Rotation
•
The basic quantity in rotation is the angular displacement
– similar role to the linear displacement studied in chapter 1
– units: radians, revolutions, degrees, …
θ
– Problem solving strategy is the same as for problems with linear
motion...
Phy1222 - Spring 2003
1
Rotation
•
•
Problems with constant angular acceleration are solved just like those
with constant linear acceleration.
Example (linear): A car can accelerate at 5.8 m/s 2. If it starts from
rest, how far does it travel before it reaches a speed of 25 m/s?
v=25m/s
v=0
∆x
•
Example (angular): A potter’s wheel can accelerate at 5.8 rad/s2. If it
starts from rest, how many revolutions does it make before it reaches
an angular velocity of 25 rad/s?
Phy1222 - Spring 2003
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1
Analogies between Rotation and Linear Motion (I)
Rotational
•
•
•
angular position: θ (radians)
angular velocity: ω (rad/sec)
angular acceleration: α (rad/s 2)
Linear
• position: x (meters)
•
•
velocity: v (m/s)
acceleration: a (m/s2)
ω = ω 0 + αt
v = v0 + at
θ − θ 0 = ω0 t + 12 αt 2
x − x0 = v0t + 12 at 2
ω 2 = ω02 + 2α (θ − θ 0 )
v 2 = v02 + 2a( x − x0 )
Phy1222 - Spring 2003
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Relation between linear and angular variables
•
s = rθ
Fundamental relations
distance traveled =
(radius from rotation center)(angular displacement [radians])
v = rω
linear velocity =(radius)(angular velocity[rad/s])
v
at = rα
at
ar θ
s
Tangential acceleration=
(radius)(angular acceleration [rad/s2])
v2
ar =
= ω 2r
r
r
Radial acceleration =
(angular velocity[rad/s]) 2(radius)
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2
Angular Variables
•
Relation between angular frequency and rotation frequency:
ω = (2πrad ) f
ω in rad/s = (2π)(frequency in Hz)
•
Reminder about frequency and period:
f =
ω=
1
T
2π
T
•
so
•
(radian units sometimes disappear in calculations)
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Angular and linear variables
•
Example: A computer hard disk rotates at 5400 rpm (rotations per
minute)
– what is its angular velocity?
– If the reading head is located 3.0 cm from the rotation axis, what is
the speed of the disk below it?
– What is the linear acceleration of that point on the disk?
– If each bit of information occupies 5.0µm of length on the disk,
how many bits per second can the writing head write?
– If the disk takes 4.0s to spin up from rest, what is the average
angular acceleration during spin up?
3.0 cm
Phy1222 - Spring 2003
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3
Kinetic energy of rotating object
•
For now, consider only a rigid body rotating about a fixed axis
v2
ω
r1
I = ∑ mi ri 2
K = 12 m1v12 + 12 m 2v22 + 12 m3 v32
K = 12 m1r12ω 2 + 12 m2 r22ω 2 + 12 m3 r32ω 2
=
v3
v1 = r1ω
v1
1
2
(m r
2
1 1
)
+ m2r22 + m3r32 ω 2
= 12 Iω 2
Rotational inertia = sum of (mass)(radius from axis)2
Phy1222 - Spring 2003
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Rotational inertia
I = ∑ mi ri2
•
Units: kg m2
plays a role in rotational motion equivalent to that of mass in linear
motion:
– resists angular acceleration
– kinetic energy is proportional to it
constructed so that the equation for kinetic energy of a rotating object
(K=½I? 2) look like that for an object in linear motion (K=½mv 2)
depends on:
•
– the mass of a body
– the location of the rotation axis
– the shape of the body
sometimes call “moment of inertia”
•
•
•
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4
Rotational Inertia
•
Example: An asymmetrical dumbbell is made from a 1kg mass and a
2kg mass connected by a rod (of negligible mass) or length 0.50m.
2.0 kg
1.0 kg
•
0.50m
What is the kinetic energy if the rod is rotated at 5 rev/s
– about a point in the center of the rod?
– about the center of the 1 kg mass?
Phy1222 - Spring 2003
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Rotational Inertia
•
For extended objects, the sum can be replace with an integral
I = ∑ ri 2mi
•
⇔
I = ∫ r 2 dm
Simplest example is a hoop rotated about an axis through its center
point, perpendicular to the plane of the hoop.
R
I = R 2 ∫ dm = R 2m
Phy1222 - Spring 2003
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5
Rotational Inertia
A few other important shapes:
• Solid disk/cylinder about center: I=½MR 2
R
•
Solid sphere about center: I=(2/5)MR2
•
Thin rod about center: I=(1/12)ML2
R
See HRW page 249 for
more extensive table
L
Phy1222 - Spring 2003
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Rotational Inertia
Composite shapes: Simply add rotational inertia, but be careful about axis
location:
• Example: Disk of mass M 1 and rod of mass M2 rotated about a
common central axis:
I = 12 M 1R 2 + 121 M 2 L2
Phy1222 - Spring 2003
R
L
12
6
Parallel Axis Theorem
•
Useful for calculating the rotational inertia about an axis parallel to an
axis through the center of mass:
h
I = I cm + Mh 2
Rotational inertia=
rotational interia about cm axis +
(total mass)(distance between axis)
•
Interesting correlary: of all the possible parallel axis, the one that gives
the smallest intertia is the one through the center of mass
Phy1222 - Spring 2003
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Parallel Axis Theorem
•
Example: A clock pendulum consists of a rod of length 20cm (of
negligible mass), with a 200 g disk of radius 5 cm at the end. What it
the rotational inertia about the pivot point?
20 cm
5 cm
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7
Torque
•
Torque plays the role in angular motion equivalent to that of Force in
linear motion
r⊥
r r r
τ = r×F
r
F⊥
φ
Torque = vector product of
displacement from axis and force
F
τ = r⊥ F = rF⊥
= rF sin φ
•
•
Magnitude of torque = (perpendicular moment arm)(Force)=
(pivot-to-force distance)(perpendicular component of force)
Units: Netwon-meters (not called “Joules” when you are talking about
torque)
Review HRW p. 46-47 for properties of vector product
Phy1222 - Spring 2003
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Torque
•
Analogy between rotational and linear:
r
r
τ = Iα
r
r
F = ma
⇔
torque = (rotational inertia)(angular acceleration)
•
Example: A rod 12 kg rod of length 4.0m is resting on a pivot which is
1.0 m from one end. What is the angular acceleration when the other
end of the rod is released?
mg
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8
Work and Power
•
Work: analogy with linear motion (constant torque, constant force):
W = τ∆θ
⇔
W = F ∆x
•
Example: A flywheel at rest is subject to a torque of 50 Nm from an
electric motor. How many turns will the flywheel make before it has a
kinetic energy of 1000 J?
•
Power: analogy with linear motion:
•
r r
P = τ ⋅ω
⇔
r r
P = F ⋅v
Example: If the flywheel in the example above has a rotational inertia
of 20 kgm 2, what is the power supplied by the motor when the
flywheel reaches 1000 J?
Phy1222 - Spring 2003
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Vectors in angular motion
•
Direction of angular velocity is perpendicular to plane of rotation with
direction given by right-hand rule:
r
ω
r
ω
•
Same with angular acceleration:
Speeding
up:
r
α
r
ω
Slowing
down:
r
α
r
ω
Phy1222 - Spring 2003
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9
Analogies between Rotation and Linear Motion (II)
Rotational
•
•
•
•
angular position: θ (radians)
angular velocity: ω (rad/sec)
angular acceleration: α (rad/s 2)
torque: τ (N· m)
Linear
• position: x (meters)
•
•
•
velocity: v (m/s)
acceleration: a (m/s2)
Force: F (N)
v = v 0 + at
ω = ω 0 + αt
θ − θ 0 = ω 0 t + αt 2
x − x 0 = v 0 t + 12 at 2
ω 2 = ω02 + 2α (θ −θ 0 )
v 2 = v 02 + 2 a ( x − x0 )
K = 2 Iω 2
r
r
τ = Iα
W = τ∆θ
r r
P = τ ⋅ω
K = 12 mv 2
r
r
F = ma
W = F∆ x
r r
P = F ⋅v
1
2
1
Phy1222 - Spring 2003
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Rolling Motion
•
Rolling is a particular combination of translational (linear) motion and
rotation motion. Consider one full revolution:
∆θ = 2π rad
∆x = 2πR
•
For rolling without slipping:
∆x = R∆θ
v = Rω
x,v,a all refer to the motion of the
center of the rolling object
a = Rα
Phy1222 - Spring 2003
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10
Rolling Motion
•
To get the velocity of any point on a rolling object, you must add the
vectors of the purely rotational velocity and the translational velocity
of the center:
=
+
A
B
C
•
•
Example: find the instantaneous speeds at points A, B, and C for a
hoop rolling at 10 m/s.
Note that the velocity at point C is zero. The point of contact is
instantaneously at rest. This is why you use static friction to evaluate
the frictional force on a rolling object.
Phy1222 - Spring 2003
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EX. A
bicycle tire rolls across the ground as shown below. What point on
the tire is moving straight down?
1. a
2. b
3. c
4. d
5. all of the above.
6. none of the above.
Phy1222 - Spring 2003
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11
Kinetic Energy for rotation + translation
•
Simply add the translational kinetic energy (of the whole mass moving
linearly with the center-of-mass velocity) plus the rotational kinetic
energy (of the purely rotational motion about the center of mass)
2
K = 12 Mv cm
+ 12 Iω 2
vcm
ω
Total kinetic energy =
translational KE +rotational KE
(true for all rigid bodies)
•
For rolling (without slipping) bodies, there is a special relation
between the translation and the rotation: v=Rω
K=
1
2
(MR
2
)
+ I ω2 =
1
2
(M + I / R )v
2
2
cm
(true for rolling objects of radius R)
Phy1222 - Spring 2003
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Kinetic energy of rolling bodies
•
Example: A solid disk rolls (without slipping) down a ramp starting
from rest to a table 2.0 meters below. What is the speed with which it
rolls along the table?
vf
2.0m
•
•
Note that the answer does not depend on
– the mass of the disk
– the angle or shape of the ramp
– the radius of the disk
It does depend on the shape (disk, hoop, sphere, etc.)
Phy1222 - Spring 2003
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12
Ex. A hollow cylinder and a solid cylinder start from rest at the same
position at the top of a ramp. Which has the larger center-of-mass
speed when it reaches the bottom of the ramp?
1. The hollow cylinder.
2. The solid cylinder.
3. The speeds are the same.
Phy1222 - Spring 2003
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Ex. Two solid cylinders with different masses start from rest at the same
position at the top of a ramp. Which has the larger center-of-mass
speed when it reaches the bottom of the ramp?
1. The heavy cylinder.
2. The light cylinder.
3. The speeds are the same.
Phy1222 - Spring 2003
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13
Forces and torques in rolling
•
Freely rolling object:
Mg sin θ − f s = Macm
N
Net force =
(mass)(acceleration of center of mass)
mg
α
a
fs R = Iα
fs
•
Net torque about cm=
(rotational inertia)(angular acceleration)
Example: If the coefficient of friction between a ball (sphere) and a
ramp is 0.20, what is the steepest angle where the ball will roll without
slipping?
Phy1222 - Spring 2003
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Yo-yo’s
•
Rolling object can drop more slowly if the mass is concentrated near
the edge of the body. Slowness characterized by
β=
•
I
MR 2
A yo-yo is a way of getting a very large β by using a small radius r0 to
fix the “rolling” and a large radius R to give a large rotational inertial:
v = ωr0
I ≈ MR
1
2
2
This is studied in detail
in the IPL (“Maxwell’s Wheel”)
R
r0
Phy1222 - Spring 2003
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14
Ex. Two brass knobs are added at the same distance from the center of
Maxwell's Wheel, which is then raised by winding the string around
the shaft. When the Wheel reaches the end of the string, its speed will
be
1. larger if the knobs are placed closer to the shaft.
2. larger if the knobs are placed farther from the shaft.
3. the same regardless of where the knobs are placed.
Phy1222 - Spring 2003
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Angular Momentum
•
Angular momentum is defined with respect to some fixed point:
r
r
p = mv
r
r
m
r r r
r r
l = r × p = m( r × v )
Angular momentum = vector product of
(vector from fixed point to object)×(linear momentum of object)
•
related to torque:
•
like momentum, this has a lot more applications when you consider a
system of particles rather than a single particle.
r
r
d l / dt = τ
Phy1222 - Spring 2003
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15
Angular Momentum
•
Example 1:
1 m/s
1 kg
A
2 kg
1m
B
1m
1 m/s
– What is the angular momentum of the system about point A?
– about point B?
Phy1222 - Spring 2003
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Angular Momentum
•
Example 2: A ball is located at position r=(3m)i+(4m)j and is has
linear momentum p=(2kgm/s)i+(1kgm/s)k. What is the angular
momentum of the ball with respect to the origin?
y
r
p
l x = ry p z − rz py
x
l y = rz p x − rx pz
l z = rx py − ry px
•
Answer: l=(4kgm 2/s)i+(-3kgm2 /s)j+(-8kgm2 /s)k
•
Check:
r r
r ⋅l = 0
r r
p⋅ l = 0
Phy1222 - Spring 2003
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16
Angular momentum
•
Rigid body rotation about fixed axis:
r
r
L = Iω
Angular momentum =
(rotational inertia)(angular velocity)
r
L
•
analogous to p=mv for linear momentum
•
Example: What is the angular momentum of the Earth due to its
rotation?
M = 6 ×1024 kg R = 6.4 ×10 6 m
E
E
I E ≈ 52 MR 2 = 9. 7 ×10 37 kgm 2
ω E = (2πrad) /(86400s) = 7 .3 × 10−5 rad/s
LE = 7.1×10 33 kgm 2 /s
Phy1222 - Spring 2003
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Conservation Angular Momentum
•
Angular momentum is conserved:
– For a system with no net external torque acting on it, the total
angular momentum is constant
– angular momentum is still conserved even if there are
• internal torques
• external forces (as long as they don’t produce torques)
•
Example: A (non-rotating) sack of rice with rotational inertia of 10
kgm 2 is dropped onto a disk rotating on at 20 rad/s. The disk has a
rotational inertia of 3.0 kgm 2. What is the rotational speed of the
system after the rice and disk couple?
Phy1222 - Spring 2003
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17
Ex. A dumbbell is rotating about its center as shown. Compared to the
dumbbell's angular momentum about its center, its angular momentum
about point B is
1. bigger.
2. the same.
3. smaller.
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Ex. A student, standing on a platform that rotates without friction, holds a
bicycle wheel that rotates counterclockwise (as seen from above).
After turning the axle of the bicycle wheel upside down, the student
and platform
1. remain stationary.
2. begin to rotate in a clockwise direction.
3. begin to rotate in a counterclockwise direction.
Phy1222 - Spring 2003
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18
Ex. A figure skater stands on one spot on the ice (assumed frictionless)
and spins around with her arms extended. When she pulls in her arms,
her angular velocity
1. remains the same.
2. increases.
3. decreases.
Phy1222 - Spring 2003
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Ex. A figure skater stands on one spot on the ice (assumed frictionless)
and spins around with her arms extended. When she pulls in her arms,
her angular momentum
1. remains the same.
2. increases.
3. decreases.
Phy1222 - Spring 2003
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19
Ex. A figure skater stands on one spot on the ice (assumed frictionless)
and spins around with her arms extended. When she pulls in her arms,
she reduces her rotational inertia and her angular speed increases so
that her angular momentum is conserved. Compared to her initial
rotational kinetic energy, her rotational kinetic energy after she has
pulled in her arms must be
1. the same.
2. larger because she's rotating faster.
3. smaller because her rotational inertia is smaller.
Phy1222 - Spring 2003
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Angular Momentum
•
Example: A meteor of mass 5x1010 kg crashes into the Earth at the
equator, moving in an eastward direction from an angle of 30 degrees
above the horizon, at a speed of 7200m/s. What is the resulting change
in the angular velocity of the Earth?
v=7200m/s
I E ≈ 52 MR 2 = 9. 7 ×10 37 kgm 2
LE = 7.1×10 33 kgm 2 /s
φ = 120 o
R
N
Lm = Rmv sin φ
(
)(
∑ L = ∑L
i
f
I Eω i + Lm = I E + mω f
)
= 6.37 ×10 6 m 5 × 1010 kg (7200m/s )(0.866 )
= 2.0 ×10 kgm /s
21
2
Phy1222 - Spring 2003
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20
Conservation of Angular Momentum
•
Special case: If two bodies have zero net angular momentum, then the
rotation angles are related by
I 1∆θ1 = − I 2 ∆ θ 2
•
This is similar to a system of two bodies at rest, where
m1∆ x1 = − m2 ∆ x2
•
Example: If you stand on a turntable and rotate a hoop of rotational
inertia 2.0kgm 2 above your head, and you turn by 30 degrees when it
makes a full turn, what is your rotational inertia?
Phy1222 - Spring 2003
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Angular Momentum and Torque
•
The fundamental relation is like that between momentum and force:
r
r dL
∑τ = dt
Sum of torques =
rate of change of angular momentum
•
average torque:
r
r ∆L
τ =
∆t
Example: The rotational inertia of the rotor
system is 3500 kgm2 , and it go from rest to its
final speed of 320 rev/min in 6.7 second.
What is the average torque from the engine?
Phy1222 - Spring 2003
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21
Conservation of L in collisions
•
Example: ball hits barbell and sticks (totally inelastic)
just after
before
later
v i/2
m
vi
m
c.m.
d
v i/3
ω=
vi
2d
v i/3
m
P = (3m)(vi / 3) = mvi
P = mvi
L = (d / 3)(mvi ) = mvi d / 3
L = Iω =
Icm = 2m(d / 3)2 + m(2d / 3)2
2
 v 
md 2  i 
3
 2d 
= mdvi / 3
= 23 md 2
Phy1222 - Spring 2003
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Analogies between Rotation and Linear Motion (III)
Rotational
•
•
•
•
•
•
angular position: θ (radians)
angular velocity: ω (rad/sec)
angular acceleration: α (rad/s 2)
rotational inertia: I (kgm 2)
torque: τ (N· m)
angular momentum: L (kgm2 /s)
Linear
• position: x (meters)
•
•
•
•
•
velocity: v (m/s)
acceleration: a (m/s2)
mass: m (kg)
Force: F (N)
linear momentum: p (kgm/s)
Phy1222 - Spring 2003
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22
Analogies between Rotation and Linear Motion (III)
Rotational
Linear
ω = ω0 + αt
v = v0 + at
θ − θ0 = ω 0t + 12 αt 2
x − x0 = v0 t + 12 at 2
K = 12 Iω 2
r
r
τ = Iα
W = τ∆ θ
r r
P = τ ⋅ω
r
r
L = Iω
r
r
∑ Li r= ∑ L f
r
τ = dL / dt
K = 12 mv 2
r
r
F = ma
W = F∆ x
r r
P = F ⋅v
r
r
p = mv
r
r
pi = ∑ p f
∑
r
r
F = dp / dt
Phy1222 - Spring 2003
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Ex. When a car accelerates forward, it tends to rotate about its center of
mass. The car will nose upward
1. when the driving force is imposed by the rear wheels (for
front-wheel drive the car would nose downward).
2. whether the driving force is imposed by the rear or the
front wheels.
Phy1222 - Spring 2003
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23
Ex. Suppose you had a car that was mostly wheels and another car that
had tiny wheels. If the cars had the same total mass and their centers of
mass are equal distances from the ground and each goes from zero to
40mph in ten seconds, which one will nose up the most?
1. Muscle Head.
2. Pip Squeak
3. Both the same.
Phy1222 - Spring 2003
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24