Physics 101Lecture 12

Angular MomentumAsist. Prof. Dr. Ali ÖVGÜN

EMU Physics Department

www.aovgun.com

April 29, 2020

Angular Momentum

❑ Torque using vectors

❑ Angular Momentum

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❑ The torque is the cross product of a force vector with the position vector to its point of application

❑ The torque vector is perpendicular to the plane formed by the position vector and the force vector (e.g., imagine drawing them tail-to-tail)

❑ Right Hand Rule: curl fingers from r to F,

thumb points along torque.

Torque as a Cross Product

Fr

=

⊥⊥ === FrFrrF sin

sum)(vector ri

i

i

inet i

allall

F

==

Superposition:

❑ Can have multiple forces applied at multiple points.

❑ Direction of net is angular acceleration axis

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i

kj

Net torque example: multiple forces at a single point

x

y

z

r

1F

3F

2F

3 forces applied at point r :ˆ ˆ ˆcos 0 sinr r + +r i j k

oˆ ˆˆ2 ; 2 ; 2 ; 3; 30r = = = = =1 2 3F i F k F j

Find the net torque about the origin:

net net 1 2 3( )

ˆ ˆ ˆˆ ˆ( ) (2 2 2 )

ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ 2 2 2 2 2 2

x z

x x x z z z

r r

r r r r r r

= = + +

= + + +

= + + + + +

τ r F r F F F

i k i j k

i × i i × j i ×k k × i k × j k ×k

.)3cos(30 )rcos( r

.)3sin(30 )rsin( r o

z

oxset

62

51

==

==

netˆ ˆ ˆˆ0 2 2 ( ) 2 2 ( ) 0x x z zr r r r= + + − + + − +τ k j j i

netˆ ˆ ˆ 3 2.2 5.2 = − − +τ i j k

Here all forces were applied at the same point.For forces applied at different points, first calculatethe individual torques, then add them as vectors,i.e., use:

sum) (vector Fr ii all

ii all

inet

==

oblique rotation axis

through origin

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Angular Momentum❑ Same basic techniques that were used in linear

motion can be applied to rotational motion.◼ F becomes

◼ m becomes I

◼ a becomes

◼ v becomes ω

◼ x becomes θ

❑ Linear momentum defined as

❑ What if mass of center of object is not moving, but it is rotating?

❑ Angular momentum

m=p v

I=L ω

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Angular Momentum I❑ Angular momentum of a rotating rigid object

◼ L has the same direction as *

◼ L is positive when object rotates in CCW

◼ L is negative when object rotates in CW

❑ Angular momentum SI unit: kg-m2/sCalculate L of a 10 kg disk when = 320 rad/s, R = 9 cm = 0.09 m

L = I and I = MR2/2 for disk

L = 1/2MR2 = ½(10)(0.09)2(320) = 12.96 kgm2/s *When rotation is about a principal axis

I=L ω

L

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Angular Momentum II❑ Angular momentum of a particle

❑ Angular momentum of a particle

◼ r is the particle’s instantaneous position vector

◼ p is its instantaneous linear momentum

◼ Only tangential momentum component contribute

◼ Mentally place r and p tail to tail form a plane, L is perpendicular to this plane

( )m= =L r×p r× v

sinsin2 rpmvrrmvmrIL ===== ⊥

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Angular Momentum of a Particle in Uniform Circular Motion

❑ The angular momentum vector points out of the diagram

❑ The magnitude is

L = rp sin = mvr sin(90o) = mvr

❑ A particle in uniform circular motion has a constant angular momentum about an axis through the center of its path

O

Example: A particle moves in the xy plane in a circular path of radius r. Find the magnitude and direction of its angular momentum relative to an axis through O when its velocity is v.

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Angular momentum III❑ Angular momentum of a system of particles

◼ angular momenta add as vectors

◼ be careful of sign of each angular momentum

net 1 2

... n i i i

all i all i

= + + + = = L L L L L r p

net 1 1 2 2 r r⊥ ⊥= + −L p p

for this case:

net 1 2 1 1 2 2= + = + L L L r p r p

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Example: calculating angular momentum for particles

Two objects are moving as shown in the figure . What is their total angular momentum about point O?

Direction of L is out of screen.

m2

m1

net 1 1 1 2 2 2

1 1 2 2

2

sin sin

2.8 3.1 3.6 1.5 6.5 2.2

31.25 21.45 9.8 kg m /s

L r mv r mv

r mv r mv

= −

= −

= −

= − =

net 1 2 1 1 2 2= + = + L L L r p r p

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❑ What would the angular momentum about point “P” be if the car leaves the track at “A” and ends up at point “B” with the same velocity ?

