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Physics 111: Mechanics Lecture 11 Dale Gary NJIT Physics Department.
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Transcript of Physics 111: Mechanics Lecture 11 Dale Gary NJIT Physics Department.
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