chapter 8: rotational motionkenahn/16spring/phys102/lecture/l19-l20.pdf · 1 1 chapter 8:...
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Chapter 8: Rotational Motion
• Rotation (rigid body) versus translation (point particle)
• Rotation concepts and variables
• Rotational kinematic quantities Angular position and displacement
Angular velocity
Angular acceleration
• Rotation kinematics formulas for constant angular acceleration
“Radian”
radians) (in r slength arc r
srad
Example: r = 10 cm, = 100 radians s = 1000 cm = 10 m.
Definition:
• 2p radians = 360 degree
ooo
.π
π
radian 357180
2
3601
s
r ’
“radian” : more convenient unit for angle than degree
( )2
2
in rads r r
p
p
2
Rigid body:
A “rigid” object, for which the position of each point relative
to all other points in the body does not change.
Rigid body can still have translational and rotational motion.
Rigid body
Example:
Solid: Rigid body
Liquid: Not rigid body
• By convention, is measured CCW
from the x-axis
• It keeps increasing past 2p, can be
negative, etc.
• Each point of the body moves around
the axis in a circle with some specific
radius x
y
rigid body rotation axis “o” fixed to body
parallel to z-axis
Reference line rotates with body
Angular position of rotating rigid body
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Angular displacement:
• Net change in the angular coordinate
rad.) in angle (aninitalfinal
Arc length: s
• Measures distance covered by a point as it moves
through (constant r) y Reference line rotating with body
x
s = r
o
f r r
arc) circular a along distance (a rs
x
y
rigid body rotation axis “o” fixed to body
parallel to z-axis
Reference line rotates with body
Angular displacement of rotating rigid body
Rigid body rotation: angular & tangential velocity
Tangential velocity vT:
• Rate at which a point sweeps out arc length along
circular path
Tv r
Angular velocity :
• Rate of change of the angular displacement
dt
d
t
tLim
tinstave
0
• Units: radians/sec. Positive in Counter-Clock-Wise sense
• Frequency f = # of complete revolutions/unit time
• f = 1/T T = period (time for 1 complete revolution
/2f /T2f ppp 2x
vT
t
r
For any point, r is the perpendicular
distance to the rotation axis
s r s
rt t
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1.1. The period of a rotating wheel is 12.57 seconds. The radius of the wheel is 3 meters. It’s angular speed is closest to:
iClicker Quiz
A. 79 rpm
B. 0.5 rad/s
C. 2.0 rad/s
D. .08 rev/s
E. 6.28 rev/s
1.2. A point on the rim of the same wheel has a tangential speed closest to:
A. 12.57 rev/s
B. 0.8 rev/s
C. 0.24 m/s
D. 1.5 m/s
E. 6.28 m/s
/T2f pp 2
rvT
rs
Rigid body rotation: angular acceleration
Angular acceleration a:
• Rate of change of the angular velocity inst0
=t
avet
dLim
t dt
a a
• Units:
• CCW considered positive
• for CONSTANT a: tf a 0
2rad/s
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1D and Angular Kinematics Equations (Same mathematical forms)
dt
dva
dt
dxv
1D motion with
constant acceleration a
x(t), v(t), a(t) (t), (t), a(t)
Angular motion with
constant angular acceleration a
dt
d
dt
da
variables
Definitions
Kinematic
Equations
atv)t(vf 0
2
2
100 attvx)t(xf
]xx[av)t(v ff 020
22
t)t(f a 0
2
2
100 tt)t(f a
][)t( ff 020
22 a
Rotational variables are vectors, having direction
The angular displacement, speed, and acceleration
( , , a )
are vectors with direction.
The directions are given by the right-hand rule:
Fingers of right hand curl along the angular direction (See Fig.)
Then, the direction of thumb is the direction of the angular quantity.
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At t = 0, a wheel rotating about a fixed axis at a constant angular acceleration has an angular velocity of 2.0 rad/s. Two seconds later it has turned through 5.0 complete revolutions. Find the angular acceleration of this wheel?
Example: Wheel rotating and accelerating
t)t(f a 0
2
2
100 tt)t(f a
][)t( ff 020
22 a
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Rigid body rotation: radial and tangential acceleration
Centripetal (radial) acceleration ac or ar
• Radial acceleration component, points toward rotation axis
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) (use v r T
Tr
va r
r r rF ma
x
vT
,a
r ac
aT
Tangential acceleration aT:
• Tangential acceleration component
• Proportional to angular acceleration α and also to radius r
• Units: length / time 2
r Ta a tangential TF ma
Rotation variables: angular vs. linear
T
rΔθ
v rω
s
22 =r T
r
va
r
Ta ra
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A ladybug sits at the outer edge of a merry-go-round, and a
gentleman bug sits halfway between her and the axis of
rotation. The merry-go-round makes a complete revolution
once each second. The gentleman bug’s angular velocity is
A. half the ladybug’s.
B. the same as the ladybug’s.
C. twice the ladybug’s.
D. impossible to determine
G
L
A ladybug sits at the outer edge of a merry-go-round, and a
gentleman bug sits halfway between her and the axis of
rotation. The merry-go-round makes a complete revolution
once each second. The gentleman bug’s velocity is
A. half the ladybug’s.
B. the same as the ladybug’s.
C. twice the ladybug’s.
D. impossible to determine
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Rotational Dynamics
• Moment of inertia – rotational analog of mass
• Torque – rotational analog of force
We want something like “F=ma” for rotational motion…..
Something like mass for rotational motion: Moment of Inertia, I
Kinetic energy of ladybug and gentlemanbug
G L
2 2
L L G GI m r m r 2 2 2
1 1 2 2 3 3 ...I m r m r m r Generally, 21
2K IKinetic energy:
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Example: Find moment of inertia for a crossed dumbbell
•Four identical balls as shown: m = 2 kg
•Connected by massless rods: length d = 1 m.
m
m
m m
d
d
d d
A B C
d 2
A) Choose rotation axis perpendicular to figure through point “A”
B) Now choose axis perpendicular to figure through point “B”
C) Let rotation axis pass through points “B” and “C”
Rotational inertia I depends on axis chosen
Calculation of Moment of inertia for continuous mass distributions requires “Integration, a kind of calculus”. We will just use the result.
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Moments of Inertia of Various Rigid Objects
Now we want to define “torque, τ”, so that “τ = I α”.
F
rp
r
FT
axis
m T TF ma m r a
Newton’s Law along tangential direction
2 TrF m r Ia a
Multiplying “r”, so that we have “I” on right side
So, let’s define torque as TrFt
It aThen we get
sinTF F Since
sinT prF rF r Ft
sinpr r and
(Torque τ could be either positive (Counter-Clockwise) or negative(Clockwise)
Line of action
Moment arm
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net It a
1 2 3 ...nett t t t
For multiple forces
F
rp
r
FT
axis
m
sinT prF rF r Ft
If r = 0, torque is zero.
If theta = 0 or 180 degree, the torque is zero.
m1=100 kg adult, m2=10 kg baby.
Distance to fulcrum point is 1 m and 11 m respectively.
The seesaw starts at horizontal position from rest.
Which direction will it rotates?
(a) Counter-Clockwise
(b) Clockwise
(c) No rotation
(d) Not enough information
m1 m2
Example: Find the net torque, moment of inertia, and
initial angular acceleration.
Choose axis of rotation through fulcrum point.