Transcript Ch33

Chapters 11 & 12: Angular
Momentum/Static Equilibrium
Chapters 11 & 12 Goals:
• we’ll do these chapters out of order
• to specialize, as in lab, to the case of no net force nor no net
torque: engineering statics
•To relax the assumption of a fixed rotation axis, and explore the
implications on how a net torque affects changes in angular
momentum
• to show how the translational and rotational versions of N2 allow
us to completely understand the motion of a body, at least
qualitatively
• to briefly understand the free symmetric top: letting the axis do
its own thing: gyroscopic motion
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The Idea of the Center of Gravity (CG)
• a system (may or may not be rigid) is acted on by gravity
because all of its masses have weight
• these forces act at different places, obviously, and we
regard them as ‘external’
• of course there may be forces beside gravity Nforces…
Fg   mi g  gM
• the total weight force is clearly
i 1
• this result assumes that g has the same value at all points
• what about torque due to weight force? Use ri = rCM + r’i
N
N
N
N
i 1
i 1
i 1
i 1
τ g   ri  mi g    mi rCM  r 'i  g    mi rCM  g    mi r 'i g 
N
 N

   mi rCM  g   mi r 'i  g  MrCM  g  0 [CM is at 0 as seen fromit]
i 1
 i 1 
τ g  rCM  Mg
• for gravity torque, gravity acts at
CMCG is same as CM
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The Science of Statics
• system feels no net force: Fext,net = 0  system momentum is
constant  CM does not accelerate, and we assume that it is
not translating either
• system feel no net force moment : text,net = 0  system
angular momentum is constant, and we assume no rotational
motion either
• This state of affairs is called static equilibrium
• upside: the right sides of the two N2 laws are zero
• downside: choice of origin is important, and one must count
equations carefully so that system is is not ‘overdetermined’
• example of an overdetermined system: 4-legged table!!
• a three-legged stool is stable… A four-legged chair will
always not quite be sure which 3 legs to use
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Example: the massive lever
250 N
100 N
50 kg
d
L=4m
S
W
FL
FR
Two forces act as shown.
The lever itself has a mass
of 50 kg, and has a length of
4 m. The whole system is
supported by the fulcrum, a
distance d from the left end.
Find support force and d.
• Take + to be UP and IN
• put all forces where they act (c.o.g.)
• Forces: – FL – W – FR + S = 0
•  S = FL + W + FR = 850 N
• Torques: put origin at (say) left end
• 0∙250 – d∙850 + 2∙500 + 4∙100 = 0
•  d = 1400/850 = 1.65 m
• a massless lever FL/FR = dR/dL
•‘The lever equation’
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A heavy sign (20 kg) hangs from a
massless bar and a support wire. Find the
Example: the building sign
tension, and find both components of the
support force S. Some kid has thrown
3
Solution bar
a pair of 3 kg shoes out at the end
0
q
• take the body to be the bar above the sign
c
• take + to be UP, to RIGHT, and IN
Fargo
m
• put all forces where they act (including CG)
• 3:4:5q  37°; sin q = .60; cos q = .80
• forces UP: S cos f – Fg + T sin q – fg = 0
40
• forces RIGHT: S sin f – T cos q = 0
cm
• put origin at left so S does not appear in equation!!
S
• torques IN: 0S + .20∙200 – .4T sin 143° – .4 ∙30 = 0
T
40  12
28
T

 117 N

f
.4 sin 143 .4.6
q
S x  S cosf  T cosq  133 N
Fg
fg
S y  S sinf  Fg  T sin q  f g  200 100 30  130 N
 Sy 
 S  S  S  (133)  130  186 N; f  T an    44.3
 Sx 
2
x
2
y
2
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2
-1
Not in book: the idea of stability in gravity
• as can tips, CG rises: stable
• once CG is directly above pivot: tipping point!
• from then on, object tips over and c.o.g. descends
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Angular momentum in terms of moment of inertia
• angular velocity vector w points along the axis by RHR
• However, L is parallel to that vector ONLY if there is no external
force moment on the body
• consider the ‘dumbbell’ comprising two masses m and a massless
stick of length L
O
• axis through the c.o.m.
perpendicular to the stick;
origin at c.o.m.

2
m
L
IC  2
4

2
m
L

2
• L = L1 + L2 and both masses contribute an identical angular
momentum; total magnitude is mL2w/2 and it points up, exactly
along w -- and it is a constant
• Therefore, there is no need for a force moment on system
• we have that L = ICM w in this case: L is parallel to w
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What happens if we alter the axis?
• put origin not at CM, but distance
O
D from it ┴
• angular velocity vector w points
along the axis by RHR
I A  I CM
1
 MD  ML2  MD 2
2
2
• we still have L = IA w and L is constant
• note now that because the CM is accelerating, there must be a net
horizontal force on the system… the thing is wobbly… the force at
this instant points to R since CM is accelerating centripetally in that
direction
• but there is still no need for a net torque since L is constant .. so L
is parallel to the angular velocity
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What happens if we alter
the axis differently?
• construct L = r x p for left mass
• p is out, so L is up and to left
• get same result for right mass
• clearly L CHANGES as τdumbell
dL

