Review #3:

Collisions, Rotational Motion

Momentum - Model

From Principia: Momentum is defined as “the quantity of motion, conjointly proportional to the mass and the velocity”.

momentum is for a system of particles

system r r definition ---p = m v

∑

i i

i

momentum is sum of individual momenta

Dynamics (change) of momentum - from F=ma

r system forces r dp

= ∑ Fi external dt i =1 Only external (through system barrier) forces !

Strategies: Momentum and

External Forces

1. Identify all forces acting on the masses

2. Select your system (make troublesome forces into internal forces; make ΣFext = 0) 3. ΣFext = 0 implies momentum is conserved 4. If there is a non-zero total external force:

r system forces r d p external = ∑ Fi dt i

Strategy - Impulse

Impulse most useful when time is short, simplifying the momentum change during the short time ∆t.

A. One large external force dominates ∆p: •Ball bouncing on floor - ignore gravity B. Finite Fext for very short ∆t --> ∆psystem = 0 •Colliding cars - ignore horizontal friction •Gun and Bullet - ignore external forces

Collisions: Momentum Conserved

When the total external force on the colliding particles is much smaller than the internal forces, the collision duration is so short that the impulse on the system is approximately zero. Then the total initial momentum of the colliding particles equals their final momentum:

r total r total

p f = p0

Gives one equation each for x, y, and z

Elastic Collisions

• Kinetic Energy does not change.

K0

total

= Kf

total

1 1 1

1 2 2 2 m1v1,0 + m2 v2,0 + ... = m1v1, f + m2 v2, f 2 + ... 2

2 2 2

• Momentum conserved also

Rotation and Translation

of Rigid Body

Motion of a thrown object

Translational Motion of the

Center of Mass

• Total momentum

r r total total p = m V cm

• External force and acceleration of center of mass r r total r r total

dp

total dV

total

cm F ext = =m = m A cm

dt dt

Rotation and Translation

of Rigid Body

• Torque produces angular acceleration about center of mass torques

∑τ

• I cm

cm,i

= I cmα cm

i is the moment of inertial about the center of mass

• α is the angular acceleration about center of mass or any other point in a rigid body. • This is really a vector relation; only z-component is non-zero if problem is planar

Fixed Axis Rotation: Kinematics

Angle variable

θ

Angular velocity

dθ ω≡ dt

Angular acceleration

d 2θ α≡ 2 dt These are exactly analogous to the variables x, vx, and ax for One dimensional motion, and obey the same equations For constant angular acceleration:

Fixed Axis Rotation: Tangential Velocity

and Tangential Acceleration Kinematics of individual mass elements:

∆mi

• Tangential velocity • Tangential acceleration • Radial Acceleration

vtan,i = r⊥ ,iω atan,i = r⊥ ,iα arad ,i =

v

2 tan,i

r⊥ ,i

= r⊥ ,iω 2

PRS: Ladybug Acceleration

A ladybug sits at the outer edge of a merry-go-round that is turning and slowing down. The tangential component of the ladybug's (Cartesian) acceleration is

QuickTime™ and a Graphics decompressor are needed to see this picture.

1.

in the +x direction.

2. 3. 4. 5.

in the -x direction. in the +y direction. in the -y direction. in the +z direction.

6. 7.

in the -z direction. zero.

Torque

Torque about axis S:

r r τ S ,i = r S ,i × Fi

r

• Counterclockwise +z

direction

• perpendicular to the plane

τ S ,i = rS ⊥,i Ftan,i = rS ⊥,i ∆mi a y = ∆mi (rS ⊥,i ) α 2

torques

∑τ i

S ,i

= I Sα cm = I Sα S

Energy of Rotating Mass

The mass is rotating: Angular velocity ω Radius R

v R ω

Speed v Speed v = ω R

Kinetic Energy = 1/2 m v^2 = 1/2 m R2ω2 = 1/2 Is ω2

K

rot S

1 2 = I Sω S 2

Moment of Inertia - Idea

Mass element

∆mi

Radius of orbit

r⊥ ,i i= N

Moment of Inertia about S

I S = ∑ ∆mi (rS ⊥ ,i )2

S

i =1

The moment of inertia takes the place of mass in the

Dynamical (F=ma) and Energy (K=mv2/2) equations

for rotational motion.

torques

∑τ i

S ,i

= I Sα cm

K

rot S

1 2 = I Sω S 2

These work about ANY NON-ACCELERATING axis

Rotational Work

r r Starting from ∆Wi = Fi ⋅ ∆r i

• tangential force

r F tan,i = Ftan,i nθö

• displacement vector

( )

r ∆rS ,i = rS ,⊥ ∆θ nö i

r r • infinitesimal work ∆Wi = Fi ⋅ ∆r i = Ftan,i rS ,⊥ i ∆θ = τ i ∆θ

( )

f

• Rotational work:

W firot = ∫ τ (θ )dθ = τ avg (θ f − θ i ) i

Note: if τ is constant, it equals τavg

General Work-Kinetic Energy Rel’n

• Fixed axis passing through the c of m of the body Wf 0

rot

1 1 2 2 = I cmω cm, f − I cm ω cm,0 = K rot , f − K rot ,0 ≡ ∆K rot 2 2

• Rotation and translation combined - General Motion

⎛1 2 ⎛1 2 1 1 2⎞ 2⎞ K f = ⎜ mvcm, f + I cmω f ⎟ = ⎜ mvcm,0 + I cmω 0 ⎟ + W fitrans + W firot = + K 0 + W fitotal 2 ⎝ 2 ⎠ ⎝ 2= ∆K W +2 ∆K ⎠ total

trans

rot

f

W firot = ∫ τ (θ )dθ = τ avg (θ f − θ i ) i

Note: if τ is constant, it equals τavg

Strategy: Moment of Inertia

Always start from a tabulated Icm plus the Parallel Axis Theorem

i= N

2 2 I = ∆m (r ) = dm (r ) Use S ∑ i S ⊥ ,i S ⊥ ,dm ∫ i =1

S

When all else fails

r⊥

Note: I cm = ∫ dm (rcm⊥ ,dm ) = α MR 2

Where α is between 0 and 1

2

Table of Icm Object Hoop Disk Sphere Rod

α 1.0 0.5

0.4 ML2/12

Rotational Dynamics & Energy - Summary

• Dynamical Equations about axis S Dynamics torques

∑

τ S , i = I Sα cm = I Sα S

i

Kinetic Energy 1 K Srot = I Sω S2 2 Requires S to be stationary

Is not true if S accelerates

Includes KE of c of m

• Most General Equations torques

∑τ

cm, i

= I cmα cm

i

1 1 2 K tot = I cmω cm + M v 2cm 2 2 KE of rotation+KE of translation

OK if c of m accelerates

PRS - kinetic energy

A disk with mass M and radius R is spinning with angular velocity ω about an axis that passes through the rim of the disk perpendicular to its plane. Its total kinetic energy is: 1. 1/4 M R2 ω2

4. 1/4M R ω2

2. 1/2M R2 ω2

5. 1/2 M R ω2

3. 3/4 M R2 ω2

6. 1/2 M R ω

Dynamics: Translational & Rotational

Translational Dynamics •Total Force

r ext ∑ Fi

ri system p

•Momentum of System

r system

r ext dp ∑ Fi = dt

i

•Dynamics of Translation

Rotational Dynamics of point mass about S

r

r r

• Torque τ S , i = ri,S × Fi

r r r • Angular momentum L S ,i = ri, S × p i • Dynamics of rotation

r ext

∑τ i

S ,i

=

r dL S dt

Rotational Angular Momentum & Energy

• Putting it Together - Spin plus Translation Kinetic Energy

Angular Momentum r system r i = N r r L S = I cmω + ∑ R cm,S × p

i =1

L of rotation+L of translation

K

tot

1 1

2 = I cmω cm + M v 2cm 2 2

KE of rotation+KE of translation

Dynamics (i.e. change) r ext

∑τ i

cf.

S ,i

=

r system dL S dt

r ext dpr ∑ Fi = dt

i

K f = K 0 + W fitrans + W firot

Physical Content of

ext τ ∑ S ,z i

dLsystem z = dt

• In the case that IS,z does not change: ext τ ∑ S ,z = i

dLsystem

z

dt

=

(

d I S , zω z dt

)= I

dω z S ,z

dt

= IS ,z

• But IS,z may change: – Spinning Skater pulls in arms – Rain falling on merry-go -round

• Conservation of Lz is richer than pz ext τ ∑ S ,z = 0 i

system Lsystem = L z, f z ,i

d 2θ = I S , zα z 2 dt

PRS - angular momentum

A disk with mass M and radius R is spinning with angular velocity ω about an axis that passes through the rim of the disk perpendicular to its plane. The magnitude of its angular momentum is: 1. 1/4 M R2 ω2

4. 1/2M R ω

2. 1/2M R2 ω2

5. 3/4 M R ω

3. 3/4 M R2 ω2

6. 3/2 M R ω

Angular Momentum is a Kepler Law

Perihelion ->

r

r

vtang

vorb

vtang

vorb Angular Momentum: Lsun = µvtang r

Area: ∆Aswept = r(vtang ∆t ) / 2 ∆Aswept ∆t

1 L = rvtang =

2 2 µ

Circular Orbital Mechanics

-class problem

A planet of mass m1 is in a circular orbit of radius R around a sun of mass m2. a. Find the period, T b. Find the ratio of kinetic to potential energy

c. As R increases, all of the physical properties of the

orbit (e.g. velocity) decrease except one; find it.