PHYS 1441 – Section 002 Lecture #20yu/teaching/spring13-1441-002/... · Wednesday, April 10, 2013 PHYS 1441-002, Spring 2013 Dr. Jaehoon Yu 14 Starting from rest, the car accelerates

PHYS 1441 – Section 002 Lecture #20

Wednesday, April 10, 2013 Dr. Jaehoon Yu

•  Equations of Rotational Kinematics •  Relationship Between Angular and Linear

Quantities •  Rolling Motion of a Rigid Body •  Torque •  Moment of Inertia

Announcements •  Second non-comp term exam

–  Date and time: 4:00pm, Wednesday, April 17 in class –  Coverage: CH6.1 through what we finish Monday, April 15 –  This exam could replace the first term exam if better

•  Special colloquium for 15 point extra credit –  Wednesday, April 24, University Hall RM116 –  Class will be substituted by this colloquium –  Dr. Ketevi Assamagan from Brookhaven National Laboratory on

Higgs Discovery in ATLAS –  Please mark your calendars!!

Wednesday, April 10, 2013

2 PHYS 1441-002, Spring 2013 Dr. Jaehoon Yu


PHYS 1441-002, Spring 2013 Dr. Jaehoon Yu

3

=fω

Rotational Kinematics The first type of motion we have learned in linear kinematics was under a constant acceleration. We will learn about the rotational motion under constant angular acceleration (α), because these are the simplest motions in both cases.

Just like the case in linear motion, one can obtain Angular velocity under constant angular acceleration:

Angular displacement under constant angular acceleration: =fθ

One can also obtain =2fω

0ω +

0θ +

( )20 02 fω α θ θ+ −

tα

0tω + 212tα

v =Linear kinematics

Linear kinematics fx = 210 2ox v t at+ +

Linear kinematics 2fv = ( )2 2o f iv a x x+ −

ov at+



4

Problem Solving Strategy •  Visualize the problem by drawing a picture. •  Write down the values that are given for any of the

five kinematic variables and convert them to SI units. –  Remember that the unit of the angle must be in radians!!

•  Verify that the information contains values for at least three of the five kinematic variables. Select the appropriate equation.

•  When the motion is divided into segments, remember that the final angular velocity of one segment is the initial velocity for the next.

•  Keep in mind that there may be two possible answers to a kinematics problem.



5

Example for Rotational Kinematics A wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the angular speed of the wheel is 2.00 rad/s at ti=0, a) through what angle does the wheel rotate in 2.00s?

θ f −θi

Using the angular displacement formula in the previous slide, one gets

=ωt

= 2.00 × 2.00 + 123.50 × 2.00( )2 = 11.0rad

= 11.02π

rev.= 1.75rev.

+ 12αt2



6

Example for Rotational Kinematics cnt’d What is the angular speed at t=2.00s?

ω f =ω i +αtUsing the angular speed and acceleration relationship

Find the angle through which the wheel rotates between t=2.00 s and t=3.00 s.

2tθ = =

srad /00.900.250.300.2 =×+=

3tθ = =

Δθ = θ3 −θ2 = 10.8rad .72.1.28.10 revrev ==π

if θθ − =ωt + 12αt 2Using the angular kinematic formula

At t=2.00s

At t=3.00s

Angular displacement

2.00× 2.00 + 123.50× 2.00 =11.0rad

2.00× 3.00 + 123.50× 3.00( )2 = 21.8rad



7

The blade is whirling with an angular velocity of +375 rad/s when the “puree” button is pushed in. When the “blend” button is pushed, the blade accelerates and reaches a greater angular velocity after the blade has 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 blade.

Ex. Blending with a Blender

θ α ω ωo t

2 2 2oω ω αθ= +ω =Which kinematic eq?

± ωo2 + 2αθ

= ± 375rad s( )2 + 2 1740rad s2( ) 44.0rad( ) = ±542rad s

Which sign? ω = 542rad s+ Why? Because the blade is accelerating in counter-clockwise!