dt
whirls around and L is NOT
parallel to w: L ≠ ICM w
• furthermore, L is changing as the
thing flies around so by virtue of
r
L
O
p
net,ext
dL
τ net,ext 
dt
• we need a torque in this case!! Fnet,ext is however zero
• thus we need a torque couple.. how do we push at this
instant?
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Example IA
O
An object is made from a thin stick that is 75 cm in length and of
mass 2.4 kg. At one end there is a solid sphere of radius 5.0 cm
and mass 4.0 kg. The object turns about its center at 8 rad/sec.
Find the c.o.m. and find the contribution to I and K for each part.
Where is center of mass of system? Put origin at left end:
1
4(0)  2.4(37.5) 90 kg - cm
XC 
mi xi 

 14.1 cm
M
4  2.4
6.4 kg

What is I for left mass? Use parallel axis theorem and IC for ball:
I AL  I CL  m  mL R  mL  
2
2
5
2
L
2
2
5
4.05  4.375
2
2
 .0040 .563  .567 kg - m 2
2
2
2



I

I

m
L

2
.
4
.
75

.
113
kg
m
CS 12 S S
What is I for stick? AS
12
1
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1
Example IB
O
I A  I AL  I AS  .57  .11  .68 kg - m 2
Question: can we regard
this K as the K of c.o.m.
+ K referred to c.o.m.?
KA 
1
2
I
w
2 A
.68(8) 2

 21.76 J
2
 KA ?  ? 12 MVC2  12 ICw 2
2

6
.
4
(
8
)
.234
1
1
2
2 2
MVC  2 Mw  
 11.21 J
2
2
2
2 RL
1
64 

L2
2
2
2 
I w  2  mL  5  .141   mS 12  .234 
2 C

 

 324.0010 .0199  2.4.0469 .0548
2

 32.084 .244  10.50 J
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
11.21 J
+ 10,50 J
= 21.71 J
YES!!
Example II
A mass of 5 kg hangs from a string that is wrapped several
times around a hoop (radius 60 cm) with a horizontal
frictionless fixed axle. The mass has a = 4 m/s2
5
Find the angular
atan 4 m/s2
2
acceleration of the atan  R    R  .60 m  6.7 rad/s
hoop, and its mass.
Make FBDs for both objects. Put
T
origin at axle. Choose UP & IN  +
S
O
Force on m : T  mg  ma
T
F
g
F
 T  m( g  a )  5(10  4)  30 N
g
T
30
Momenton M : RT  I C  MR   M 

 7.5 kg
R .6(6.7)
2
Forceon M :  T  S  Mg  0  S  Mg  T  105N
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A taste of angular momentum conservation
By Example
We model an ice skater as a
cylinder (head/torso/legs) of
mass 50 kg and radius 25
cm) along with two
cylindrical arms, each of
mass 4 kg, radius 5 cm and
length 80 cm)
a) Find moment of inertia with arms DOWN
and with arms OUT
b) Argue that angular momentum should be
conserved when a skater spins, and then
assuming an initial rotation rate of 2
rotations/sec with arms OUT, find the final
rotation rate with arms DOWN.
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I C,axial  2 MR 2
1
example
II
 I C,DOWN
I A,normal@end  4 MR 2  3 ML2
1
1
  .05 m 2
2
 50 kg .25 m   24 kg  2  .30 m  


1
2
2
 1.56 kg - m 2  .73 kg - m 2  2.29 kg - m 2
• first term is torso axial c.o.m. moment of inertia
• factor of 2 x mass is for two arms of mass 4 kg
• first term in big parenthesis is arm’s axial c.o.m. IC
• second term in big parenthesis is parallel-axis
part
2
2
 I C,OUT 
1
2
50 kg.25 m2  24 kg .054m 


.80 m 
3

2
 .25 m  


 1.56 kg - m 2  2.21 kg - m 2  3.77 kg - m 2
• L is conserved because the external force moment is zero!!
 Linit  Lfin  I initwinit  I fin wfin  wfin 

winit  2 rev
2
s
rad
rev
 4
rad
 wfin
s
I
init w
init
I
fin
 2.29  4 rad   6.6
3.77

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rev 
rad
s
How does torque affect angular momentum,
in general?
• consider a wheel (a hoop,
say), that is rotating with
M, R
d
L
a large angular velocity
(and hence a large angular
O
momentum
• the wheel is supported at one point only at the end of its axis a
distance d from wheel’s CM
• put an origin at the support point
• the net external torque tnet,ext on the wheel is therefore of
magnitude Mgd, and it points which way? This is the same as
the (instantaneous) direction of dL/dt, of course!!
• right! into the page!! So L’s change is into the page, which
means that L moves around on a big circle
• The wheel does NOT fall over.. It precesses
dL
dw
dw gd
 Mgd φˆ  MR 2
Mgd 
 2  rate of precession
dt
dt
dt
R
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τ net,ext 