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2.004 Dynamics and Control II Spring 2008

For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008

Lecture 141

Reading:

• Class Handout: Modeling Part 1: Energy and Power Flow in Linear Systems Sec. 3.

• Class Handout: Modeling Part 2: Summary of One-Port Primitive Elements

1 The Modeling of Rotational Systems.

With the the modeling framework as we defined it in Lecture 13, we have seen that in each energy domain we need to define

(a) Two power variables, an across variable, and a through variable. the product of these variables is power.

(b) Two ideal sources, and across variable source, and a through variable source.

(c) Three ideal modeling elements, two energy storage elements (a T-type element, and a A-Type element), and a dissipative (D-Type) element.)

(d) A pair of interconnection laws.

We now address modeling of rotational mechanical systems.

(a) Definition of Power variables: In a rotational system we consider the motion of a system around an axis of rotation:

W , T

Consider the rotary motion resulting from a force F applied at a radius r from the rotational axis 1copyright �c D.Rowell 2008

14–1 D x F r

q

W

The work done by the force F in moving an infinitesimal distance Δx is

ΔW = F Δx = F rθ

and the power P is dΔW dθ P = = F r = T Ω dt dt where T = F r is the applied torque (N.m), and Ω = dθ/ dt is the angular velocity (rad/s). We note that if T and Ω have the same sign, then P > 0 and power is flowing into the system or element that is being rotated. Similarly, if T and Ω have the opposite signs, then P < 0 and power is flowing from the system or element, in other words the system is doing work on the source. Note that the angular velocity Ω can be different across an element, but that torque T is transmitted through an element:

W2 , T

A ngular velocities W 1 a n d W 2 c a n b e d i f f e r e n t a c r o s s an elem ent. The torque T transm itted t h r o u g h a n elem ent is constant.

W1 , T

We therefore define our power variables as torque T and angular velocity Ω, where • T is chosen as the through variable • Ω is chosen as the across variable.

(b) Ideal Sources: With the choice of modeling variables we can define our pair of ideal sources

The Angular Velocity Source: Ωs(t) By definition the angular velocity source is an across variable source.The ideal angular velocity source will maintain the rotational speed regardless of the torque it must generate to do so:

14–2 W a W

arrow show s direction angular velocity W i s of assum ed ang. vel. independent of torque T d r o p . W Wa > W b

W b 0 T t o r q u e

The Torque Source: Ts(t) By definition the torque source is a through variable source. The ideal torque source will maintain the applied torque regardless of the angular velocity it must generate to do so:

W T o r q u e T is independent of angular velocity W

arrow show s direction T o f t o r q u e .

0 T t o r q u e

(c) Ideal Modeling Elements: 1 The Moment of Inertia: Consider a mass element m rotating at a fixed radius R about the axis of rotation.

The stored energy is m 1 1 E = m(rΩ)2 = JΩ2 r 2 2 where J = mr2 is defined to be the mo­ ment of inertia of the particle. W

For a collection of n mass particles mi at radii ri, i = 1, . . . , n, the moment of inertia is �n 2 J = miri . i=1

14–3 For a continuous distribution of mass about the axis of rotation, the moment of inertia is

W

� R r J = r 2dm d m m a s s a t 0 r a d i u s r

Examples:

W m L 2 A uniform rod of length L J = 1 2 rotating about its center. w h e r e m is the m ass L o f t h e r o d

1 J 2 r = m r A uniform disc with radius r 2 rotating about its center. w h e r e m is the m ass of the disk

The elemental equation for the moment of inertia J is

dΩ T = J J J dt

We note that the energy stored in a rotating mass is E = JΩ2/2, that is it is a function of the across variable, defining the moment of inertia as an A-type element. As in the case of a translational mass element, the angular velocity drop associated with a rotary inertia J is always measured with respect to a non-accelerating reference frame. Elemental Impedance: By definition

ΩJ (s) 1 ZJ = = TJ (s) Js from the elemental equation.