+44.0rad +375rad/s ? +1740rad/s2



8

Relationship Between Angular and Linear Quantities What do we know about a rigid object that rotates

about a fixed axis of rotation?

When a point rotates, it has both the linear and angular components in its motion. What is the linear component of the motion you see?

v

Every particle (or masslet) in the object moves in a circle centered at the same axis of rotation with the same angular velocity.

Linear velocity along the tangential direction. How do we related this linear component of the motion with angular component?

l = rθThe arc-length is So the tangential speed v is

What does this relationship tell you about the tangential speed of the points in the object and their angular speed?

Although every particle in the object has the same angular speed, its tangential speed differs and is proportional to its distance from the axis of rotation. The farther away the particle is from the center of rotation, the higher the tangential speed.

The direction of ω follows the right-hand rule.

= ΔlΔt

=Δ rθ( )Δt

= r ΔθΔt

= rω



9

Is the lion faster than a horse? A rotating carousel has one child sitting on a horse near the outer edge and another child on a lion halfway out from the center. (a) Which child has the greater linear speed? (b) Which child has the greater angular speed?

(a)  Linear speed is the distance traveled divided by the time interval. So the child sitting at the outer edge travels more distance within the given time than the child sitting closer to the center. Thus, the horse is faster than the lion.

(b) Angular speed is the angle traveled divided by the time interval. The angle both the children travel in the given time interval is the same. Thus, both the horse and the lion have the same angular speed.



10

How about the acceleration?

vt =

Two How many different linear acceleration components do you see in a circular motion and what are they?

Total linear acceleration is

Since the tangential speed v is

What does this relationship tell you?

Although every particle in the object has the same angular acceleration, its tangential acceleration differs proportional to its distance from the axis of rotation.

Tangential, at, and the radial acceleration, ar.

atThe magnitude of tangential acceleration at is

The radial or centripetal acceleration ar is raWhat does this tell you?

The father away the particle is from the rotation axis, the more radial acceleration it receives. In other words, it receives more centripetal force.

a

=

vtf − vt0

Δt =

rω f − rω0

Δt

rv2= =

rω( )2r

2ωr=

= at2 + ar

2 = rα( )2 + rω 2( )2 = r α 2 +ω 4

rω

= r

ω f −ω0

Δt= rα



11

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.

Ex. A Helicopter Blade

ω =

Tv =

Ta =

α =

rev 2 rad6.50 s 1 rev

π⎛ ⎞⎛ ⎞ =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠40.8 rad s

rϖ = ( )( )3.00 m 40.8rad s 122m s=

2

rev 2 rad1.30 s 1 rev

π⎛ ⎞⎛ ⎞ =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠28.17 rad s

( )( )23.00 m 8.17rad s = 224.5m srα =



12

Rolling Motion of a Rigid Body What is a rolling motion?

To simplify the discussion, let’s make a few assumptions

Let’s consider a cylinder rolling on a flat surface, without slipping.

A more generalized case of a motion where the rotational axis moves together with an object

Under what condition does this “Pure Rolling” happen?

The total linear distance the CM of the cylinder moved is

Thus the linear speed of the CM is

A rotational motion about a moving axis 1.  Limit our discussion on very symmetric

objects, such as cylinders, spheres, etc 2.  The object rolls on a flat surface

R θ s

s=Rθ

s =

vCM =

ΔsΔt

The condition for a “Pure Rolling motion”

RtθΔ=Δ ωR=

Rθ



13

More Rolling Motion of a Rigid Body

As we learned in rotational motion, all points in a rigid body moves at the same angular speed but at different linear speeds.

At any given time, the point that comes to P has 0 linear speed while the point at P’ has twice the speed of CM

The magnitude of the linear acceleration of the CM is

A rolling motion can be interpreted as the sum of Translation and Rotation

CMa

Why?? P

P’

CM vCM

2vCM

CM is moving at the same speed at all times.