Angular Momentum for a Car

A) 5.0 102

B) 5.0 106

C) 2.5 104

D) 2.5 106

E) 5.0 103 )sin( pr rp pr L === ⊥⊥

P

A

B

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Recall: Linear Momentum and Force

❑ Linear motion: apply force to a mass

❑ The force causes the linear momentum to change

❑ The net force acting on a body is the time rate of change of its linear momentum

net

d dm m

dt dt= = = =

v pF F a

L

t

=

m=p v

net t= = I F p

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Angular Momentum and Torque

❑ Rotational motion: apply torque to a rigid body

❑ The torque causes the angular momentum to change

❑ The net torque acting on a body is the time rate of change of its angular momentum

❑ and are to be measured about the same origin

❑ The origin must not be accelerating (must be an inertial frame)

net

d

dt= =

pF F net

d

dt= =

Lτ τ

τ L

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Demonstration

❑ Start from

❑ Expand using derivative chain rule

)()( vrdt

dmpr

dt

d

dt

Ld

==

dt

Ldnet

==

dt

pdFFnet

==

arvvmdt

vdrv

dt

rdmvr

dt

dm

dt

Ld

+=

+== )(

netnetFramrarmarvvmdt

Ld

====+= )(

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What about SYSTEMS of Rigid Bodies?

• i = net torque on particle “i”

• internal torque pairs are

included in sum

i = LLsys

• individual angular momenta Li

• all about same origin

==i

i

sys

dt

Ld

dt

Ld

i

BUT… internal torques in the sum cancel in Newton 3rd law

pairs. Only External Torques contribute to Lsys

Nonisolated System: If a system interacts with its environment in the sense that there is an external torque on the system, the net external torque acting on a system is equal to the time rate of change of its angular momentum.

dt

Ld i i :body single a forlaw Rotational

nd

=2

Total angular momentum

of a system of bodies:

net external torque on the systemnet,

==i

exti

sys

dt

Ld

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a

a

Example: A Non-isolated SystemA sphere mass m1 and a block of mass m2 are connected by a light cord that passes over a pulley. The radius of the pulley is R, and the mass of the thin rim is M. The spokes of the pulley have negligible mass. The block slides on a frictionless, horizontal surface. Find an expression for the linear acceleration of the two objects. gRmext 1=

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Masses are connected by a light cord. Find the

linear acceleration a.

• Use angular momentum approach• No friction between m2 and table

• Treat block, pulley and sphere as a non-isolated system rotating about pulley axis. As sphere falls, pulley rotates, block slides• Constraints:

for pulley

Equal 's and 's for block and sphere

/

v a

v ωR α d dt

a αR dv / dt

= =

= =• Ignore internal forces, consider external forces only

• Net external torque on system:

• Angular momentum of system:

(not constant)ωMRvRmvRmIωvRmvRmLsys

2

2121 ++=++=

gRmτMR)aRmR(mαMRaRmaR mdt

dLnet

sys

121

2

21 ==++=++=

1 about center of wheelnet m gR =

21

1 mmM

gma

++= same result followed from earlier

method using 3 FBD’s & 2nd law

I

a

a

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Isolated System

❑ Isolated system: net external torque acting on a system is ZERO

◼ no external forces

◼ net external force acting on a system is ZERO

constant or tot i f= =L L L

0totext

d

dt = =

Lτ

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Angular Momentum Conservation

❑ Here i denotes initial state, f is the final state

❑ L is conserved separately for x, y, z direction

❑ For an isolated system consisting of particles,

❑ For an isolated system that is deformable

1 2 3 constanttot n= = + + + =L L L L L

constant== ffii II

constant or tot i f= =L L L

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First Example

❑ A puck of mass m = 0.5 kg is attached to a taut cord passing through a small hole in a frictionless, horizontal surface. The puck is initially orbiting with speed vi = 2 m/s in a circle of radius ri = 0.2 m. The cord is then slowly pulled from below, decreasing the radius of the circle to r = 0.1 m.

❑ What is the puck’s speed at the smaller radius?

❑ Find the tension in the cord at the smaller radius.

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Angular Momentum Conservation

❑ m = 0.5 kg, vi = 2 m/s, ri = 0.2 m, rf = 0.1 m, vf = ?

❑ Isolated system?

❑ Tension force on m exert zero torque about hole, why?

i f=L L

0.22 4m/s

0.1

if i

f

rv v

r= = =

( )m= = L r p r v

iiiii vmrvmrL == 90sinfffff vmrvmrL == 90sin

N 801.0

45.0

22

====f

f

cr

vmmaT

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constant0 == L τ axis about z - net

==final

ff

initial

ii ωI ωI L

Moment of inertia changes

Isolated System

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Controlling spin () by changing I (moment of inertia)

In the air, net = 0L is constant

ffii IIL ==

Change I by curling up or stretching out- spin rate must adjust

Moment of inertia changes

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Example: A merry-go-round problem

A 40-kg child running at 4.0 m/s jumps tangentially onto a stationary circular merry-go-round platform whose radius is 2.0 m and whose moment of inertia is 20 kg-m2. There is no friction.

Find the angular velocity of the platform after the child has jumped on.

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❑ The moment of inertia of the system = the moment of inertia of the platform plus the moment of inertia of the person.◼ Assume the person can be

treated as a particle❑ As the person moves toward

the center of the rotating platform the moment of inertia decreases.

❑ The angular speed must increase since the angular momentum is constant.

The Merry-Go-Round

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Solution: A merry-go-round problem

I = 20 kg.m2

VT = 4.0 m/s

mc = 40 kg

r = 2.0 m

0 = 0

rv mrvmII L TcTciii =+== 0

fcfff ωrmIωIL )( 2+==

rvmωrmI Tcfc =+ )( 2

rad/s 78.124010

244022=

+

=

+=

rmI

rvmω

c

Tcf

tot i i f f I I= = L ω ω

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April 29, 2020