physics 1501: lecture 22, pg 1 physics 1501: lecture 22 today’s agenda l announcements çhw#8: due...
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Physics 1501: Lecture 22, Pg 1
Physics 1501: Lecture 22Physics 1501: Lecture 22TodayToday’’s Agendas Agenda
AnnouncementsHW#8: due Oct. 28
Honors’ studentssee me Wednesday at 2:30 in P-114
TopicsRolling motionAngular MomentumFigure Skating Day !!! + other fun demos …
Physics 1501: Lecture 22, Pg 2
Torque, Work, Kinetic EnergyTorque, Work, Kinetic Energy We can define torque as: = r r x FF
= r F sin X = y FZ - z FY
Y = z FX - x FZ
Z = x FY - y FX
We find the work : W = Kinetic Energy of rotation: K = ½ I
Recall the Work Kinetic-Energy Theorem: K = WNET
So for an object that rotates about a fixed axis:
rr
FF
x
y
z
Physics 1501: Lecture 22, Pg 3
Connection with CM motion...Connection with CM motion... So for a solid object which rotates about its center of mass and whose CM is moving:
VCM
Physics 1501: Lecture 22, Pg 4
Rolling MotionRolling Motion Cylinders of different I rolling down an inclined plane:
h
v = 0
= 0
K = 0
R
K = - U = Mgh
v = R
M
Physics 1501: Lecture 22, Pg 5
Rolling...Rolling...
Use v= R and I = cMR2 .
So:
The rolling speed is always lower than in the case of simple
sliding since the kinetic energy is shared between CM motion
and rotation.
hoop: c=1
disk: c=1/2
sphere: c=2/5
etc...cc
cc
Physics 1501: Lecture 22, Pg 6
Lecture 22, Lecture 22, ACT 1ACT 1Rolling MotionRolling Motion
A race !!
Two cylinders are rolled down a ramp. They have the same radius but different masses, M1 > M2. Which wins the race to the bottom ?
A) Cylinder 1
B) Cylinder 2
C) It will be a tie
M1
h
M?
M2
Active Figure
Physics 1501: Lecture 22, Pg 7
Example :Example : Rolling MotionRolling Motion A cylinder is about to roll down an inclined plane. What is its
speed at the bottom of the plane ?
M
h
Mv ?
Ball has radius R
M M
M
M
M
Physics 1501: Lecture 22, Pg 8
Example :Example : Rolling MotionRolling Motion Use conservation of energy.
Ei = Ui + 0 = Mgh
Ef = 0 + Kf = 1/2 Mv2 + 1/2 I 2
= 1/2 Mv2 + 1/2 (1/2MR2)(v/R)2
Mgh = 1/2 Mv2 + 1/4 Mv2
v2 = 4/3 g h
v = ( 4/3 g h ) 1/2
Physics 1501: Lecture 22, Pg 9
Consider a roller coaster.
We can get the ball to go around the circle without leaving the loop.
Note:Radius of loop = RRadius of ball = r
Example :Example : Roller CoasterRoller Coaster
Physics 1501: Lecture 22, Pg 10
How high do we have to start the ball ?Use conservation of energy.Also, we must remember that the minimum
speed at the top is vtop = (gR)1/2
E1 = mgh + 0 + 0E2 = mg2R + 1/2 mv2 + 1/2 I2
= 2mgR + 1/2 m(gR) + 1/2 (2/5 mr2)(v/r)2
= 2mgR + 1/2 mgR + (2/10)m (gR) = 2.7 mgR
1
2
Physics 1501: Lecture 22, Pg 11
How high do we have to start the ball ?
E1 = mgh + 0 + 0E2 = 2.7 mgRmgh = 2.7 mgRh = 2.7 Rh = 1.35 D
(The rolling motion added an extra 2/10 R to the height: without it, h = 2.5 R)
1
2
Physics 1501: Lecture 22, Pg 12
Chap. 11: Angular MomentumChap. 11: Angular Momentum When we write = I we are really talking about
the z component of a more general vector equation. (we normally choose the z-axis to be the the rotation axis.)
z = Izz
We usually omit the
z subscript for simplicity.
z
z
z
Iz
Physics 1501: Lecture 22, Pg 13
Angular MomentumAngular Momentum
: angular momentum
r
mF
An object of mass m is rotating in a circular path under the action of a constant torque:
Physics 1501: Lecture 22, Pg 14
Angular MomentumAngular Momentum
The torque acting on an object is equal to the time rate of change of the object’s angular momentum
The angular momentum of an system is conserved when the net external torque acting on the system is zero. That is, when = 0, the initial angular momentum equals the final angular momentum.
Lf = Li
Physics 1501: Lecture 22, Pg 15
Angular Momentum:Angular Momentum:Definitions & DerivationsDefinitions & Derivations
We have shown that for a system of particles
Momentum is conserved if
What is the rotational version of this ??
The rotational analogue of force F F is torque
Define the rotational analogue of momentum pp to be
angular momentum
p=mv
Physics 1501: Lecture 22, Pg 16
Definitions & Derivations...Definitions & Derivations...
