Download - Angular Momentum
R. Field 10/22/2013 University of Florida
PHY 2053 Page 1
Angular Momentum• The vector angular momentum of the point
mass m about the point P is given by:
prL
perprprpL sin
x-axis
r
P
m
p
y-axis pperp
The position vector of the mass m relative to the point P is:
(units = kg∙m2/s)
yyxxr ˆˆ
The momentum vector of the mass m is:
ymvxmvypxpp yxyx ˆˆˆˆ
The magnitude of the angular momentum is:
xyzyx ypxpLLL 00The components of the angular momentum are:
Distance from the Point P to the mass m times the perpendicular component of the momentum.
R. Field 10/22/2013 University of Florida
PHY 2053 Page 2
Torque• The torque vector about the point P due to
the force F acting at r is given by:
Fr
perprFrF sin
x-axis
r
P
F
y-axis Fperp
The position vector of the mass m relative to the point P is:
(units = N∙m)
yyxxr ˆˆ
The force acting on the mass m is:
yFxFF yx ˆˆ
The magnitude of the torque is:
xyzyx yFxF 00The components of the torque are:
Distance from the Point P to the mass m times the perpendicular
component of the force.
R. Field 10/22/2013 University of Florida
PHY 2053 Page 3
Rotation: Angular Variables
The arc length s is related to the angle (in radians = rad) as follows:
• Arc Length:
rs r
s
if • Angular Displacement and Angular Velocity:
rdt
dsvv tt
dt
d
tt
0
lim
• Tangential Velocity and Angular Velocity:
(radians/s2)
• Angular Acceleration:
(360o = 2 rad)
Tangential Velocity
rvt
dt
d
tt
0
lim
(radians/second)
R. Field 10/22/2013 University of Florida
PHY 2053 Page 4
Rolling Without Slipping: Rotation & Translation
R
v
s x-axis
x
• If a cylinder of radius R rolls without slipping along the x-axis then:
rsx
R
dt
dR
dt
dxv
Translational Speed
Rotational Speed
R. Field 10/22/2013 University of Florida
PHY 2053 Page 5
Translation vs Rotation• Translation:
Mass: m
• Rotation:Moment of Inertia: I
Force: Torque: F
Position: x Angular Position:
Velocity: vx Angular Velocity:
Acceleration: ax Angular Acceleration:
dt
pdamF
dt
LdI
vmp
IL
221 mvKEtrans
221 IKErot
0 F
0If p
then L
If constant
then constant
Momentum Conservation! Angular Momentum Conservation!
R. Field 10/22/2013 University of Florida
PHY 2053 Page 6
Exam 2 Fall 2010: Problem 11
• A non-uniform cylinder with mass M and radius R rolls without sliding along the floor. If its translational kinetic energy is three times greater than its rotational kinetic energy about the rotation axis through its center of mass (i.e. the central axis of the cylinder), what is its moment of inertia about the central axis?Answer: MR2/3 % Right: 44%
221 MvKE ntranslatio
Ra
Rv
Rs
221 IKErotation
2
22
212
21
2
3)(33
R
IvIKEMvKE rotationalntranslatio
231 MRI
R. Field 10/22/2013 University of Florida
PHY 2053 Page 7
Example: Rolling without Slipping R
h
• If a cylinder with moment of inertia I and radius R starts from rest at a height h above the ground and rolls without slipping down an incline. What is its translational speed when it reaches the ground?
)( RhMgEi
MgRMRIMvMgRIMvE f ))/(1( 22212
212
21
Rv
MgRMRIMvRhMg ))/(1()( 2221
fi EE
))/(1/(2 2MRIghv • Example: I = MR2/2 (solid cylinder), h =
9.8 m then
smsmghv /3.11)/8.9( 34
34
R. Field 10/22/2013 University of Florida
PHY 2053 Page 8
Example: Rolling without Slipping R
h
d
• If a cylinder with moment of inertia I and radius R starts from rest at a height h above the ground and rolls without slipping down an incline. If the cylinder starts from rest at t = 0, when does it reach the ground?
))/(1/(2 2MRIghv adv 22 d
va
2
2
d
hsin
))/(1(
sin
))/(1( 22 MRI
g
MRId
gha
221 atd
a
dt
2
sin
hd
2
2
sin
))/(1(2
g
MRIht
• Example: I = MR2/2 (solid cylinder), h = 9.8 m, = 45o then
sg
h
g
ht 45.2
6
sin
32
R. Field 10/22/2013 University of Florida
PHY 2053 Page 9
Exam 2 Fall 2010: Problem 14
x
P
Fwall
Mg
L
r
• A 100-N uniform plank leans against a frictionless wall as shown. What is the magnitude of the torque (about the point P) applied to the plank by the wall?
Answer: 150 N∙m % Right: 42%
3 m
4 m
P 0 gravitywall
mNmNWxL
xWL
MgL
gravitywall
150)3)(100(2
sin2
21
21
yyxxr ˆˆ21
21
yMgF ˆ
xWxMgyFxF xyz 21
21
R. Field 10/22/2013 University of Florida
PHY 2053 Page 10
Exam 2 Fall 2011: Problem 13
hinge • A thin stick with mass M, length L, and moment
of inertia ML2/3 is hinged at its lower end and allowed to fall freely as shown in the figure. If its length L = 2 m and it starts from rest at an angle = 20o, what is the speed (in m/s) of the free end of the stick when it hits the table?Answer: 7.43 m/s % Right: 14% cos
2
LMgMghEi 2
2
212
21
L
vIIE f
ff
smmsm
gLmL
mgL
I
mgLv f
/43.7)20cos()2)(/8.9(3
cos3coscos
2
231
33
fi EE
cos2
Lh