speed as the time interval approaches zero

q SI unit: radian per second (rad/s)q Angular speed positive if rotating in counterclockwiseq Angular speed will be negative if rotating in clockwise

f iavg

f it t tq q qw - D

= =- D

lim0 t

dt dtq qw D ®

Dº =

D

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Average Angular Acceleration

q The average angular acceleration, a, of an object is defined as the ratio of the change in the angular speed to the time it takes for the object to undergo the change:

f iavg

f it t tw w wa - D

= =- D

t = ti: wi t = tf: wf

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Instantaneous Angular Acceleration

q The instantaneous angular acceleration is defined as the limit of the average angular acceleration as the time goes to 0

q SI Units of angular acceleration: rad/s² q Positive angular acceleration is counterclockwise (RH

rule – curl your fingers in the direction of motion).n if an object rotating counterclockwise is speeding upn if an object rotating clockwise is slowing down

q Negative angular acceleration is clockwise.n if an object rotating counterclockwise is slowing downn if an object rotating clockwise is speeding up

lim0 t

dt dtw wa D ®

Dº =

D

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Rotational Kinematicsq A number of parallels exist between the equations

for rotational motion and those for linear motion.

q Under constant angular acceleration, we can describe the motion of the rigid object using a set of kinematic equationsn These are similar to the kinematic equations for

linear motionn The rotational equations have the same

mathematical form as the linear equations

tx

ttxx

vif

ifavg D

D=

-

-= f i

avgf it t tq q qw - D

= =- D

Analogy with Linear Kinematics

qStart with angular acceleration:q Integrate once:q Integrate again:

qJust substitute symbols, and all of the old equations apply:

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ddtwa =

f idt tw a w a= = +ò( ) 21

2f i i it dt t tq w a q w a= + = + +ò21

2f i ix x v t at= + +

f iv v at= +

xva

qwa

ÞÞÞ

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Comparison Between Rotational and Linear

Equations

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Ex:1 A Rotating Wheel

stsradsradi

0.2/ 5.3/ 0.2

2

==

=

a

w

q A wheel rotates with a constant angular acceleration of 3.5 rad/s2. If the angular speed of the wheel is 2.0 rad/s at t = 0

(a) through what angle does the wheel rotate between t = 0 and t = 2.0 s? Given your answer in radians and in

revolutions.(b) What is the angular speed of the wheel at t = 2.0 s?

??

=

=-

f

if

w

qq

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Relationship Between Angular and Linear Quantities

q Every point on the rotating object has the same angular motion

q Every point on the rotating object does not have the same linear motion

q Displacement

q Speeds

q Accelerations

s rq=

v rw=a ra=

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Speed Comparisonq The linear velocity is always

tangent to the circular pathn Called the tangential velocity

q The magnitude is defined by the tangential speed

ts

rtrs

t DD

=DD

=DD 1q

rsD

=Dq

ww rvorrv

==

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Acceleration Comparisonq The tangential

acceleration is the derivative of the tangential velocity

aw rt

rtv

=DD

=DD

wD=D rv

arat =

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Speed and Acceleration Note

q All points on the rigid object will have the same angular speed, but not the same tangential speed

q All points on the rigid object will have the same angular acceleration, but not the same tangential acceleration

q The tangential quantities depend on r, and r is not the same for all points on the object

ww rvorrv

== arat =

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Centripetal Accelerationq An object traveling in a circle,

even though it moves with a constant speed, will have an accelerationn Therefore, each point on a

rotating rigid object will experience a centripetal acceleration

222 )( ww r

rr

rvar ===

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Resultant Accelerationq The tangential component of

the acceleration is due to changing speed

q The centripetal component of the acceleration is due to changing direction

q Total acceleration can be found from these components

2 2 2 2 2 4 2 4t ra a a r r ra w a w= + = + = +

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Rotational Kinetic Energyq An object rotating about z axis with

an angular speed, ω, has rotational kinetic energy

q Each particle has a kinetic energy ofKi = ½ mivi

2

q Since the tangential velocity depends on the distance, r, from the axis of rotation, we can substitute

vi = wri => Ki = ½ miw2ri2

December 21, 2018

Rotational Kinetic Energyq There is an analogy between the kinetic

energies associated with linear motion (K = ½ mv2) and the kinetic energy associated with rotational motion (KR = ½ Iw2)

q Rotational kinetic energy is not a new type of energy, the form is different because it is applied to a rotating object

q Units of rotational kinetic energy are Joules (J)

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Work-Energy Theorem for pure Translational motion

q The work-energy theorem tells us

q Kinetic energy is for point mass only, ignoring rotation.

