Post on 03-Jun-2020
transcript
Physics 101Lecture 13
Angular MomentumAsist. Prof. Dr. Ali ÖVGÜN
EMU Physics Department
www.aovgun.com
February 16, 2020
Angular Momentum
❑ Torque using vectors
❑ Angular Momentum
February 16, 2020
❑ 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
February 16, 2020
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 3
F 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
February 16, 2020
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
February 16, 2020
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
February 16, 2020
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 ===== ⊥
February 16, 2020
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.
February 16, 2020
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
February 16, 2020
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
February 16, 2020
❑ 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
February 16, 2020
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
February 16, 2020
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
February 16, 2020
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
====+= )(
February 16, 2020
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
February 16, 2020
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=
February 16, 2020
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
February 16, 2020
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
February 16, 2020
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
February 16, 2020
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.
February 16, 2020
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 4 m/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
February 16, 2020
constant0 == L τ axis about z - net
==final
ff
initial
ii ωI ωI L
Moment of inertia changes
Isolated System
February 16, 2020
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
February 16, 2020
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.
February 16, 2020
❑ 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
February 16, 2020
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