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Physics 100
Fall 2005
Lecture 11
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Physics 100
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Radians and Other Angular
Units
Define radian
q R
R
q = 1 radian (rad.)
360o = 2p rad
180o = p rad
90o = p/2 rad
etc.
1 rad  57.3o
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Vector or cross products
C  A B
Note that this yields another vector
(actually an axial vector)
Contrast with the scalar product:
q
C  A B
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A B
yielding a scalar
A  B  AB cos q  Ax Bx  Ay By  Az Bz
iˆ
C  Ax
Bx
ˆj
Ay
By
C x  Ay Bz  Az B y
kˆ
Az
Bz
C y  Az Bx  Ax Bz
C z  Ax B y  Ay Bx
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C  A B
C  AB sin q
 To the plane of AB
x
C
Right Hand Rule
‛x A
Direction of advance of a
right hand screw
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C  A B
Note that the vector product is not commutative
A  B  B  A
Again look at Right Hand Rule
x
C
A×B
‛x A
B×A
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Relation between Angular and
Linear Kinematics
v
s
r
s  rq arc length
v = rw
a  ra
Different points on a rigid body do not
Have the same s, v, a but do have the
same q, w, a
Direction of w is given by the right hand rule
z
Right hand
Fingers are direction of q
w
Note: product is not
y
Thumb is direction of
commutative
v  wr
q
r
v
x
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Point rotating a distance r from an axis
v
r
P
w out
v  wr
Apply Right Hand Rule: Rotate w into r, right thumb
points in direction of v
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Magnitude
Direction
v
w
r
Vector vs. Axial Vector
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Rotational Dynamics
y
axis
F4
O
x
F1
F2
F3
Forces pointing thru the axis (F1, F4) do not affect rotation
Forces  x, with lever arm ≠ 0, ( F2, F3), have maximal
effect and the effect is proportional to the length of the
lever arm.
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Torque
N  r F
(m·N)
N   in book
N = r F sin q
= r F when r  F
Torque is the rotational analog of force.
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Into paper
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N  rF sin q
q
Into the paper
F
r
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Example: Compound Wheel, R1 =30 cm, R2=50 cm
N  R1F1 sin 90  R2 F2 sin 60
 (0.30 N)(50 N)  (0.5 m)(50 N)sin 60
 6.7 mN
Net torque acts to accelerate rotation
clockwise. Right Hand Rule
N points into the page
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Moment of Inertia
a. k. a. rotational inertia
Consider a collection of particles each having kinetic energy
Ki = ½ mi vi2 = ½ mi ri2 w2
Rigid body
Then the total kinetic energy of the rigid body is
K=½
Define I =
(S m r ) w
(S m r )
2
i i
i i
2
2
Moment of inertia (kg m2)
Rotational analog of mass
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Newton’s First Law for Rotation
N  r F
F 0 N 0
massless rod
An object at rest or in uniform
rotational motion, in the absence of
externally imposed torques, remains at
rest or in uniform rotational motion.
If you don’t do anything, nothing happens
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Statics and Equilibrium
Statics: A special case of dynamics in which bodies remain at
rest in static equilibrium.
For a rigid body to be in equilibrium both linear and angular
accelerations must be 0.
F   Fi  0 (net force = 0)
Translational equilibrium:
or the 3 scalar equations
i
Fx   Fix  0
3  axes, all zero
i
Fy   Fiy  0
i
Fz   Fiz  0
i
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Rotational equilibrium:
N   Ni  0 (net torque = 0)
i
again 3 scalar equations for the
torques around 3  axes.
Center of Gravity and Center of Mass:
For g uniform, i . e. small heights above Earth’ surface the
C. M. and the C. G. are coincident.
i mi gxi i mi xi
xCG 

 xCM
 mi g  mi
i
i
similarly for y and z
A translational force like gravity acts like the total force is
F  g  mi acting downward at the C. G./C. M.
i
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Experimental Determination of the C. G.:
Suspend an object. Extending the line of suspension
downward will pass thru the C. G.
A
B
Hang from another point
b
a
Dashed lines intersect at the C. G.
C. G.
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Examples of Equilibrium:
F1
F2
m
F1 = -F2 translational
F=0
F1
F2
r
x
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N1 = -N2 = F1 r = -F2 r
N=0
Physics 100
rotational
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Stable, Unstable and Neutral Equilibrium
Earth’s gravitational field g
Potential energy as a function of position
U
unstable
unstable
neutral
neutral
stable
stable
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For an extended body
F or F
F or F
F
or
F

C. M. is raised
stable
Pendulum with rigid
cable
C. M. falls
unstable
C. M. neither rises
nor falls
neutral
unstable
stable
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Tip Over Angle:
mg
mg
Slides
q
q
Tips over
Arrows () refer to direction of mg from the CM not the
magnitude.
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Newton’s Second Law for Rotation
N  r F
F  ma
N  mar
(translational)
 everything is 
massless rod
Recall from rotational kinematics
a  a  r or in the  case a  a r
N   mr 2  a
but, I  mr 2 is just the moment of inertia
 N  Ia or vectorially N  Ia
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Angular momentum
N=Ia
Or
N = DL / Dt
Where L = I w is called the angular
momentum
Lrp
L is another conserved quantity
e. g. ice skater, piano stool, bicycle stability,
gyroscopes, stellar collapse, etc.
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Conservation of angular momentum
No net torque, decrease I => w must increase ad vice versa
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Kepler’s Laws
I.
Planetary orbits are ellipses with the sun at one focus.
f
•
f’
II. Equal fractional areas of the ellipses are swept out in equal
times.
Conservation of L
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Newton’s 3rd Law for rotation
Conservation of angular momentum
I f w f  I iwi
Action vs. reaction
Twist something, something twists on you
Action reaction pairs
couples
A pair of forces with equal magnitudes, opposite
directions and different lines of action.
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What is the torque produced
about the 3 axes A, B and C?
A
F
N  N A  NB
L
2
B
F
L
2
C
At A,
N  0  FL   FL
At C,
N   FL  0   FL
L
L
N   F  F   FL
2
2
At B,
A couple produces a torque that does not depend on the
location of the axis.
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Example: Stellar Collapse
Inner core of a larger star collapses into a neutron star of very small radius
r = rsun = 7 x 105 km, m = 2 msun,, T = 10 d, rn-star = 10 km
Assume no mass is lost in collapse. What is n-star’ rate of rotation?
Conservation of L:





 Ii
I f w f  I iwi  w f  
I
 f

 wi

2

mr 2 
2


I
r
5
wi  f n  star   i  fi  2
fi



2
rn  star
 If 
mrn2 star 
5

1
r 2 1 7 105 km
1
f 
so f n  star 

T
rn  star T
10 km (10d)(24 h /d)(3600 s /h)
 6 103 Hz
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Rotational energy and work
Work: Exert a torque through an angular displacement
W  Nq
Potential energy is increased by the doing of work
Kinetic energy of rotation
1 2
Kq  I w
2
This makes conservation of the total energy a little more complicated
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Wheel rolling down incline
m, I
Wheel is at rest at top. Initial energy
is all potential. U = mgh
h
v, w v  wr
At bottom wheel is rolling and has lost all its potential energy
w.r.t. ground. Total energy is all kinetic.
Kinetic energy has two parts, translational and rotational
1 2 1 2
K  mv  I w
2
2
Conservation of total energy here requires
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1 2 1 2
mgh  mv  I w
2
2
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Displacement
Rotation
Velocity
Angular velocity
Acceleration
Angular acceleration
Mass
Moment of inertia
Force
Torque
Momentum
Angular momentum
Energy
Rotational energy
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