P

P’

CM vCM

vCM

vCM

+ P

P’

CM

v=Rω

v=0

v=Rω

= P

P’

CM

2vCM

vCM

CMvt

Δ=Δ

RtωΔ=Δ

αR=



14

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?

Ex. An Accelerating Car

α =

θ α ω ωo t

θ =( ) ( )22 21

2 2.42rad s 20.0 s= −

ar=

220.800m s 2.42rad s

0.330 m=

otω 212 tα+

484 rad= −

-2.42 rad/s2 0 rad/s 20.0 s ?



15

Torque Torque is the tendency of a force to rotate an object about an axis. Torque, τ, is a vector quantity.

τ ≡r

Magnitude of torque is defined as the product of the force exerted on the object to rotate it and the moment arm.

F φ

l1

The line of Action

Consider an object pivoting about the point P by the force F being exerted at a distance r from P.

P

r

Moment arm

The line that extends out of the tail of the force vector is called the line of action. The perpendicular distance from the pivoting point P to the line of action is called the moment arm.

When there are more than one force being exerted on certain points of the object, one can sum up the torque generated by each force vectorially. The convention for sign of the torque is positive if rotation is in counter-clockwise and negative if clockwise.

l2

F2

1 2τ τ τ= +∑1 1 2 2Fl F l= −

( )( )sinF r φ= = Fl

( )Magnitude of the Force

( )Lever Arm×

Unit? N m⋅



16

The tendon exerts a force of magnitude 790 N on the point P. Determine the torque (magnitude and direction) of this force about the ankle joint which is located 3.6x10-2m away from point P.

Ex. The Achilles Tendon

τ =

cos55 =o

( )( )2720 N 3.6 10 m cos55−= × o

790 N =l

F

23.6 10 m−×l

23.6 10 cos55−× =o

( ) ( )2 23.6 10 sin 90 55 2.1 10 m− −= × − = ×o o

First, let’s find the lever arm length

So the torque is

Since the rotation is in clock-wise τ =

3.6x10-2m

= 720 N( ) 3.6×10−2 m( )sin35 =15 N ⋅m

−15N ⋅m



17

Moment of Inertia Rotational Inertia:

What are the dimension and unit of Moment of Inertia?

I ≡

kg ⋅m2ML2⎡⎣ ⎤⎦

Measure of resistance of an object to changes in its rotational motion. Equivalent to mass in linear motion.

Determining Moment of Inertia is extremely important for computing equilibrium of a rigid body, such as a building.

I ≡For a group of objects

For a rigid body

miri2

i∑

r 2∫ dm

Dependent on the axis of rotation!!!



18

Two particles each have mass m1 and m2 and are fixed at the ends of a thin rigid rod. The length of the rod is L. Find the moment of inertia when this object rotates relative to an axis that is perpendicular to the rod at (a) one end and (b) the center.

Ex. The Moment of Inertia Depends on Where the Axis Is.

(a) I =

1 0r =1 2m m m= =

I =

(b) I =

1 2r L=1 2m m m= =

I =

2r L=

mL2

m 0( )2

+ m L( )2=

mr 2( )∑ = m1r1

2 + m2r22

mr 2( )∑ = m1r1

2 + m2r22

2 2r L=

m L 2( )2

+ m L 2( )2=

12 mL2

Which case is easier to spin? Case (b)

Why? Because the moment of inertia is smaller



19

Example for Moment of Inertia In a system of four small spheres as shown in the figure, assuming the radii are negligible and the rods connecting the particles are massless, compute the moment of inertia and the rotational kinetic energy when the system rotates about the y-axis at angular speed w.

I

Since the rotation is about y axis, the moment of inertia about y axis, Iy, is

RKThus, the rotational kinetic energy is

Find the moment of inertia and rotational kinetic energy when the system rotates on the x-y plane about the z-axis that goes through the origin O.

x

y

This is because the rotation is done about y axis, and the radii of the spheres are negligible. Why are some 0s?