q Work

q Power

òò®®

×== sdFdWWnet

22

21

21

ififnet mvmvKEKEKEW -=-=D=

®®®

®

×=×== vFdtsdF

dtdWP

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Mechanical Energy Conservation

q Energy conservation

q When Wnc = 0,

q The total mechanical energy is conserved and remains the same at all times

q Remember, this is for conservative forces, no dissipative forces such as friction can be present

f f i iK U U K+ = +

ffii mgymvmgymv +=+ 22

21

21

ncW K U= D +D

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Total Energy of a Systemq A ball is rolling down a rampq Described by three types of energy

n Gravitational potential energy

n Translational kinetic energy

n Rotational kinetic energy

q Total energy of a system 2 21 12 2CME Mv Mgh Iw= + +

212rK Iw=

U Mgh=21

2t CMK Mv=

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Work done by a pure rotation

q Apply force F to mass at point r, causing rotation-only about axis

q Find the work done by F applied to the object at P as it rotates through an infinitesimal distance ds

q Only transverse component of F does work – the same component that contributes to torque qtddW =

cos(90 )sin sin

dW F d s F dsF ds Fr d

jj j q

® ®

= × = -= =

!

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Work-Kinetic Theorem pure rotation

q As object rotates from qi to qf , work done by the torque

q I is constant for rigid object

q Powertwqt ===

dtd

dtdWP

òòòòò =====f

i

f

i

f

i

f

i

f

i

dIddtdIdIddWW

q

q

q

q

q

q

q

q

q

q

wwqwqaqt

22

21

21

if IIdIdIWf

i

f

i

wwwwwwq

q

q

q

-=== òò

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q An motor attached to a grindstone exerts a constant torque of 10 N-m. The moment of inertia of the grindstone is I = 2 kg-m2. The system starts from rest.n Find the kinetic energy after 8 s

n Find the work done by the motor during this time

n Find the average power delivered by the motor

n Find the instantaneous power at t = 8 s

1600 200 watts8avg

dWPdt

= = =

2 21 1600 40 rad/s 5 rad/s2f f f iK I J t

Itw w w a a= = Ü = + = Ü = =

JdW if

f

i

160016010)( =´=-== ò qqtqtq

q21( ) 160 rad

2f i it tq q w a- = + = 2 21 1 1600 J2 2f i f iW K K I Iw w= - = - =

10 40 400 wattsP tw= = ´ =

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Work-Energy Theoremq For pure translation

q For pure rotation

q Rolling: pure rotation + pure translation

2 2, ,

1 12 2net cm cm f cm i f iW K K K mv mv= D = - = -

2 2, ,

1 12 2net rot rot f rot i f iW K K K I Iw w= D = - = -

, , , ,

2 2 2 2

( ) ( )

1 1 1 12 2 2 2

net total rot f cm f rot i cm i

f f i i

W K K K K K

I mv I mvw w

= D = + - +

æ ö æ ö= + - +ç ÷ ç ÷è ø è ø

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Energy Conservationq Energy conservation

q When Wnc = 0,

q The total mechanical energy is conserved and remains the same at all times

q Remember, this is for conservative forces, no dissipative forces such as friction can be present

, , , ,rot f cm f f rot i cm i iK K U K K U+ + = + +

fffiii mgymvImgymvI ++=++ 2222

21

21

21

21 ww

nc totalW K U= D +D

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Total Energy of a Rolling System

q A ball is rolling down a rampq Described by three types of energy

n Gravitational potential energy

n Translational kinetic energy

n Rotational kinetic energy

q Total energy of a system 22

21

21 wIMghMvE ++=

212rK Iw=

U Mgh=

212tK Mv=

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Problem Solving Hintsq Choose two points of interest

n One where all the necessary information is givenn The other where information is desired

q Identify the conservative and non-conservative forces

q Write the general equation for the Work-Energy theorem if there are non-conservative forcesn Use Conservation of Energy if there are no non-

conservative forcesq Use v = rw to combine termsq Solve for the unknown

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A Ball Rolling Down an Incline

q A ball of mass M and radius R starts from rest at a height of h and rolls down a 30° slope, what is the linear speed of the ball when it leaves the incline? Assume that the ball rolls without slipping.

22

210

2100 ff IMvMgh w++=++

2222

21

21

21

21

fffiii ImgymvImgymv ww ++=++

222

222

51

21

52

21

21

fff

f MvMvRv

MRMvMgh +=+=

2

52MRI =

Rv f

f =w

2/1)710( ghv f =

December 21, 2018

Rotational Work and Energyq A ball rolls without slipping down incline A,

starting from rest. At the same time, a box starts from rest and slides down incline B, which is identical to incline A except that it is frictionless. Which arrives at the bottom first?

q Ball rolling:

q Box sliding

2222

21

21

21

21

fffiii ImgymvImgymv ww ++=++

2 21 12 2f fmgh mv Iw= +

ffii mgymvmgymv +=+ 22

21

21

21sliding: 2 fmgh mv=

2 2 2 21 1 2 7( / )2 2 5 10f f fmv mR v R mvæ ö= + =ç ÷

è ø

27rolling: 10 fmgh mv=

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Blocks and Pulleyq Two blocks having different masses m1 and

m2 are connected by a string passing over a pulley. The pulley has a radius R and moment of inertia I about its axis of rotation. The string does not slip on the pulley, and the system is released from rest.

q Find the translational speeds of the blocks after block 2 descends through a distance h.

q Find the angular speed of the pulley at that time.

