Rotational Motion
• Rotation (rigid body) versus translation (point particle)
• Rotation concepts and variables
• Rotational kinematic quantities
ƒ
ƒ
Angular position and displacement
Angular velocity
ƒ Angular acceleration
• Rotation kinematics formulas for constant angular acceleration
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“Radian”
“radian” : more convenient unit for angle than degree
Definition:
• 2
π radians = 360 degree 1 radian =
360
2 π o
=
180 o
π
= 57 .
3 o arc length ≡ s = r θ (in radians) θ rad
≡ r s
= 2 π r
×
θ
2 π
= r
θ
θ ’ r s
Example: r = 10 cm, θ = 100 radians Æ s = 1000 cm = 10 m.
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Rigid body
Rigid body:
A “rigid” object, for which the position of each point relative to all other points in the body does not change.
Example:
Solid: Rigid body
Liquid: Not rigid body
Rigid body can still have translational and rotational motion.
Angular position of rotating rigid body
• By convention, θ is measured CCW from the x-axis
• It keeps increasing past 2 π , can be y
• Each point of the body moves around the axis in a circle with some specific radius rotation axis “o” fixed to body parallel to z-axis
θ x
Reference line rotates with body rigid body
2

Angular displacement of rotating rigid body y
Angular displacement:
• Net change in the angular coordinate
Δ θ ≡ θ final
− θ inital
(an angle in rad.)
Arc length: Δ s
• Measures distance covered by a point as it moves through Δθ (constant r)
Δ s ≡ r Δ θ
(a distance along a circular arc) rotation axis “o” fixed to body parallel to z-axis
θ x
Reference li t t with body rigid body
Reference line rotating with body y
Δ s = r Δθ r
θ f
θ o x r
Rigid body rotation: angular & tangential velocity
Angular velocity ω :
• Rate of change of the angular displacement
ω ave
≡
Δ
Δ
θ t
ω inst
≡
Δ
Lim t → 0
Δ
Δ
θ t
≡ d θ dt
• Units: radians/sec. Positive in Counter-Clock-Wise sense
For any point, r is the perpendicular distance to the rotation axis v
T
• Frequency f = # of complete revolutions/unit time
• f = 1/T T = period (time for 1 complete revolution r
θ = ωΔτ
ω = 2 π f = 2 π /T f = ω /2 π x
Tangential velocity v
T:
• Rate at which a point sweeps out arc length along circular path s r
θ
Æ
Δ
Δ t s
= r
Δ
Δ
θ t
Æ v
T
= r
ω
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iClicker Quiz
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:
A.
79 rpm
B.
d/
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 v
Δ s ≡ r Δ θ
T
= ω r
ω = 2 π f = 2 π /T
Rigid body rotation: angular acceleration
Angular acceleration α:
• Rate of change of the angular velocity
• Units: rad/s 2
• CCW considered positive
• for CONSTANT α: ω f
= ω
0
+ α Δ t
α ave
≡
Δ ω
Δ t
α inst
≡
Lim t 0
Δ ω
Δ t
= d
ω dt
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1D and Angular Kinematics Equations
(Same mathematical forms)
1D motion with constant acceleration a x(t), v(t), a(t)
Angular motion with constant angular acceleration α variables θ( t ), ω( t ), α( t ) v = dx dt a = dv dt
Definitions ω = d θ dt
α = d ω dt v f
( t ) = v
0
+ at v x f
( t ) = f f
2 ( ( t ) = v x
0
+ v
0 t +
2
0
+ 2 a [ [ x f f
1
2 at 2
− x ] ]
Kinematic
Equations
ω f
( t ) = ω
0
+ α t
θ f
( t ) = θ
0
+ ω
0 t +
1
2
α t 2 f f
2
( ( t t ) ) = 2
0
2 α [ f f
− ] ]
Rotational variables are vectors, having direction
The angular displacement, speed, and acceleration
( θ, ω , α ) 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 h di quantity.
f h l
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Example:
A grindstone is rotating with constant angular acceleration about a fixed axis in space.
Initial conditions at t = 0:
α = 0.35
rad/s
2 ω
0
= 4.6
rad/s
When is Δ
θ
= 0 again in addition to t=0?
Positive directions: right hand rule
Example: Wheel rotating and accelerating
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?