M M l l

m

m

b

b O

I RK

= mii∑ ri

2 = Ml2 = 2Ml2

=12Iω 2=

122Ml2( )ω 2 = Ml2ω 2

= mii∑ ri

2 = Ml2 = 2 Ml2 + mb2( ) =

12Iω 2=

122Ml2 + 2mb2( )ω 2= Ml2 + mb2( )ω 2

+Ml2 +m ⋅02 +m ⋅02

+Ml2 +mb2 +mb2



20

Check out Figure 8 – 21 for moment of

inertia for various shaped

objects



21

Torque & Angular Acceleration Let’s consider a point object with mass m rotating on a circle.

What does this mean?

The tangential force Ft and the radial force Fr

The tangential force Ft is

What do you see from the above relationship?

m r

Ft

Fr What forces do you see in this motion?

t tF ma=The torque due to tangential force Ft is tF rτ =

Iτ α=Torque acting on a particle is proportional to the angular acceleration.

What law do you see from this relationship? Analogs to Newton’s 2nd law of motion in rotation.

How about a rigid object?

r

dFt

dm

O

The external tangential force δFt is tFδ =

δτ =∑The torque due to tangential force Ft is The total torque is

δτ =

What is the contribution due to radial force and why?

Contribution from radial force is 0, because its line of action passes through the pivoting point, making the moment arm 0.

mrα=

tma r= = mr 2α Iα=

tmaδ = mrδ α

tF rδ = r 2δm( )α

α r 2δm∑ = Iα



22

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.

Ex. Hoisting a Crate

2

'1 1 2T T Iτ α= − =∑ l l

2y yF T mg ma′= − =∑2

'yT mg ma= +

2ya α= l

( )1 1 2yT mg ma Iα− + =l l( )1 1 2 2T mg m Iα α− + =l l l

( )( ) ( )( )( )( )( )

221 1 2

22 22

2150 N 0.600 m 451 kg 9.80m s 0.200 m6.3rad s

46.0 kg m 451 kg 0.200 mT mgI m

α−−= = =

+ ⋅ +l l

l



23

Rotational Kinetic Energy What do you think the kinetic energy of a rigid object that is undergoing a circular motion is?

Since a rigid body is a collection of masslets, the total kinetic energy of the rigid object is

Since moment of Inertia, I, is defined as

Kinetic energy of a masslet, mi, moving at a tangential speed, vi, is

ri mi

θ

O x

y vi

iK

RK

∑=i

iirmI 2

RK =The above expression is simplified as

2

21

iivm= 2= ω221

iirm

∑=i

iK ∑ 2=i

iirm ω221 2⎟

⎠⎞⎜

⎝⎛= ∑ ω

iiirm2

21

12Iω 2



24

Example for Moment of Inertia In a system of four small spheres as shown in the figure, assuming the radii are negligible and the rods connecting the particles are massless, compute the moment of inertia and the rotational kinetic energy when the system rotates about the y-axis at angular speed w.

I

Since the rotation is about y axis, the moment of inertia about y axis, Iy, is

RKThus, the rotational kinetic energy is

Find the moment of inertia and rotational kinetic energy when the system rotates on the x-y plane about the z-axis that goes through the origin O.

x

y

This is because the rotation is done about y axis, and the radii of the spheres are negligible. Why are some 0s?

M M l l

m

m

b

b O

I RK

= mii∑ ri

2 = Ml2 = 2Ml2

=12Iω 2=

122Ml2( )ω 2 = Ml2ω 2

= mii∑ ri

2 = Ml2 = 2 Ml2 + mb2( ) =

12Iω 2=

122Ml2 + 2mb2( )ω 2= Ml2 + mb2( )ω 2

+Ml2 +m ⋅02 +m ⋅02

+Ml2 +mb2 +mb2



25

Kinetic Energy of a Rolling Sphere

Since vCM=Rω

Let’s consider a sphere with radius R rolling down the hill without slipping.

=K

R

h q

vCM

ω

212

CMCM

vIR

⎛ ⎞= ⎜ ⎟⎝ ⎠

Since the kinetic energy at the bottom of the hill must be equal to the potential energy at the top of the hill

What is the speed of the CM in terms of known quantities and how do you find this out?

