Transcript 2053_Lecture_10-22-13

Angular Momentum
• The vector angular momentum of the point
mass m about the point P is given by:
y-axis
  
Lrp
r
The position vector of the mass m relative to the point P
is:

r  xxˆ  yyˆ
The momentum vector of the mass m is:

p  p x xˆ  p y yˆ  mvx xˆ  mvy yˆ
P
p
pperp

m
x-axis
Distance from the Point P to the
mass m times the perpendicular
component of the momentum.
The magnitude of the angular momentum is:
L  rp sin   rp perp
(units = kg∙m2/s)
The components of the angular momentum are:
Lx  0 Ly  0 Lz  xpy  yp x
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PHY 2053
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Torque
• The torque vector about the point P due to
the force F acting at r is given by:
y-axis
  
  r F
F
Fperp

r
The position vector of the mass m relative to the point P
is:

r  xxˆ  yyˆ

F  Fx xˆ  Fy yˆ
The force acting on the mass m is:
P
x-axis
Distance from the Point P to the
mass m times the perpendicular
component of the force.
The magnitude of the torque is:
  rF sin   rFperp (units = N∙m)
The components of the torque are:
 x  0  y  0  z  xFy  yFx
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PHY 2053
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Rotation: Angular Variables
• Arc Length:
The arc length s is related to the angle  (in radians = rad)
as follows:
s
(360o = 2p rad)

s  r
r
• Angular Displacement and Angular Velocity:
   f   i
 d
  lim

t 0 t
dt
(radians/second)
• Tangential Velocity and Angular Velocity:
 ds
vt  vt 
 r
dt
vt  r
Tangential
Velocity
• Angular Acceleration:
 d
  lim

t 0 t
dt
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(radians/s2)
PHY 2053
Page 3
Rolling Without Slipping:
Rotation & Translation
• If a cylinder of radius R rolls without
slipping along the x-axis then:
v

x  s  r
R
s
x-axis
x
dx
d
v
R
 R
dt
dt
Translational
Speed
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Rotational
Speed
PHY 2053
Page 4
Translation vs Rotation
• Translation:
• Rotation:
Mass: m
Position: x
Velocity: vx
Acceleration: ax
Force:
Moment of Inertia: I
Angular Position: 
Angular Velocity: 
Angular Acceleration: 

F
Torque:


L  I
KErot  12 I 2


p  mv
If
KEtrans  mv


 F  0 then p 
2
constant
If
Momentum Conservation!
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

 dL

  I  dt


 dp
 F  ma  dt
1
2

PHY 2053

  0 then

L  constant
Angular Momentum
Conservation!
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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%
s  R
v  R
a  R
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KErotation  12 I 2
KEtranslation  12 Mv 2
KEtranslation
2
3
Iv
 12 Mv  3KErotational  3( 12 I ) 
2R 2
2
2
I  13 MR 2
PHY 2053
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Example: Rolling without Slipping
• 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?
Ei  Mg (h  R)
R
h

v  R
Ei  E f
E f  Mv  I  MgR  Mv (1  I /( MR ))  MgR
Mg (h  R)  12 Mv 2 (1  I /( MR ))  MgR
1
2
2
1
2
2
v  2 gh /(1  I /( MR 2 ))
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2
1
2
2
2
• Example: I = MR2/2 (solid cylinder), h =
9.8 m then
v
PHY 2053
4
3
gh  (9.8m / s)
4
3
 11.3m / s
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Example: Rolling without Slipping
R
• If a cylinder with moment of inertia I
and radius R starts from rest at a
height h above the ground and rolls h
without slipping down an incline. If
h
the cylinder starts from rest at t = 0,
sin  
when does it reach the ground?
d
2
v
2
2
a

v  2ad
v

2
gh
/(
1

I
/(
MR
))
2d
gh
g sin 
2
1
d

at
a

2
2
2
d (1  I /( MR )) (1  I /( MR ))
2h(1  I /( MR 2 ))
t
g sin 2 
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d

h
d
sin 
2d
t
a
• Example: I = MR2/2 (solid cylinder), h =
9.8 m,  = 45o then
t
PHY 2053
3h
6h

 2.45s
2
g sin 
g
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Exam 2 Fall 2010: Problem 14
• 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%



 wall   gravity  0
P
3m

L
L x
 wall   gravity  Mg sin   W
2
2 L
 12 Wx  12 (100 N )(3m)  150 N  m
 1
r  2 xxˆ  12 yyˆ

F   Mgyˆ
 z  xFy  yFx   12 xMg   12 xW
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4m
PHY 2053
Fwall

L
r

Mg
P
x
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Exam 2 Fall 2011: Problem 13
• 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%
L
Ei  Mgh  Mg cos 
2
E f  12 I 2f  12 I
hinge

L
h  cos 
2
v 2f
2
L
Ei  E f
mgL3 cos 
mgL3 cos 
vf 

 3gL cos 
2
1
I
3 mL
 3(9.8m / s 2 )( 2m) cos( 20 )  7.43m / s
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University of Florida
PHY 2053
Page 10