Two-Dimensional Rotational Kinematics

8.01W09D1

Young and Freedman: 1.10 (Vector Products) 9.1-9.6, 10.5

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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

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.

DemoCenter of Mass and Rotational

Motion of Baton

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

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

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

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

Cylindrical Coordinate System

Coordinates

Unit vectors

(r,θ,z)

ˆˆ ˆ( , , )θr z

Rotational Kinematicsfor

Point-Like Particle

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

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

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.

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.

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

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

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

∫ .

⇒

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.

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.

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?

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

Rigid Body Kinematicsfor Fixed Axis Rotation

Kinetic Energy and Moment of Inertia

Rigid Body Kinematicsfor Fixed Axis Rotation

Body rotates with angular velocity and angular acceleration

rω

rα

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

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

∑

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

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

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.

Concept Question

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

Concept Question

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.

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?

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

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

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

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.

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

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?

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

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

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

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.