K

222

1CM

CM vMRI

⎟⎠⎞⎜

⎝⎛ +=

222

1CM

CM vMRI

⎟⎠⎞⎜

⎝⎛ += Mgh=

2/12MRI

ghvCM

CM +=

212 CMI ω 2 21

2MR ω+

212 CMMv+



26

Example for Rolling Kinetic Energy For solid sphere as shown in the figure, calculate the linear speed of the CM at the bottom of the hill and the magnitude of linear acceleration of the CM. Solve this problem using Newton’s second law, the dynamic method.

∑ xF

Gravitational Force,

Since the forces Mg and n go through the CM, their moment arm is 0 and do not contribute to torque, while the static friction f causes torque

M h

θ

αRaCM =

CMτ

We know that

What are the forces involved in this motion?

Mg

f

Newton’s second law applied to the CM gives Frictional Force, Normal Force

n

2

52MRICM =

We obtain f

Substituting f in dynamic equations CMMaMg

57sin =θ

fMg −= θsin CMMa=

∑ yF θcosMgn−= 0=

= fR αCMI=

RICMα= ⎟

⎠⎞⎜

⎝⎛=Ra

R

MRCM

2

52

CMMa52=

θsin75 gaCM =



27

Work, Power, and Energy in Rotation Let’s consider the motion of a rigid body with a single external force F exerting on the point P, moving the object by Δs. The work done by the force F as the object rotates through the infinitesimal distance Δs=rΔθ is

What is Fsinϕ? The tangential component of the force F.

ΔW

Since the magnitude of torque is rFsinϕ,

F φ

O r Δθ Δs

What is the work done by radial component Fcosϕ?

Zero, because it is perpendicular to the displacement. ΔW

The rate of work, or power, becomes P How was the power defined in linear motion?

The rotational work done by an external force equals the change in rotational Kinetic energy. ∑τ

The work put in by the external force then

= F⋅ Δs

= τΔθ

=ΔWΔt

=τ ΔθΔt

τω=

αI= = I ΔωΔt

⎛⎝⎜

⎞⎠⎟

ΔW 22

21

21

if II ωω −=

= rF sinφ( )Δθ

= F sinφ( )rΔθ

τΔθ∑ = IωΔω



28

Angular Momentum of a Particle If you grab onto a pole while running, your body will rotate about the pole, gaining angular momentum. We’ve used the linear momentum to solve physical problems with linear motions, the angular momentum will do the same for rotational motions.

L = mvr = mr2ϖ = Iϖ

Let’s consider a point-like object ( particle) with mass m located at the vector location r and moving with linear velocity v

L≡ r× pThe angular momentum L of this

particle relative to the origin O is

What do you learn from this? If the direction of linear velocity points to the origin of rotation, the particle does not have any angular momentum.

What is the unit and dimension of angular momentum? 2 /kg m s⋅

Note that L depends on origin O. Why? Because r changes The direction of L is +z. What else do you learn?

Since p is mv, the magnitude of L becomes

If the linear velocity is perpendicular to position vector, the particle moves exactly the same way as a point on a rim.

2 1[ ]ML T −

z

x

y



29

Angular Momentum of a System of Particles The total angular momentum of a system of particles about some point is the vector sum of the angular momenta of the individual particles

L=

Since the individual angular momentum can change, the total angular momentum of the system can change.

τ

ext∑ =ΔL

Δt

Thus the time rate change of the angular momentum of a system of particles is equal to only the net external torque acting on the system

Let’s consider a two particle system where the two exert forces on each other.

Since these forces are the action and reaction forces with directions lie on the line connecting the two particles, the vector sum of the torque from these two becomes 0.

Both internal and external forces can provide torque to individual particles. However, the internal forces do not generate net torque due to Newton’s third law.

