Physics 111: Mechanics Lecture 11

Dale Gary

NJIT Physics Department

April 19, 2023

Angular Momentum Vectors Cross Product Torque using vectors Angular Momentum

April 19, 2023

Vector Basics We will be using vectors a lot in

this course. Remember that vectors have both

magnitude and direction e.g. You should know how to find the

components of a vector from its magnitude and direction

You should know how to find a vector’s magnitude and direction from its components

amF

amF

m

aF

Ways of writing vector notation

y

x

a

sin

cos

aa

aa

y

x

xy

yx

aa

aaa

/tan 1

22

,a

ax

ay

April 19, 2023

y

x

a

z

Projection of a Vector in Three Dimensions

Any vector in three dimensions can be projected onto the x-y plane.

The vector projection then makes an angle from the x axis.

Now project the vector onto the z axis, along the direction of the earlier projection.

The original vector a makes an angle from the z axis.

y

x

a

z

April 19, 2023

Vector Basics (Spherical Coords)

You should know how to generalize the case of a 2-d vector to three dimensions, e.g. 1 magnitude and 2 directions

Conversion to x, y, z components

Conversion from x, y, z components

Unit vector notation:

y

x

a

z

, ,a

cos

sinsin

cossin

aa

aa

aa

z

y

x

xy

z

zyx

aa

aa

aaaa

/tan

/cos1

1

222

kajaiaa zyxˆˆˆ

sina

April 19, 2023

A Note About Right-Hand Coordinate Systems

A three-dimensional coordinate system MUST obey the right-hand rule.

Curl the fingers of your RIGHT HAND so they go from x to y. Your thumb will point in the z direction.

y

x

z

April 19, 2023

Cross Product

The cross product of two vectors says something about how perpendicular they are.

Magnitude:

is smaller angle between the vectors Cross product of any parallel vectors = zero Cross product is maximum for

perpendicular vectors Cross products of Cartesian unit vectors:

sinABBAC

BAC

A

B

sinA

sinB

0ˆˆ ;0ˆˆ ;0ˆˆ

ˆˆˆ ;ˆˆˆ ;ˆˆˆ

kkjjii

ikjjkikji

y

x

z

ij

k

i

kj

April 19, 2023

Cross Product Direction: C perpendicular

to both A and B (right-hand rule) Place A and B tail to tail Right hand, not left hand Four fingers are pointed

along the first vector A “sweep” from first vector

A into second vector B through the smaller angle between them

Your outstretched thumb points the direction of C

First practice ? ABBA

? ABBA

A B B A

April 19, 2023

More about Cross Product The quantity ABsin is the area of

the parallelogram formed by A and B

The direction of C is perpendicular to the plane formed by A and B

Cross product is not commutative

The distributive law

The derivative of cross product obeys the chain rule Calculate cross product

dt

BdAB

dt

AdBA

dt

d

CABACBA

)(

A B B A

kBABAjBABAiBABABA xyyxzxxzyzzyˆ)(ˆ)(ˆ)(

April 19, 2023

Derivation How do we show that ? Start with

Then

But

So

kBjBiBB

kAjAiAA

zyx

zyx

ˆˆˆ

ˆˆˆ

)ˆˆˆ(ˆ)ˆˆˆ(ˆ)ˆˆˆ(ˆ

)ˆˆˆ()ˆˆˆ(

kBjBiBkAkBjBiBjAkBjBiBiA

kBjBiBkAjAiABA

zyxzzyxyzyxx

zyxzyx

0ˆˆ ;0ˆˆ ;0ˆˆ

ˆˆˆ ;ˆˆˆ ;ˆˆˆ

kkjjii

ikjjkikji

jBkAiBkA

kBjAiBjAkBiAjBiABA

yzxz

zyxyzxyx

ˆˆˆˆ

ˆˆˆˆˆˆˆˆ

kBABAjBABAiBABABA xyyxzxxzyzzyˆ)(ˆ)(ˆ)(

zyx

zyx

BBB

AAA

kji

BA

ˆˆˆ

April 19, 2023

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

ii

inet iallall

F

Superposition:

Can have multiple forces applied at multiple points.

Direction of net is angular acceleration axis

April 19, 2023

Calculating Cross Products

mjirNjiF )ˆ5ˆ4( )ˆ3ˆ2(

BA

Solution: i

kj

jiBjiA ˆ2ˆ ˆ3ˆ2

kkkijji

jjijjiii

jijiBA

ˆ7ˆ3ˆ40ˆˆ3ˆˆ40

ˆ2ˆ3)ˆ(ˆ3ˆ2ˆ2)ˆ(ˆ2

)ˆ2ˆ()ˆ3ˆ2(

Calculate torque given a force and its location

(Nm) ˆ2ˆ10ˆ120ˆ2ˆ5ˆ3ˆ40

ˆ3ˆ5ˆ2ˆ5ˆ3ˆ4ˆ2ˆ4

)ˆ3ˆ2()ˆ5ˆ4(

kkkijji

jjijjiii

jijiFr

Solution:

Find: Where:

ˆˆ ˆ

4 5 0

2 3 0

i j k

A B

April 19, 2023

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

oz

oxset

6251

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 i

i alli

i allinet

oblique rotation axisthrough origin

April 19, 2023

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

mp v

IL ω

April 19, 2023

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 mL = I and I = MR2/2 for diskL = 1/2MR2 = ½(10)(0.09)2(320) = 12.96 kgm2/s

*When rotation is about a principal axis

IL ω

L

April 19, 2023

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

April 19, 2023

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.

April 19, 2023

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 iall 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

April 19, 2023

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

April 19, 2023

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

April 19, 2023

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

Lt

mp v

net t I F p

April 19, 2023

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

April 19, 2023

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

)(

April 19, 2023

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

isys

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 momentumof a system of bodies:

net external torque on the system

net,

i

extisys

dt

Ld

April 19, 2023

a

a

Example: A Non-isolated System

A 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

April 19, 2023

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 av 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)

ωMRvRmvRmIωvRmvRmLsys2

2121

gRmτMR)aRmR(mαMRaRmaR mdt

dLnet

sys121

221

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

April 19, 2023

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

April 19, 2023

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

April 19, 2023

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.

April 19, 2023

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 fL L

0.22 4 m/s

0.1i

f if

rv v

r

( )m L r p r v

iiiii vmrvmrL 90sin fffff vmrvmrL 90sin

N 801.0

45.0

22

f

fc r

vmmaT

April 19, 2023

constant0 L τ axis about z - net

final

ffinitial

ii ωI ωI L

Moment of inertia changes

Isolated System

April 19, 2023

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

In the air, net = 0

L is constant

ffii IIL

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

Moment of inertia changes

April 19, 2023

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.

April 19, 2023

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

April 19, 2023

Solution: A merry-go-round problem

I = 20 kg.m2

VT = 4.0 m/smc = 40 kgr = 2.0 m0 = 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