READING QUIZ Torque primarily depends on: angular acceleration. angular velocity. angular mass.
Angular position and displacement Angular velocity Angular
Transcript of Angular position and displacement Angular velocity Angular
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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
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
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
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
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.
at t = 0:
Example:
A grindstone is rotating with constant angular
acceleration about a fixed axis in space.
Initial conditions
Positive directions:
right hand rule
rad/s 4.6 - rad/s 0.35 2 a 0
When is = 0 again in addition to t=0?
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
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
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
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 2 2
2 2 2 2 2 2 2 2
1 1 1 1( ) ( )
2 2 2 2
1 1 1 1
2 2 2 2
L L G G L L G G
L L G G L L G G
K m v m v m r m r
m r m r m r m r I
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:
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.
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
(could be positive or negative)
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.
PP10606-49*: When she is launched from a springboard, a diver's angular speed
about her center of mass changes from zero to 6.20 rad/s in 220 ms. Her rotational
inertia about her center of mass is constant at 12.0 kg·m2. During the launch, what
are the magnitudes of (a) her average angular acceleration and (b) the average
external torque on her from the board?
Example: second law for rotation
I totnet at
F=5 N
5 N tangential force is applied at the edge of a uniform disk
of radius 2 m and mass of 8 kg.
Find angular acceleration.
Formula: 21
2I MR
Axis of rotation
1m
F=5 N
5 N tangential force is applied as in the figure, on a uniform disk
of radius 2 m and mass of 8 kg.
Find angular acceleration.
Formula: 21
2I MR
Axis of rotation
F=5 N
5 N force is applied at 1 m from the center of a uniform disk of
radius 2 m and mass of 8 kg.
Find torque.
Axis of rotation
45 degree
Torque on extended object by gravitational force
Assume that the total gravitational force effectively
acts at the center of mass.
Gravitational potential energy of extended object
M g H, where H is the height of the center of mass and M
is the total mass.
Axis of rotation
Horizontal uniform rod
of length L & mass M
iClicker Q
Find the torque by gravitational force.
A. LMg
B. (L/2)Mg
C. 2LMg
D. (3/2)LMg
E. None of the above
Find the angular acceleration. 2
,
1
3end rodI ML
Axis of rotation
Horizontal uniform rod of length L & mass M is released from rest.
Example of energy conservation
Find its angular speed at the lowest point, assuming no friction
between axis of rotation and the rod.
Initial
Final
A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is
pivoted about a horizontal, frictionless pin through one
end of the rod. (The moment of inertia of the rod about
this axis is ML2/3.) The rod is released when it makes an
angle of 37° with the horizontal. What is the angular
acceleration of the rod at the instant it is released?
Example
Example: Torque and Angular Acceleration of a Wheel
r
a
mg
•Cord wrapped around disk, hanging weight • Cord does not slip or stretch • Let m = 1.2 kg, M = 2.5 kg, r =0.2 m • Find acceleration of mass m, find angular acceleration a for disk, tension, and torque on the disk
2
2
1MrI Formula:
Rolling : motion with translation and rotation about center of mass
vcm
total rot cmK K K
21
2rot cmK I
21
2cm cmK Mv
gravity cmU Mgh
mech totE K U
For rolling without slipping
s R
cmv R
mech non conservativeE W
Example: Use energy conservation to find the speed of the bowling ball as it rolls w/o slipping to the bottom of the ramp
Rotation accelerates if there is friction between the sphere and the ramp Friction force produces the net torque
and angular acceleration.
There is no mechanical energy change because the contact point is always at rest relative to the surface, so no work is done against friction
Formula: For a solid sphere
2
5
2MRIcm
Hint:
Given: h=2m
iClicker Q: A solid sphere and a spherical shell of the same radius r and same mass M roll to the bottom of a ramp without slipping from the same height h. True or false? : “The two have the same speed at the bottom.”
A) True
B) False. Shell is faster.
C) False. Solid sphere is faster.
D) Not enough information.
I_(cm, spherical shell) = (2/3) MR^2
I_(cm, solid sphere)=(2/5) MR^2