+L

2 ......+ +L

n = L

i∑ L

1

τ ext∑ =ΔLzΔt

=Δ Iϖ( )Δt

=IΔϖΔt

= IαFor a rigid body, the external torque is written

F

∑ =Δ p

ΔtJust like



30

Example for Rigid Body Angular Momentum A rigid rod of mass M and length l is pivoted without friction at its center. Two particles of mass m1 and m2 are attached to either end of the rod. The combination rotates on a vertical plane with an angular speed of w. Find an expression for the magnitude of the angular momentum.

I

The moment of inertia of this system is

α

First compute the net external torque

θτ cos21lgm=1

m1 g

x

y

O

l

m1

m2

θ m2 g

If m1 = m2, no angular momentum because the net torque is 0. If θ=+/-π/2, at equilibrium so no angular momentum.

⎟⎠⎞⎜

⎝⎛ ++== 21

2

31

4mmMlIL ωω

Find an expression for the magnitude of the angular acceleration of the system when the rod makes an angle θ with the horizon.

2τττ += 1ext

Thus α becomes

21 mmrod III ++= =

112

Ml2 +

⎟⎠⎞⎜

⎝⎛ ++= 21

2

31

4mmMl

cos222 θτ lgm−=

( )2

cos 21 mmgl −= θ

Iext∑=

τ ( )1 2

2

1 2

1 cos21

4 3

m m gl

l M m m

θ−=

⎛ ⎞+ +⎜ ⎟⎝ ⎠

( )1 2

1 2

2 cos13

m m glM m m

θ−=⎛ ⎞+ +⎜ ⎟⎝ ⎠

21

14m l + 2

214m l



31

Conservation of Angular Momentum Remember under what condition the linear momentum is conserved?

Linear momentum is conserved when the net external force is 0.

i fL L=r r

Three important conservation laws for isolated system that does not get affected by external forces

Angular momentum of the system before and after a certain change is the same.

By the same token, the angular momentum of a system is constant in both magnitude and direction, if the resultant external torque acting on the system is 0.

iLr

p const=ur

extτ =∑r

What does this mean?

Mechanical Energy

Linear Momentum

Angular Momentum

L=

F

∑ = 0 =Δ p

Δt

ΔL

Δt=0

fL=r

constant=

i fp p=r ri i f fK U K U+ = +

const



32

Example for Angular Momentum Conservation A star rotates with a period of 30 days about an axis through its center. After the star undergoes a supernova explosion, the stellar core, which had a radius of 1.0x104km, collapses into a neutron star of radius 3.0km. Determine the period of rotation of the neutron star.

ω =

What is your guess about the answer? The period will be significantly shorter, because its radius got smaller.

fi LL =

Let’s make some assumptions:

1.  There is no external torque acting on it 2.  The shape remains spherical 3.  Its mass remains constant

The angular speed of the star with the period T is

Using angular momentum conservation

Thus fω

ffi II ωωι =

f

i

II ιω=

if

i

Tmrmr π2

2

2

=

fTfωπ2= i

i

f Trr

⎟⎟⎠

⎞⎜⎜⎝

⎛= 2

2

days30100.10.3 2

4 ×⎟⎠⎞⎜

⎝⎛

×= days6107.2 −×= s23.0=

2π T



33

Similarity Between Linear and Rotational Motions All physical quantities in linear and rotational motions show striking similarity.

Quantities Linear Rotational Mass Mass Moment of Inertia

Length of motion Distance Angle (Radian)

Speed Acceleration

Force Force Torque Work Work Work Power

Momentum Kinetic Energy Kinetic Rotational

2I mr=

rvt

Δ=Δ t

θω Δ=Δ

vat

Δ=Δ t

ωα Δ=Δ

F= ma

τ= Iα

W F d= ⋅rr

P = F⋅ v τω=P

2

21 mvK = 2

21 ωIKR =

LM

θ

W τθ=

p= mv

L= Iω

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PHYS 1441 – Section 002 Lecture #20yu/teaching/spring13-1441-002/... · Wednesday, April 10, 2013 PHYS 1441-002, Spring 2013 Dr. Jaehoon Yu 14 Starting from rest, the car accelerates