14–4 (2) The Torsional Spring:

T

Wb , q b K

W = W a - W b

T

W a , q a

Let θa and θb be the angular displacements of the two ends from their rest positions. Hooke’s law for a torsional spring is

T = K(θa − θb). where K is defined to be the torsional stiffness. Differentiation gives dT d(θ − θ ) = K a b dt dt dT = KΩ dt ˙ ˙ where Ω = (θa − θb) is the angular velocity drop across the spring. Torsional stiffness may result from the material properties of a “long” shaft

J 1 J 2

W 1 W 2

shaft can tw ist w ith different angular velocities at the t w o e n d s or may be intentional, for example in a coil (“hair”) spring in a mechanical watch.

T

W

coil spring

14–5 The energy stored in a torsional spring is � t 1 E = T Ω dt = T 2 ∞ 2K which is a function of the through variable, defining the spring as a T-type element. Elemental Impedance: By definition

ΩK (s) s ZK = = TK (s) K

from the elemental equation. (3) The Rotational Damper: We look for an algebraic relationship between T and Ω of the form T = BΩ

viscous frictional T c o n t a c t

W b

W = W a - W b T

W a

which is approximated as viscous rotational friction:

external viscous s h a f t frictional contact T , W T b e a r i n g s ( f r i c t i o n )

W a

Notice that P = T Ω = BΩ2 > 0, which defines the damper as a D-type element. Elemental Impedance: By definition

ΩB (s) 1 ZB = = TB(s) B

from the elemental equation.

(d) Interconnection Laws: Consider an inertial element J subject to n external torques T1, T2, . . . , Tn, for example

14–6 J

T T 3 4 TT1 2

r e f e r e n c e W d i r e c t i o n then dΩ J = T − T 2 + T + T dt 1 3 4 and in general �n dΩ T = J i dt i=1

As in the translational case, we consider a “fictitious” d’Alembert torque Tj and write �n Ti − TJ = 0 i=1 as the torque balance (continuity condition) at a node.

2 3

W J 1 arrow on an inertial branch J alw ays points to the reference node W J is alw ays m easured w ith repect to an inertial W r e f = 0 reference fram e

For an “inertia-less” node (J = 0), �n Ti = 0 i=1 which states that the external torques sum to zero, for example at node (a) below, TB − TK = 0.

K W 2

B K B ( a )

W a

T

W1

14–7 Continuity Condition: The sum of torques (including a d’Alembert torque associated with an inertia a element) at any node on a system graph is zero.

Nodes represent points of distinct angular velocity in a rotational system, and by analogy with translational systems, the compatibility condition is

Compatibility Condition: The sum of angular velocity drops around any closed loop on a system graph is zero.

For example, on the graph:

J

W b K B ( a ) ( b ) J K 1

2 B

W a W = 0

two compatibility equations are:

ΩK + ΩJ − Ωs = 0 (Loop 1),

ΩB − ΩJ = 0 (Loop 2).

2 Updated Tables of Generalized Elements to Include Rotational Elements:

The tables presented in Lecture 13 are now updated to include rotational systems.

A-Type Elements:

Element Elemental equation Energy

dv 1 Generalized A-type f = C E = Cv2 dt 2 dv 1 Translational mass F = m E = mv 2 dt 2 dΩ 1 Rotational inertia T = J E = JΩ2 dt 2 dv 1 Electrical capacitance i = C E = Cv2 dt 2

14–8 T-Type Elements :

Element Elemental equation Energy

1 Generalized T-type v = Ldf/dt E = Lf2 2 1 dF 1 Translational spring v = E = F 2 K dt 2K 1 dT 1 Torsional spring Ω = E = T 2 K dt 2K di 1 Electrical inductance v = L E = Li2 dt 2

D-Type Elements:

Element Elemental equations Power dissipated

1 1 Generalized D-type f = v v = Rf P = v 2 = Rf2 R R 1 1 Translational damper F = Bv v = F P = Bv2 = F 2 B B 1 1 Rotational damper T = BΩ ω = T P = BΩ2 = T 2 B B 1 1 Electrical resistance i = v v = Ri P = v 2 = Ri2 R R

Generalized Impedances:

A-Type T-Type D-Type

1 Generalized sL R Cs 1 1 1 Translational s sm K B 1 1 1 Rotational s sJ K B 1 Electrical sL R Cs

14–9