Newton-Euler Equations - University of Pennsylvaniameam535/cgi-bin/pmwiki/uploads/... · MEAM 535...

MEAM 535

University of Pennsylvania 1

Newton-Euler Equations

From Dynamics of Systems of Particles to Rigid Body Dynamics

MEAM 535


Outline 1.  Newton’s 2nd Law for a particle - recap 2.  Newton’s 2nd Law for a system of particles - recap 3.  Center of mass, F=ma for the system of particles 4.  Angular momentum (relative to O or the center of mass) 5.  Newton’s 2nd Law for Rigid Body Motion

-  F=ma -  Rotational equation of motion

6.  Principal axes and moments of inertia 7. Newton-Euler equations of motion

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Newton’s Second Law for a Particle Newton’s Second Law for a particle P

  Position vector p in an inertial frame A   F is the force acting on the particle, mass m

€

A d Ldt

=…

O a1

a3

m a2

p

P

Angular Momentum of the system S relative to O in A

€

AHOP = p × m

A d pdti=1

N

∑

Linear momentum in A

Rate of change of angular and linear momentum in A

€

A d AHOP

dt=…

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Newton’s Second Law for a System of Particles Newton’s Second Law for a particle P

  Position vector p in an inertial frame A   F is the force acting on the particle, mass m

Linear momentum in A

Extend to many particles   Mass mi at Pi

  Fi is the external force acting on Pi

  Fij is the force exerted by Pj on Pi

O a1

a3

mi a2

pi

Pi

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Center of Mass

O

a2

a1

a3

mi

pi

Define Newton’s Second Law

i Fi

Fij k Fik

j

Pi

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Newton’s Second Law for a System of Particles

The center of mass for a system of particles, S, accelerates in an inertial frame (A) as if it were a single particle with mass m (equal to the total mass of the system) acted upon by a force equal to the net external force.

Center of mass

Conservation of Linear Momentum The linear momentum of a system of particles stays constant in an inertial frame (A) if the net external force equals zero.

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Angular Momentum of a System of Particles

O a1

a3

mi

pi

Angular Momentum of Pi relative to O in A

Angular Momentum of Pi relative to C in A Note: C is not fixed in A


Angular Momentum of the system S relative to C in A

ri

rC

Pi

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Rate of Change of Angular Momentum relative to O

O a1

a3

mi

pi


ri

rC

Resultant moment of all external forces acting on the system S about (relative to) O

The rate of change of angular momentum of the system S relative to O in A is equal to the resultant moment of all external forces acting on the system relative to O

Note Notes: (1) A is inertial; and

(2) O is fixed in A

Pi

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Rate of Change of Angular Momentum relative to C

O a1

a3

mi

pi

Angular Momentum of the system S relative to C in A

ri

rC

Resultant moment of all external forces acting on the system S about (relative to) C

The rate of change of angular momentum of the system S relative to C in A is equal to the resultant moment of all external forces acting on the system relative to C

0 0 Pi

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Principles of Conservation of Angular Momentum for a System of Particles

The angular momentum of S relative to O is constant in A

The angular momentum of S relative to C is constant in A

(1)

(2)

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Examples 1  Particles A and B, each of mass m, are attached

to massless rods of length l which are pivoted at A and O in the horizontal plane. Solve for the velocities as a function of positions, assuming θ and φ are zero at t=0, and the rate of change of θ and φ are known to be zero and ω0 respectively at t=0.

Answer

€

˙ φ =ω 03 − 2cosφ1+ sin2 φ

€

˙ θ =1− cosφ( )ω 0

1+ sin2 φ( ) 3 − 2sinφ( )

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Examples 2  Each of the three balls has a mass m and is

welded to rigid angular frame of negligible mass. The assembly rests on a smooth horizontal surface. If a force F is applied to one bar as shown, determine (a) the acceleration of the point O and the (b) the angular acceleration of the frame.

Answer

r r

r

b

F m

m m

O

€

Fb =ddt

−3mr2ω( )

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Angular Momentum for a System of Particles that are Rigidly Connected (relative to O)

O a1

a3

mi

C b1

b2

b3

B is a reference frame that is attached to the rigid body of mass m

pi €

AHOS = ri × mi ˙ p i

i=1

N

∑

= pi × mi

A ddti=1

N

∑ rC + ri( )

Angular momentum of a single particle of mass m (equal to the sum of the N masses) translating as if it were attached to the center of mass C.

