Department of Physics and Applied PhysicsPHYS.1410 Lecture 21 Danylov
Lecture 21
Chapter 12
Cross ProductAngular Momentum
Physics I
Course website:http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI
Department of Physics and Applied PhysicsPHYS.1410 Lecture 21 Danylov
Today we are going to discuss:
Chapter 12:
Cross Product/Torque: Section 12.10 Angular Momentum: Section 12.11
IN THIS CHAPTER, you will continue discussing rotational dynamics
Department of Physics and Applied PhysicsPHYS.1410 Lecture 21 Danylov
Torque is a turning force (the rotational equivalent of force).
It should depend on force, lever arm, angle:
(Refreshing)
rF sinF
Torque
Axis of rotation
Sign of a torque convention:A positive torque tries to rotate an object in a CCW directionA negative torque tries to rotate an object in a CW direction
r
Department of Physics and Applied PhysicsPHYS.1410 Lecture 21 Danylov
TorqueExampleThe body shown in the figure is pivoted at O. Three forces act on it in the direction shown in the figure: FA=10 N at point A, 8.0 m from O; FB=10 N at point B, 4.0 m from O and FC=19 N at point C, 3.0 m from O. Calculate the net torque about point O.
Department of Physics and Applied PhysicsPHYS.1410 Lecture 21 Danylov
Torque causes angular acceleration:
Force causes linear acceleration: (Translational N.2nd law)
Newton’s 2nd law of rotation
I
I is the Moment of Inertia(rotational equivalent of mass)
amF
Angular accelerationTorque
(rotational equivalent of force)
(Rotational N.2nd law)
Department of Physics and Applied PhysicsPHYS.1410 Lecture 21 Danylov
Pulley and massExample
An object of mass m is hung from a cylindrical pulley of radius R and mass M and released from rest. What is the acceleration of the object?
Department of Physics and Applied PhysicsPHYS.1410 Lecture 21 Danylov
Notation for Vectors Perpendicular to the Page
Physics requires a three-dimensional perspective, but two-dimensional figures are easier to draw. We will use the following notation:
Department of Physics and Applied PhysicsPHYS.1410 Lecture 21 Danylov
Vector Cross Product
rF sinThis construction is quite common in Physics, so mathematicians decided to abbreviate it and give a nice name -
Department of Physics and Applied PhysicsPHYS.1410 Lecture 21 Danylov
The Cross Product
=(ab sin θ, direction given by the right-hand rule)
From now on, some equations will have a cross product
The cross product of vectors a and b is a vector perpendicular to both a and b.
[I] Point fingers in the direction of the 1st (a) vector, then bend them in the direction of the 2nd one (b).The outstretched thumb will give a direction of the cross product
[II] Use three fingers as shown in the figure
Order is important in the cross product:
Department of Physics and Applied PhysicsPHYS.1410 Lecture 21 Danylov
Cross product
The cross product vector increases from 0 to AB as θ increases from 0 to 90
A
B
C
A
B
C A
B ABsin
0BA
θ=30
θ=0
A
B
θ=90A
B
The vector product is zero when vectors are parallel
The vector product increases The vector product is max when vectors are perpendicular
ABBA
ABBA 21
Department of Physics and Applied PhysicsPHYS.1410 Lecture 21 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)
Department of Physics and Applied PhysicsPHYS.1410 Lecture 21 Danylov
Angular MomentumWe will introduce angular momentum of
• A point mass m
• A rigid object
Department of Physics and Applied PhysicsPHYS.1410 Lecture 21 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 21 Danylov
Angular Momentum of a single particle
L
r p rpSinL
xz
y
O
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
OCarefull: Let’s calculate angular momentum of m about point O’
prL sinpr
r
,since pr
0
0sin,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 21 Danylov
Angular momentum (about the origin) of an object of mass m dropped from rest.
Example
Department of Physics and Applied PhysicsPHYS.1410 Lecture 21 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 21 Danylov
Two definitions of Angular Momentum
L
r
p
L
L I
L
r p
Rigid symmetrical bodySingle particle
Department of Physics and Applied PhysicsPHYS.1410 Lecture 21 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)
Department of Physics and Applied PhysicsPHYS.1410 Lecture 21 Danylov
Translational – vs- Rotational N. 2nd law
amF
dtpdF
I
Translational N.2nd law Rotational N.2nd law
dtLd