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# DYNAMIC SYST AND CONTROL MEEN 364

Texas A&M

GPA 3.85

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This 12 page Class Notes was uploaded by Kameron Hyatt Sr. on Wednesday October 21, 2015. The Class Notes belongs to MEEN 364 at Texas A&M University taught by Bryan Rasmussen in Fall. Since its upload, it has received 38 views. For similar materials see /class/225988/meen-364-texas-a-m-university in Mechanical Engineering at Texas A&M University.

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Date Created: 10/21/15

MEEN 364 Parasuram Lecture 67 August 7 2001 HANDOUT E7 EXAIVTPLES ON MODELLING OF lVlECHANICAL SYSTEMS MIXED ROTATIONAL AND TRANSLATIONAL Note that the time dependence of variables is ignored for all manipulations Example 1 One DOF system Consider the system shown below y The disc is of radius R and has a moment of inertia I There is friction between the disc and the block of mass m Kinematics stage From the above gure it can be seen that there are two rigid bodies The coordinates representing their independent movement is given by X and 9 respectively But since there exists a relation between these two coordinates which is given by x R0 the number of degrees of freedom of the system is one Let the degree of freedom be X The velocity and the acceleration are given by x x 9 9 respectively This completes the kinematics stage Kinetics stage Free body diagram of block F kx cx I Ff Note that the gravity force is not considered as the reaction force of the disc on the block balances this force Writing the Newton s second law of motion we get MEEN 364 Parasuram Lecture 67 August 7 2001 2Fma gtF kx cx Ff mx gtmxcxkxFf F 1 Free body diagram of the disc gtFf Note that the gravity force is ignored as the reaction force due to the block on the disc balances the gravity force Taking moments about the center of the disc we have 2M 19 gtFfR19 gtFf 9 Since xR0we get Ff x Substituting the value of Ff in equation 1 we get mxcxkxFf F gtmxcxkx xF gtmRI 2xcxkxF 2 Equation 2 represents the governing equation of motion for the system de ned MEEN 364 Parasuram Lecture 67 August 7 2001 State space representation Let the states of the system be x x1 3 x From the above relations we get x1 x2 4 Substituting the relations given by equation 3 in equation 2 we have I mFxcxkx F I gt mR 2xzcx2 kx1 F F k xZ x1 x2 5 I I I m l39F m l39F m l39F Rewriting the equations 4 and 5 in matrix format we get 0 l 0 k c l H 1 1x1 1F 6 x2 mR 2 mF x2 mR 2 If the output of the system is the displacement of the block then the output relation can be expressed in matrix format as follows y 1 org 7 x2 Equations 6 and 7 represent the statespace form of the system defined MEEN 364 Parasuram Lecture 67 August 7 2001 Example 2 One DOF system PKG X Consider the system shown above Assume no friction Kinematics stage There is one rigid body The number of degrees of freedom of the system is one Let the degree of freedom be represented as X The relation between the rotation of the disc and the linear displacement moved by the disc is given by x R0 Assume rolling without slipping The velocity and the acceleration of the disc are x x respectively Kinetics stage Free body diagram of the disc 10 10 Writing the Newton s second law of motion we get 2F ma gt 2kx m x m x 2kx 0 8 Substituting the above relation in equation 8 we get me 2kR9 0 9 MEEN 364 Parasuram Lecture 67 August 7 2001 Equation 9 represents the governing differential equation of motion State space representation Let the states of the system be de ned as 9 x1 10 9 x2 From the above relations the following equation can be written x1x2 11 Substituting the relations given by equation 10 in equation 9 we have mR 9 2kR9 0 2 m XZ kal 0 gtxz x1 12 m Rewriting equations 1 l and 12 in matrix format we have 0 l EH Lh m m x2 If the output of the system is the linear displacement of the disc then the output relation can be expressed in the matrix format as yxR9 Rx1 y R org 14 x2 Equations 13 and 14 represent the statespace form of the system defined MEEN 364 Parasuram Lecture 67 August 7 2001 Example 3 Two DOF system 37 direction X m Kinematics stage There are two rigid bodies and the number of degrees of freedom of the system is two Let the two degrees of freedom be represented by the angular displacement of the bob 9 and the linear displacement of the block X Therefore the linear velocity and the linear acceleration of the block are x x