lecture 22 ch12 f16 angular momentum -...
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Department of Physics and Applied PhysicsPHYS.1410 Lecture 22 Danylov
Lecture 22
Chapter 12
Angular MomentumConservation of Angular Momentum
Physics I
Course website:http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI
Department of Physics and Applied PhysicsPHYS.1410 Lecture 22 Danylov
Today we are going to discuss:
Chapter 12:
Angular Momentum: Section 12.11 Rotational Newton’s 2nd Law (general form): Section 12.11 Conservation of Angular Momentum: Section 12.11
IN THIS CHAPTER, you will continue discussing rotational dynamics
Department of Physics and Applied PhysicsPHYS.1410 Lecture 22 Danylov
Torque is a turning force (the rotational equivalent of force).
rF sin
F
Now we can write Torque as a vector product
Axis of rotation
r
r
F
Now, with the vector product notation we can rewrite torque as
Torque direction – out of page (right hand rule)
ConcepTest Figure SkaterA) Steadily increases
B) Increases for awhile, then holds steady
C) Holds steady
D) Decreases for awhile, then holds steady
A student gives a quick push to a puck that can rotate in a horizontal circle on a frictionless table. After the push has ended, the puck’s angular speed
A torque causes angular acceleration which leads to changes of the angular velocity. With no torque, the angular velocity stays the same.
221 IKrot
Department of Physics and Applied PhysicsPHYS.1410 Lecture 22 Danylov
Angular velocity as a vector
The magnitude of the angular velocityvector is ω.
The angular velocity vector points along the axis of rotation in the direction given by the right-hand rule as illustrated.
A more general description of rotationalmotion requires us to replace the scalars ω and τ with the vector quantities and
Department of Physics and Applied PhysicsPHYS.1410 Lecture 22 Danylov
Angular Momentum
We will introduce angular momentum of
• A point mass m
• A rigid object
Department of Physics and Applied PhysicsPHYS.1410 Lecture 22 Danylov
Angular momentum is the rotational equivalent of linear momentum
?L
vmp
For translational motion we needed the concepts of
force, Flinear momentum, p
mass, m
For rotational motion we needed the concepts of
torque, angular momentum, L
moment of inertia, I
Department of Physics and Applied PhysicsPHYS.1410 Lecture 22 Danylov
Angular Momentum of a single particle
L
r prpSinL
x
z
yO
r pm
L
r p Suppose we have a particle with-linear momentum -positioned at r
p
Then, by definition: Angular momentum of a particle about point O is
O
Carefull: Let’s calculate angular momentum of m about point O’
prL sinpr
r
,since pr
00sin,0 so
Thus, angular momentum of m 0OL but 0OLAngular Momentum is not an intrinsic property of a particle.
It depends on a choice of origin
So, never forget to indicate which origin is being used
Department of Physics and Applied PhysicsPHYS.1410 Lecture 22 Danylov
Angular momentum (about the origin) of an object of mass m dropped from rest.
Example
(The shortest distance between the origin and the line of motion)
ConcepTest Traffic light/carA car of mass 1000 kg drives away from atraffic light h=10 m high, as shown below, at aconstant speed of v=10 m/s. What is theangular momentum of the car with respect tothe light?
A) B) C)
skgmk 2 )ˆ(000,100
x
y
z
h
skgmi 2 ˆ000,100
v
prL
skgmk 2 )ˆ(000,10
r
)ˆ)(( krSinmv )ˆ( kmvh )ˆ(000,100 k
Department of Physics and Applied PhysicsPHYS.1410 Lecture 22 Danylov
Angular Momentum of a rigid body
L I
points towardsL
For the rotation of a symmetrical object about the symmetry axis, the angular momentum and the angular velocity are related by (without a proof)
IL
IL
IL
I – moment of inertia of a body
Department of Physics and Applied PhysicsPHYS.1410 Lecture 22 Danylov
Two definitions of Angular Momentum
L
r
p
L
L I
L
r p
Rigid symmetrical bodySingle particle
Summary
Department of Physics and Applied PhysicsPHYS.1410 Lecture 22 Danylov
Rotational N. 2nd lawLet’s rewrite our rotational Newton’s 2nd Law in terms of angular momentum:
dtLd
Torque causes the particle’s angular momentum to change
Rotational N. 2nd lawwritten in terms of L.
dtLd
IdtdI
dtId )(
dtLd
(We use the angular momentum expression for a rigid body but it can also be shown for a point mass. See the end of the presentation)
Department of Physics and Applied PhysicsPHYS.1410 Lecture 22 Danylov
Translational – vs- Rotational N. 2nd law
amF
dtpdF
I
Translational N.2nd law Rotational N.2nd law
dtLd
End of the class
Department of Physics and Applied PhysicsPHYS.1410 Lecture 22 Danylov
Angular momentum (about the origin) of an object of mass m dropped from rest (cont.).
Example
(cont.)
Department of Physics and Applied PhysicsPHYS.1410 Lecture 22 Danylov
Conservation of Angular Momentum
dtLd
Angular momentum is an important concept because, under certain conditions, it is conserved.
If the net external torque on an object is zero, then the total angular momentum is conserved.
,0 extIf
constL
In Ch11 we derived the linear momentum conservation, where we showed that internal forces cancel each other out and cannot change the total linear momentum.
Similar, torques produced by the internal forces cancel each other out and cannot change the total angular momentum. So, only external torques are left to play the game.
0int ernal
extdtLd
0dtLdthen
, so
Department of Physics and Applied PhysicsPHYS.1410 Lecture 22 Danylov
Angular Momentum Conservation helps to solve many problems
I11 I22
IL For a rigid body21 LL
Launch: leg and hands are out to make I large
Flight: leg and hands are in to make large
Landing: leg and hands are out to “dump” large
1212 )( II
large1 I mallI s2
Example Figure Skater’s Jump
Angular Momentum stays constant throughout the whole jump
1 2
small-1 large-2
ConcepTest Figure SkaterA) the same
B) larger because she’s rotating faster
C) smaller because her rotational inertia is smaller
A figure skater spins with her arms extended. When she pulls in her arms, she reduces her rotational inertia and spins faster so that her angular momentum is conserved. Comparedto her initial rotational kinetic energy, her rotational kinetic energy after she pulls in her arms must be:
KErot = I 2 = (I ) = L (used L = I ). Because L is conserved, larger means larger KErot. The “extra” energy comes from the work she does on her arms.
12
12
12
221 IKrot
Department of Physics and Applied PhysicsPHYS.1410 Lecture 22 Danylov
Example Bullet strikes cylinder edge
A bullet of mass m moving with velocity vstrikes and becomes embedded at the edge ofa cylinder of mass M and radius R. Thecylinder, initially at rest, begins to rotateabout its symmetry axis, which remains fixedin position. Assuming no frictional torque,what is the angular velocity of the cylinderafter this collision? Is kinetic energyconserved?
Department of Physics and Applied PhysicsPHYS.1410 Lecture 22 Danylov
Thank youSee you on Monday
Department of Physics and Applied PhysicsPHYS.1410 Lecture 22 Danylov
Rotational N. 2nd law
L
r p dtLd
dtLd
Let’s find relationship between angular momentum and torque for a point particle:
dtpdFlawndN
2.
vmp
dtLd
Torque causes the particle’s angular momentum to change
Rotational N. 2nd lawwritten in terms of L.
p
dtrd
dtpdr
vmv Fr
Read if only if you want