ω f
( t ) = ω
0
+ α t
θ f
( t ) = θ
0
+ ω
0 t +
1
2
α t
2
ω f
2
( t ) = ω 2
0
+ 2 α [ θ f
− θ
0
]
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Rigid body rotation: radial and tangential acceleration
Centripetal (radial) acceleration a c or a r
• Radial acceleration component, points toward rotation axis a r
= v
T
2 r
= ω 2 r (use v
T
= ω r
)
F r
= ma r a c v
ω,α a
T r x
Tangential acceleration a
T
:
• Tangential acceleration component
• Proportional to angular acceleration α and also to radius r
• Units: length / time 2 a
T
= F tangential
= ma
T
Rotation variables: angular vs. linear s
= r Δθ v
T
= r ω a
T
= r
a r
v
T
2
2
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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
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
We want something like “F=ma” for rotational motion…..
• Moment of inertia – rotational analog of mass
• Torque – rotational analog of force
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Something like mass for rotational motion: Moment of Inertia, I
G
L
Kinetic energy of ladybug and gentlemanbug
K
=
1
2
=
2
2 +
1
2
2 ω 2 +
2
2 =
2 ω 2
1
2 m r
L
ω ) 2
=
2
(
+
1
2 m r
G
ω ) 2 m r
2 + m r
2
)
ω 2 =
I
m r
2
L L
m r
2
G G
Generally, I
= m r
2 + m r
2 + m r
2 + ...
2
I
ω 2
Kinetic energy: K
=
1
2
I
ω 2
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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
B d d
A d
2 d
C m
Rotational inertia I depends on axis chosen m
A) Choose rotation axis perpendicular to figure through point “A” h i di l t fi th h i t “B”
C) Let rotation axis pass through points “B” and “C”
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 ”.
Since
F
T
F
θ m
Newton’s Law along tangential direction
F
T
= ma
T
= m r
α
Multiplying “r”, so that we have “I” on right side axis r p
F
T
=
F sin θ rF
T
= m r
2
So, let’s define torque as
α =
I
α
≡ rF
T
Then we got
τ =
I
α and r p
r sin
τ = rF
T
= rF sin θ = r F p
(could be positive or negative)
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F
T
F
θ m r r p axis
τ = rF
T
= rF sin θ =
If r = 0, torque is zero.
If theta = 0 or 180 degree, the torque is zero.
For multiple forces
net
3
net
=
I
...
m
1 m
2 m1=100 kg adult, m2=10 kg baby.
Distance to fulcrum point is 1 m and 11 m respectively.
t t t h i t l
Which direction will it rotates?
iti f t
(a) Counter-Clockwise
(b) Clockwise
(c) No rotation
(d) Not enough information
Example: Find the net torque, moment of inertia, and initial angular acceleration.
Choose axis of rotation through fulcrum point.
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Example: second law for rotation
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·m 2 . During the launch, what are the magnitudes of (a) her average angular acceleration and (b) the average external torque on her from the board?
G net
= I tot
G
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.
F=5 N
Formula:
=
1
I MR
2
2
Axis of rotation
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5 N tangential force is applied at 1 m from the center of a uniform disk of radius 2 m and mass of 8 kg.
Find angular acceleration.
Formula:
=
1
I MR
2
2
Axis of rotation
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.
45 degree
F=5 N
Axis of rotation
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Torque on extended object by gravitational force
Æ Assume that the total gravitational force effectively acts at the center of mass.
Gravitational potential energy of extended object
Æ M g H, where H is the height of the center of mass and M is the total mass.
iClicker Q
Axis of rotation l if of length L & mass M d
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.
I
=
1
3
ML
2
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Example of energy conservation
Axis of rotation
Horizontal uniform rod of length L & mass M is released from rest.
Find its angular speed at the lowest point, assuming no friction between axis of rotation and the rod.
Example
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 ML 2 /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?
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Example: Torque and Angular Acceleration of a Wheel
•Cord wrapped around disk, hanging weight
• Cord does not slip or stretch g, g, m
• Find acceleration of mass m, find angular acceleration α for disk, tension, and torque on the disk a
Formula: I =
2
Mr mg r
Rolling : motion with translation and rotation about center of mass
ω
K total
=
K rot
+
K cm
K
=
1
I
2 v cm
U
K cm
1
2
Mv
2 cm gravity
=
Mgh cm
E mech
=
K tot
+
U
Δ
E mech
=
W
For rolling without slipping s R v cm
=
R
ω
θ
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Example: Use energy conservation to find the speed of the bowling ball as it rolls w/o slipping to the bottom of the ramp
Given: h=2m
Hint:
ƒ Formula: For a solid sphere
I cm
= 2
5
MR
2
ƒ 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 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
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