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q Find the translational speeds of the blocks after block 2 descends through a distance h.

q Find the angular speed of the pulley at that time.

ghmghmvRImm f 12

2221 )(

21

-=++

000)()21

21

21( 21

222

21 ++=-+++ ghmghmIvmvm fff w

, , , ,rot f cm f f rot i cm i iK K U K K U+ + = + +

2/1

221

12

/)(2

úû

ùêë

é++

-=

RImmghmmv f

2/1

221

12

/)(21

úû

ùêë

é++

-==

RImmghmm

RRv f

fw

qExample:

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qExample:q A roll of toilet paper is held by the first piece and

allowed to unfurl as shown in the diagram to the right. The roll has an outer radius R = 6.0 cm, an inner radius r = 1.8 cm, a mass m = 200 g, and falls a distance s = 3.0 m. Assuming the outer diameter of the roll does not change significantly during the fall, determine…the final angular speed of the roll

q the final translational speed of rollq the angular acceleration of the rollq the translational acceleration of the rollq the tension in the sheets

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qExample:q The top shown below consists of a cylindrical spindle of negligible mass attached

to a conical base of mass m = 0.50 kg. The radius of the spindle is r = 1.2 cm and the radius of the cone is R = 10 cm. A string is wound around the spindle. The top is thrown forward with an initial speed of v0 = 10 m/s while at the same time the string is yanked backward. The top moves forward a distance s = 2.5 m, then stops and spins in place.

q Using energy considerations determine…q the tension T in the string

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qExample:q Calculating Helicopter Energiesq A typical small rescue helicopter has four blades: Each is 4.00 m

long and has a mass of 50.0 kg. The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. (a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter when it flies at 20.0 m/s, and compare it with the rotational energy in the blades.

q (a) Sketch of a four-blade helicopter. (b) A water rescue operation featuring a helicopter from the Auckland Westpac Rescue Helicopter Service.

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qExample:q Energy in a Boomerangq A person hurls a boomerang into the air with a velocity

of 30.0 m/s at an angle of 40.0° with respect to the horizontal It has a mass of 1.0 kg and is rotating at 10.0 rev/s. The moment of inertia of the boomerang is given as I = 1/12 mL2 where L = 0.7 m. (a) What is the total energy of the boomerang when it leaves the hand? (b) How high does the boomerang go from the elevation of the hand, neglecting air resistance?

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Moment of Inertia of Point Mass

q For a single particle, the definition of moment of inertia is

n m is the mass of the single particlen r is the rotational radius

q SI units of moment of inertia are kg.m2

q Moment of inertia and mass of an object are different quantities

q It depends on both the quantity of matter and its distribution (through the r2 term)

2mrI =

December 21, 2018

Moment of Inertia of Point Mass

q For a composite particle, the definition of moment of inertia is

n mi is the mass of the ith single particlen ri is the rotational radius of ith particle

q SI units of moment of inertia are kg.m2

q Consider an unusual baton made up of four sphere fastened to the ends of very light rods

q Find I about an axis perpendicular to the page and passing through the point O where the rods cross

...244

233

222

211

2 ++++=å= rmrmrmrmrmI ii

2222222 22 mbMaMambMambrmI ii +=+++=å=

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The Baton Twirlerq Consider an unusual baton made

up of four sphere fastened to the ends of very light rods. Each rod is 1.0m long (a = b = 1.0 m). M = 0.3 kg and m = 0.2 kg.

q (a) Find I about an axis perpendicular to the page and passing through the point where the rods cross. Find KR if angular speed is �

q (b) The majorette tries spinning her strange baton about the axis y, calculate I of the baton about this axis and KR if angular speed is �

December 21, 2018

Moment of Inertia of Extended Objects

q Divided the extended objects into many small volume elements, each of mass Dmi

q We can rewrite the expression for I in terms of Dm

q Consider a small volume such that dm = r dV. Then

q If r is constant, the integral can be evaluated with known geometry, otherwise its variation with position must be known

2 2

0limi

i im iI r m r dm

D ®= D =å ò

2I r dVr= ò

Densitiesq You know the density (volume density) as

mass/unit volumen r = M/V = dm/dV => dm = rdV

q We can define other densities such as surface density (mass/unit area)n s = M/A = dm/dA => dm = sdV

q Or linear density (mass/unit length)n l = M/L = dm/dx => dm = ldV

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Moment of Inertia of a Uniform Rigid Rod

q The shaded area has a mass n dm = l dx

q Then the moment of inertia is

/22 2

/2

2112

L

y L

MI r dm x dxL

I ML

-= =

=

ò ò

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Moment of Inertia for some other common shapes

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P1:

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P2:

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P3:

P4: