two-dimensional rotational kinematics 8.01 w09d1 young and freedman: 1.10 (vector products) 9.1-9.6,...
TRANSCRIPT
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Two-Dimensional Rotational Kinematics
8.01W09D1
Young and Freedman: 1.10 (Vector Products) 9.1-9.6, 10.5
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Announcements
No Math Review Night Next Week
Pset 8 Due Nov 1 at 9 pm, just 3 problems
W09D2 Reading Assignment Young and Freedman: 1.10 (Vector Product) 10.1-10.2, 10.5-10.6 ; 11.1-11.3
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Rigid Bodies
A rigid body is an extended object in which the distance between any two points in the object is constant in time.
Springs or human bodies are non-rigid bodies.
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DemoCenter of Mass and Rotational
Motion of Baton
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Overview: Rotation and Translationof Rigid Body
Demonstration: Motion of a thrown baton
Translational motion: external force of gravity acts on center of mass
Rotational Motion: object rotates about center of mass
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Recall: Translational Motion of the Center of Mass
Total momentum of system of particles
External force and acceleration of center of mass
rp
sys=msys
rVcm
rF
exttotal =
drpsys
dt=msys
drVcm
dt=msys
rAcm
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Main Idea: Rotational Motion about Center of Mass
Torque produces angular acceleration about center of mass
is the moment of inertial about the center of mass
is the angular acceleration about center of mass
I
cm
τcm
total =Icmαcm
αcm
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Two-Dimensional Rotational Motion
Fixed Axis Rotation:
Disc is rotating about axis passing through the center of the disc and is perpendicular to the plane of the disc.
Motion Where the Axis Translates:
For straight line motion, bicycle wheel rotates about fixed direction and center of mass is translating
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Cylindrical Coordinate System
Coordinates
Unit vectors
(r,θ,z)
ˆˆ ˆ( , , )θr z
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Rotational Kinematicsfor
Point-Like Particle
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Rotational Kinematicsfor Fixed Axis Rotation
A point like particle undergoing circular motion at a non-constant speed has
(1) an angular velocity vector
(2) an angular acceleration vector
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Fixed Axis Rotation: Angular Velocity
Angle variable SI unit:
Angular velocitySI unit:
Vector: Component
magnitude
direction
θ
rω ≡ω z k̂ ≡
dθdt
k̂ [rad]
1rad s−⎡ ⎤⋅⎣ ⎦
ω ≡ω z ≡
dθdt
ω z ≡dθ / dt> 0, direction + k̂
ω z ≡dθ / dt< 0, direction −k̂
ω z ≡
dθdt
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Concept Question: Angular Speed
Object A sits at the outer edge (rim) of a merry-go-round, and object B sits halfway between the rim and the axis of rotation. The merry-go-round makes a complete revolution once every thirty seconds. The magnitude of the angular velocity of Object B is
1. half the magnitude of the angular velocity of Object A . 2. the same as the magnitude of the angular velocity of Object A . 3. twice the the magnitude of the angular velocity of Object A . 4. impossible to determine.
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Concept Question: Angular Speed
Object A sits at the outer edge (rim) of a merry-go-round, and object B sits halfway between the rim and the axis of rotation. The merry-go-round makes a complete revolution once every thirty seconds. The magnitude of the angular velocity of Object B is
• Answer: 2. All points in a rigid body rotate with the same angular velocity.
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Example: Angular Velocity
Consider point-like object rotating with velocity tangent to the circle of radius r as shown in the figure below with
The angular velocity vector points in the direction, given by
−k̂
rω =ω zk̂ =(vθ / r) k̂
rv =vθθ̂
vθ < 0
ω z < 0
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Fixed Axis Rotation: Angular Acceleration
Angular acceleration:SI unit
Vector:Component:
Magnitude:
Direction:
α z ≡
d2θdt2
≡dω z
dt
2rad s−⎡ ⎤⋅⎣ ⎦
α ≡α z ≡
dω z
dt
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Rotational Kinematics: Integral Relations
The angular quantities
are exactly analogous to the quantities
for one-dimensional motion, and obey the same type of integral relations
Example: Constant angular acceleration
θ(t) =θ0 +ω z,0 t+
12α z t2
ω z(t) =ω z,0 +α z t
ω z(t)
2 =ω z,02 + 2α z(θ(t) −θ0 )
x, vx, and a
x
θ,ω z, andα z
ω z(t) −ω z,0 = α z( ′t )d ′t
0
t
∫ , θ(t) −θ0 = ω z( ′t )d ′t
0
t
∫ .
⇒
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Concept Question: Rotational Kinematics
The figure shows a graph of ωz and αz versus time for a particular rotating body. During which time intervals is the rotation slowing down?
1. 0 < t < 2 s
2. 2 s < t < 4 s
3. 4 s < t < 6 s
4. None of the intervals.
5. Two of the intervals.
6. Three of the intervals.
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Concept Question: Rotational Kinematics
The figure shows a graph of ωz and αz versus time for a particular rotating body. During which time intervals is the rotation slowing down?
1. 0 < t < 2 s
2. 2 s < t < 4 s
3. 4 s < t < 6 s
4. None of the intervals.
5. Two of the intervals.
6. Three of the intervals.
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Table Problem: Rotational Kinematics
A turntable is a uniform disc of mass m and a radius R. The turntable is initially spinning clockwise when looked down on from above at a constant frequency f . The motor is turned off and the turntable slows to a stop in t seconds with constant angular deceleration.
a) What is the direction and magnitude of the initial angular velocity of the turntable?
b) What is the direction and magnitude of the angular acceleration of the turntable?
c) What is the total angle in radians that the turntable spins while slowing down?
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Summary: Circular Motion for Point-like Particle
Use plane polar coordinates: circle of radius r
Unit vectors are functions of time because direction changes
Position
Velocity
Acceleration
rr(t) =r r̂(t)
rv(t) =r
dθdt
θ̂(t) =rω zθ̂(t) ≡vθθ̂(t)
ra(t) =ar r̂(t) + at θ̂(t)
a
t=rα z, ar =−vθ ω z =−rω 2 =−(v2 / r)
v ≡ r
v(t)
a ≡
ra(t) =(ar
2 + at2 )1/ 2
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Rigid Body Kinematicsfor Fixed Axis Rotation
Kinetic Energy and Moment of Inertia
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Rigid Body Kinematicsfor Fixed Axis Rotation
Body rotates with angular velocity and angular acceleration
rω
rα
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Divide Body into Small Elements
Body rotates with angular velocity, angular acceleration
Individual elements of mass
Radius of orbit
Tangential velocity
Tangential acceleration
Radial Acceleration
Δmi
vθ ,i =r⊥,iω z
aθ ,i =r⊥,iα z
r⊥,i
rω ≡ω z k̂
rα ≡α z k̂
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Rotational Kinetic Energy and Moment of Inertia
Rotational kinetic energy about axispassing through S
Moment of Inertia about S :
SI Unit:
Continuous body:
Rotational Kinetic Energy:
21
2cm cm cmK I ω=
Krot
= Krot,ii∑ =
12Δmi r⊥,i( )2
i∑⎛
⎝⎜⎞
⎠⎟ω 2 =
12
dm(r⊥,dm)2
body∫
⎛
⎝⎜
⎞
⎠⎟ω 2 =
12
ISω2
K
rot,i=
12Δmivi
2 =12Δmi (r⊥,i
2 )ω 2
I
S= Δmi (r⊥,i )
2
i=1
i=N
∑2kg m⎡ ⎤⋅⎣ ⎦
IS= dm(r⊥,dm)2
body∫
Δmi → dm r⊥,i → r⊥,dm
→body∫
i=1
i=N
∑
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Discussion: Moment of Inertia
How does moment of inertia compare to the total mass and the center of mass?
Different measures of the distribution of the mass.
Total mass: scalar
Center of Mass: vector (three components)
Moment of Inertia about axis passing through S:
IS= dm(r⊥,dm)2
body∫
mtotal = dmbody∫
cm totalbody
1dm
m= ∫R r
r r
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Concept Question
All of the objects below have the same mass. Which of the objectshas the largest moment of inertia about the axis shown?
(1) Hollow Cylinder (2) Solid Cylinder (3)Thin-walled Hollow Cylinder
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Concept Question
All of the objects below have the same mass. Which of the objectshas the largest moment of inertia about the axis shown?
Answer 3. The mass distribution for the thin-hollow walled cylinder is furthest from the axis, hence it’s moment of inertia is largest.
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Concept Question
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Concept Question
Answer 3. The mass of the sphere is on average closer to an axis of rotation passing through its center then in the case of the ring or the disc
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Concept Question
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Concept Question
Answer 2. The mass of the ring is furthest on average from an axis passing perpendicular to the plane of the disc and passing through a point on the edge of the disc.
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Worked Example: Moment of Inertia for Uniform Disc
Consider a thin uniform disc of radius R and mass m. What is the moment of inertia about an axis that pass perpendicular through the center of the disc?
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Strategy: Calculating Moment of Inertia
Step 1: Identify the axis of rotation
Step 2: Choose a coordinate system
Step 3: Identify the infinitesimal mass element dm.
Step 4: Identify the radius, , of the circular orbit of the infinitesimal mass element dm.
Step 5: Set up the limits for the integral over the body in terms of the physical dimensions of the rigid body.
Step 6: Explicitly calculate the integrals.
r⊥,dm
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Worked Example: Moment of Inertia of a Disc
Consider a thin uniform disc of radius R and mass m. What is the moment of inertia about an axis that pass perpendicular through the center of the disc?
da =r dr dθ
σ =
dmda
=mtotal
Area=
MπR2
dm =σ r dr dθ =
MπR2
r dr dθ
Icm
= (r⊥,dm)2 dmbody∫ =
MπR2
r3 dθθ=0
θ=2π
∫r=0
r=R
∫ dr
I
cm=
MπR2
dθθ=0
θ=2π
∫⎛⎝
⎞⎠r3dr
r=0
r=R
∫ =M
πR22πr3dr
r=0
r=R
∫ =2MR2
r3drr=0
r=R
∫
Icm
=2MR2
r3drr=0
r=R
∫ =2MR2
r4
4r=0
r=R
=2MR2
R4
4=
12
MR2
r⊥,dm =r
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Parallel Axis Theorem
• Rigid body of mass m. • Moment of inertia about
axis through center of mass of the body.
• Moment of inertia about parallel axis through point S in body.
• dS,cm perpendicular distance between two parallel axes.
I
S=Icm + mdS,cm
2
I
cm
IS
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Table Problem: Moment of Inertia of a Rod
Consider a thin uniform rod of length L and mass M.
Odd Tables : Calculate the moment of inertia about an axis that passes perpendicular through the center of mass of the rod.
Even Tables: Calculate the moment of inertia about an axis that passes perpendicular through the end of the rod.
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Summary: Moment of Inertia
Moment of Inertia about S:
Examples: Let S be the center of mass
• rod of length l and mass m
• disc of radius R and mass m
Parallel Axis theorem:
IS= Δmi (r⊥,S,i )
2
i=1
i=N
∑ = r⊥,S2
body∫ dm
I
cm=
112
ml2
I
cm=
12
mR2
I
S=Icm + mdS,cm
2
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Table Problem: Kinetic Energy of Disk
A disk with mass M and radius R is spinning with angular speed ω about an axis that passes through the rim of the disk perpendicular to its plane. The moment of inertia about the cm is (1/2)M R2. What is the kinetic energy of the disk?
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Concept Question: Kinetic Energy
A disk with mass M and radius R is spinning with angular speed ω about an axis that passes through the rim of the disk perpendicular to its plane. Moment of inertia about cm is (1/2)M R2. Its total kinetic energy is:
4. (1/4)M Rω2
5. (1/2)M Rω2
6. (1/4)M Rω
1. (1/4)M R2 ω2
2. (1/2)M R2 ω2
3. (3/4)M R2 ω2
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Concept Question: Kinetic Energy
Answer 3. The parallel axis theorem states the moment of inertia about an axis passing perpendicular to the plane of the disc and passing through a point on the edge of the disc is equal to
The moment of inertia about an axis passing perpendicular to the plane of the disc and passing through the center of mass of the disc is equal to
Therefore
The kinetic energy is then
I
edge=Icm + mR2
Icm=(1 / 2)mR2
I
edge=(3 / 2)mR2
K =(1 / 2)Iedgeω
2 =(3 / 4)mR2ω 2
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Summary: Fixed Axis Rotation Kinematics
Angle variable
Angular velocity
Angular acceleration
Mass element
Radius of orbit
Moment of inertia
Parallel Axis Theorem
θ
ω z ≡dθ / dt
α z ≡dω z / dt≡d2θ / dt2
Δmi
r⊥,i
IS= Δmi (r⊥,i )
2
i=1
i=N
∑ → dm(r⊥ )2
body∫
IS=Md2 + Icm
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Table Problem: Pulley System and Energy
Using energy techniques, calculate the speed of block 2 as a function of distance that it moves down the inclined plane using energy techniques. Let IP denote the moment of inertia of the pulley about its center of mass. Assume there are no energy losses due to friction and that the rope does slip around the pulley.