respectively Similarly the angular velocity and the angular acceleration of the bob are 9 9 respectively Assume x gt 0 Kinetics stage Free body diagram of the bob kx L9 The linear acceleration of the bob is towards the right and is equal to L9 Therefore writing the Newton s law of motion we get ZFX ma gt kx L9 Tsin9 mLO 15 MEEN 364 Parasuram Lecture 67 August 7 2001 Similarly ZFy ma gt T cos9 mg Substituting the value of tension T in equation 15 we have kx L9 mgtan9 mL9 16 For small angles we have tan0 z 0 Therefore equation 16 reduces to kx L9 mg9 mL9 17 Free body diagram of block kx L9 F 1 Note that the gravity force is neglected as the normal reaction force of the ground balances the gravity force Writing the Force balance equation we get 2F ma gtF kx L9 Mx gtMxkx L9 F 18 Equations 17 and 18 represent the governing differential equation State space representation Let the states of the system be defined as 9 x1 9 x 19 xx3 xx MEEN 364 Parasuram Lecture 67 August 7 2001 From the above relations the following two equations can be derived X1 X2 20 X3 X4 Substituting the relations given by equation 19 in equation 17 we get kx L9 mg9 mLO gt kx3 Lx1 mgx1 mLxZ kLmgx1 k m gtx x 21 2 mL L 3 Similarly substituting the relations given by equation 19 in equation 18 we get Mxkx L9 F gtankx3 Lx1F F gtX4 HHxl Hx3 22 Rewriting equations 20 21 and 22 in matrix format we have 0 l 0 0 x 0 x1 kL mg k 1 x L 0 7 0 x 0 2 m m 2 0 0 0 1 x 0 F 23 X3 CL 3 l 0 0 x X4 M 4 M If the output of the system is the linear displacement of the block then the output relation can be expressed in the matrix format as yxx39 x1 x2 y0 0 1 0 x 24 3 X Equations 23 and 24 represent the statespace form of the abovedefined system MEEN 364 Parasuram Lecture 67 August 7 2001 Example 4 Two DOF system Consider the system shown below 37 J zmn X Moment of inertia of the drum is equal to I Kinematics stage There are two rigid bodies and they can move independent of each other Therefore the number of degrees of freedom of the system is two Let the degrees of freedom be chosen as the linear displacement of the block y and the angular displacement of the drum 9 Therefore the linear velocity and the linear acceleration of the block is given by y y respectively Similarly the angular velocity and the angular acceleration of the drum is given by 9 9 respectively Assume y gt9 Kinetics stage Free body diagram of the block R CyRQ 16939 MEEN 364 Parasuram Lecture 67 August 7 2001 Note that the displacements chosen are from the static equilibrium position Hence spring force due to the initial elongation of the spring balances the gravity force Writing the Newton s second law of motion we have ZFy ma gtcy R9 ky 9 my R mycy R9 ky 39 0 25 Free body diagram of the drum R CyRQ My 39 Taking moments about the center of the drum we have 2M 1 9 gt Tcy R9 Rky 9 g kRQ R c59 g 19 2 Z 2 19 cR2d 9 cRykR2k9 k yT 26 Equations 25 and 26 represent the governing differential equations of motion State space representation yx19 yx29 27 9 x3 9 x MEEN 364 Parasuram Lecture 67 August 7 2001 From the above relations the following equations can be deduced X1 x2 28 X3 x4 Substituting the relations given by equation 27 in equation 25 we get R mycy R9 ky 39 0 R gt mxzcx2 Rx4kx1 3x3 0 gtxZ x1 x2 Ex3 299 29 m m 2m m Similarly substituting the relations given by equation 27 in equation 26 we get cR2 kR2 R 19 cR2 9 cRykR2 9 k yT 4 4 2 Z 2 gt 1964 SCR x4 cRx2 Skf x3 k x1 T gtx Zk R x 5kR2 x SCR2 x 30 4121x112 413 41 Rewriting equations 28 29 and 30 in matrix format we have 0 l 0 0 0 M i LR 2 x1 0 X2 m 2m 2m x2 0 0 0 1 x3 0 T 31 XS kR cR SkRZ 5ch 1 x4 x4 2 I 4 4 I If the output of the system is the displacement moved by the block then the output equation can be expressed in the matrix format as Yyx 1 5 N 32 88 u 4 Equations 3 l and 32 represent the statespace form of the system defined MEEN 364 Parasuram Lecture 67 August 7 2001 Assignment 1 Derive the governing differential equation of motion for the following system Mass m I L2 39 I L2 2k k Hinged support The bar can rotate about This point 2 Consider the system shown below O m The above gure consists of a drum to the center of which a simple pendulum is attached The bob of the pendulum is displaced slightly to the right Derive the governing differential equation of motion Assume small angle motion Recommended Reading Feedback Control of Dynamic Systems 43911 Edition by Gene F Franklin etal 7 pp 24 45

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