Angular momentum of the rigid body due to its rotation

ri

rC €

= rC × mAvC + ri × mi

Aω B × ri( )i=1

N

∑

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Angular Momentum for a System of Particles that are Rigidly Connected (relative to C)

O a1

a3

mi

C b1

b2

b3

ri

B is a reference frame that is attached to the rigid body of mass m

pi

€

AHCS = ri × mi ˙ p i

i=1

N

∑

= ri × mi

A ddti=1

N

∑ rC + ri( )

= ri × miAω B × ri( )

i=1

N

∑rC

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Inertia Dyadic

€

IC ⋅AωB = mi ri ⋅ ri( )U− riri[ ]

i=1,N∑ ⋅ AωB

= mi ri ⋅ ri( ) AωB − ri ri ⋅AωB( )[ ]

i=1,N∑

= miri ×AωB × ri( )

i=1,N∑

€

AHCS = ri ×mi

AωB × ri( )i=1

N

∑Definition

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Rotational equations of motion

The rate of change of angular momentum of the system S relative to C in A is equal to the resultant moment of all external forces acting on the system relative to C

What is the difficulty?

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Principal Axes and Principal Moments

Principal axis of inertia u is a unit vector along a principal axis if I . u is parallel to u There are 3 independent principal axes!

Principal moment of inertia The moment of inertia with respect to a principal axis, u . I . u, is called a principal moment of inertia.

Physical interpretation

HO and AωB are not parallel!

O

B AωB

O B

AωB

HO and AωB are parallel!

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Properties of the inertia matrix

With respect to any origin, and any reference triad!

1.  The inertia matrix is Symmetric 2.  The inertia matrix is Positive Definite 3.  Moments of inertia are always positive 4.  Moments of inertia are the least when C is the reference point 5.  It is always possible to find three distinct mutually orthogonal

principal axes of inertia. [may not be unique] 6.  If the reference triad is aligned with the principal axes, the

products of inertia are zero

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Newton-Euler Equations

C

b1

b2

b3 B

Pi

ρi

R

MC

Newton’s second law

If b1, b2, b3, are along principal axes and AωB = ω1 b1 + ω2 b2 + ω3b3

Note C can be replaced by any point that is fixed in A

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Euler’s Equations

C

b1

b2

b3 B

Pi

ρi

R

MC

Let b1, b2, b3, be along principal axes and AωB = ω1 b1 + ω2 b2 + ω3b3

1

2

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Example

Problem

The rectangular homogeneous plate is welded to the shaft and the assembly rotates about a vertical axis at uniform speed (ignore gravity). Find the forces and moments that the weld at O must be subject to. What are the forces and moments that the bearings must support?

F1 , M1

F2 , M2

Weld

45 deg

2b

2a

P

p2

p1

a1

a2

a3

θ

with reference to pi

O

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Newton-Euler Equations (point O fixed in A)

C

b1

b2

b3

B

R

MC

A

b1

b2

b3

O

Newton’s equations

If b1, b2, b3, are along principal axes and AωB = ω1 b1 + ω2 b2 + ω3b3

Euler’s equations 1

2

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Rotation of an axisymmetric body about a principal axis

Suppose bi are along principal axes, and

And, suppose the rigid body is axisymmetric with b3 as an axis of symmetry

f1

f2

b3 f3

b1

b2 OR

Ωt

Option 2 With respect to fi

Option 1 With respect to bi

Which expression is simpler?

Note: The angular velocity is

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Example: Precession of a Spinning Top Three simple angular velocities

  Spin   Precession   Nutation

Three rotations   Rotation about a3 through φ   Rotation about f1 through θ   Rotation about f3 through ψ

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Example (continued)

B

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Example (continued)

B

Transverse moments of inertia

Axial moment of inertia

mgl sinθ

Let |OG|=l

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Problem: Rate Gyro

A rate gyro produces a measurable nutation angle θ proportional to the precessional or yaw angular velocity. Assume the precession angular velocity is constant and is directed along the vertical axis, the linear spring holds the angle θ to small values and the linear damper keeps the angular rate small. Derive the equation of motion and show when accelerations and velocities are small, the angle θ is proportional to the yaw velocity.

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Problem: Dynamics of a Thrown Football

A football is thrown horizontally at an angle θ with respect to the horizontal axis with a spin about its axial symmetry axis. The air drag can be assumed to act at a distance e from the center of mass in the horizontal direction. Because the ball is not thrown perfectly there are small rotational velocity components in the transverse directions.

Write down equations of motion for the football and show that a sufficiently large spin velocity stabilizes the motion for θ=0.

vC

FD

mg

e

b1 b2

θ

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Problem A thin homogeneous disk of mass 800 grams and radius 100 mm rotates at a constant rate of ω2 = 20 rads/sec with respect to the arm ABC, which itself rotates at ω1 = 10 rads/sec in an inertial frame. For the position shown, determine the dynamic reactions at the bearings D and E.

Problem Two particles of mass m are attached to a massless rod at an angle α as shown in the figure in such a way that the center of mass is on the horizontal shaft of length 2l. Calculate the bearing reactions if the horizontal shaft rotates with a constant angular velocity ω.

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