First consider the rate of change of LL:
So (so what...?)
Physics 1501: Lecture 22, Pg 17
Definitions & Derivations...Definitions & Derivations...
Recall that
Which finally gives us:
Analogue of !!
EXT
Physics 1501: Lecture 22, Pg 18
What does it mean?What does it mean?
where and
In the absence of external torquesIn the absence of external torques
Total angular momentum is conservedTotal angular momentum is conserved
Active torque Active angular momentum
Physics 1501: Lecture 22, Pg 19
i
j
Angular momentum of a rigid bodyAngular momentum of a rigid bodyabout a fixed axis:about a fixed axis:
Consider a rigid distribution of point particles rotating in the x-y plane around the z axis, as shown below. The total angular momentum around the origin is the sum of the angular momentum of each particle:
rr1
rr3
rr2
m2
m1
m3
vv2
vv1
vv3
We see that LL is in the z direction.
Using vi = ri , we get
L=I
(since ri , vi , are
perpendicular)
Analogue of p = mv !!
Physics 1501: Lecture 22, Pg 20
Angular momentum of a rigid bodyAngular momentum of a rigid bodyabout a fixed axis:about a fixed axis:
In general, for an object rotating about a fixed (z) axis we can write LZ = I
The direction of LZ is given by theright hand rule (same as ).
We will omit the ”Z” subscript for simplicity,and write L = I
z
LZ = I
Physics 1501: Lecture 22, Pg 21
Example: Two DisksExample: Two Disks
A disk of mass M and radius R rotates around the z axis with angular velocity 0. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity F.
0
z
F
z
Physics 1501: Lecture 22, Pg 22
Example: Two DisksExample: Two Disks
First realize that there are no external torques acting on the two-disk system.Angular momentum will be conserved !
Initially, the total angular momentum is due only to the disk on the bottom:
0
z
2
1
Physics 1501: Lecture 22, Pg 23
Example: Two DisksExample: Two Disks
First realize that there are no external torques acting on the two-disk system.Angular momentum will be conserved !
Finally, the total angular momentum is dueto both disks spinning:
F
z
21
Physics 1501: Lecture 22, Pg 24
Example: Two DisksExample: Two Disks
Since LINI = LFIN
0
z
F
z
LINI LFIN
An inelastic collision,since E is not
conserved (friction) !
1/2 MR2 MR2 F
F = 1/2
Physics 1501: Lecture 22, Pg 25
Demonstration ofDemonstration ofConservation of Angular MomentumConservation of Angular Momentum Figure Skating :
A
z
B
z
Arm Arm
IA < IB
A > B
LA = LB
Physics 1501: Lecture 22, Pg 26
Angular Momentum ConservationAngular Momentum Conservation
A freely moving particle has a definite angular momentum about any given axis.
If no torques are acting on the particle, its angular momentum will be conserved.
In the example below, the direction of LL is along the z axis, and its magnitude is given by LZ = pd = mvd.
y
x
vv
d
m
Physics 1501: Lecture 22, Pg 27
Example: Bullet hitting stickExample: Bullet hitting stick
A uniform stick of mass M and length D is pivoted at the center. A bullet of mass m is shot through the stick at a point halfway between the pivot and the end. The initial speed of the bullet is v1, and the final speed is v2.
What is the angular speed F of the stick after the collision? (Ignore gravity)
v1 v2
M
F
before after
mD
D/4
Physics 1501: Lecture 22, Pg 28
Example: Bullet hitting stick...Example: Bullet hitting stick...
Conserve angular momentum around the pivot (z) axis! The total angular momentum before the collision is due
only to the bullet (since the stick is not rotating yet).
v1
D
M
before
D/4m
Physics 1501: Lecture 22, Pg 29
Example: Bullet hitting stick...Example: Bullet hitting stick...
Conserve angular momentum around the pivot (z) axis! The total angular momentum after the collision has contributions
from both the bullet and the stick.
where I is the moment of inertia
of the stick about the pivot.
v2
F
after
D/4
Physics 1501: Lecture 22, Pg 30
Example: Bullet hitting stick...Example: Bullet hitting stick...
Set LBEFORE = LAFTER using
v1 v2
M
F
before after
mD
D/4
Physics 1501: Lecture 22, Pg 31
Example: Throwing ball from stoolExample: Throwing ball from stool
A student sits on a stool which is free to rotate. The moment of inertia of the student plus the stool is I. She throws a heavy ball of mass M with speed v such that its velocity vector passes a distance d from the axis of rotation. What is the angular speed F of the student-stool
system after she throws the ball ?
top view: before after
d
vM
I I
F
Physics 1501: Lecture 22, Pg 32
Example: Throwing ball from stool...Example: Throwing ball from stool...
Conserve angular momentum (since there are no external torques acting on the student-stool system):LBEFORE = 0 , Lstool = IF
LAFTER = 0 = Lstool - Lball , Lball = Iballball = Md2 (v/d) = M d v
0 = IF - M d v
top view: before after
d
vM
I I
F
F = M v d / I