Mathematical Modeling and response analysis of mechanical systems are the subjects of this chapter.


 Valerie Barker
 1 years ago
 Views:
Transcription
1 Chapter 3 Mechanical Systems Dr. A.Aziz. Bazoune 3.1 INTRODUCTION Mathematical Modeling and response analysis of mechanical systems are the subjects of this chapter. 3. MECHANICAL ELEMENTS Any mechanical system consists of mechanical elements. There are three types of basic elements in mechanical systems: Inertia elements Spring elements Dampers elements Inertia Elements. Mass and moment of inertia. Inertia may be defined as the change in force (torque) required to make a unit change in acceleration (angular acceleration). That is, change in force N inertia (mass) = or kg change in acceleration m/s change in torque Nm inertia (moment of inertia) = or kg change in ang. accel. rad/s Spring Elements. A linear spring is a mechanical element that can be deformed by eternal force or torque such that the deformation is directly proportional to the force or torque applied to the element. Translational Springs For translational motion, (Fig 31(a)), the force that arises in the spring is proportional to and is given by F = k (31) where is the elongation of the spring and k is a proportionality constant called the spring constant and has units of [force/displacement]=[n/m] in SI units. 1/34
2 At point P, the spring force F acts opposite to the direction of the force F applied at point P. X X+ P f f Q X+ 1 X + 1 P f Figure 31 (a) One end of the spring is deflected; (b) both ends of the spring are deflected. (X is the natural length of the spring) Figure 31(b) shows the case where both ends P and Q of the spring are deflected due to the forces f applied at each end. The net elongation of the spring is 1. The force acting in the spring is then F = k (3) 1 Notice that the displacement X+ 1 and of the ends of the spring are measured relative to the same reference frame. Practical Eamples. Pictures of various types of realworld springs are found below. /34
3 Torsional Springs Consider the torsional spring shown in Figure 3 (a), where one end is fied and a torque τ is applied to the other end. The angular displacement of the free end is θ. The torque T in the torsional spring is T k θ = T (33) where θ is the angular displacement and k T is the spring constant for torsional spring and has units of [Torque/angular displacement]=[nm/rad] in SI units. θ τ τ θ θ 1 τ Figure 3 (a) A torque τ is applied at one end of torsional spring, and the other end is fied; (b) a torque τ is applied at one end, and a torque τ, in the opposite direction, is applied at the other end. At the free end, this torque acts in the torsional spring in the direction opposite to that of the applied torque τ. For the torsional spring shown in Figure 3(b), torques equal in magnitude but opposite in direction, are applied to the ends of the spring. In this case, the torque T acting in the torsional spring is T T = k θ θ (34) 1 At each end, the spring acts in the direction opposite to that of the applied torque at that end. For linear springs, the spring constant k may be defined as follows spring constant k for translational spring spring constant k T for torsional spring change in force N = change in displacement of spring m change in torque = change in angular displacement of spring Nm rad Spring constants indicate stiffness; a large value of k or k T corresponds to a hard spring, a small value of k or k T to a soft spring. The reciprocal of the spring 3/34
4 constant k is called compliance or mechanical capacitance C. Thus C 1 k Compliance or mechanical capacitance indicates the softness of the spring. Practical Eamples. Pictures of various types of realworld springs are found below. =. Practical spring versus ideal spring. Figure 33 shows the force displacement characteristic curves for linear and nonlinear springs. All practical springs have inertia and damping. F An ideal spring has neither mass nor damping (internal friction) and will obey the linear force displacement law. o Figure 33 Forcedisplacement characteristic curves for linear and nonlinear springs. Damper Elements. A damper is a mechanical element that dissipates energy in the form of heat instead of storing it. Figure 34(a) shows a schematic diagram of a translational damper, or a dashpot that consists of a piston and anoilfilled cylinder. Any relative motion between the piston rod and the cylinder is resisted 4/34
5 by oil because oil must flow around the piston (or through orifices provided in the piston) from one side to the other. 1 f & & 1 f τ θ & θ & 1 τ Figure 34 (a) Translational damper; (b) torsional (or rotational) damper. Translational Damper In Fig 34(a), the forces applied at the ends of translation damper are on the same line and are of equal magnitude, but opposite in direction. The velocities of the ends of the damper are & 1 and &. Velocities & 1 and & are taken relative to the same frame of reference. In the damper, the damping force F that arises in it is proportional to the velocity differences & 1 & of the ends, or F = b & & = b& (35) 1 & = & & and the proportionality constant b relating the damping force F where 1 to the velocity difference & is called the viscous friction coefficient or viscous friction constant. The dimension of b is [force/velocity] = [Ns/m] in SI units. Torsional Damper For the torsional damper shown in Figure 34(b), the torques τ applied to the ends of the damper are of equal magnitude, but opposite in direction. The angular velocities of the ends of the damper are & θ 1 and & θ and they are taken relative to the same frame of reference. The damping torque T that arises in the damper is proportional to the angular velocity differences & θ1 & θ of the ends, or T ( 1 ) T = b & θ & θ = b & θ (36) & θ = & θ & θ and the proportionality b T relating the damping torque T to the angular velocity difference & θ is where, analogous to the translation case, 1 constant T 5/34
6 called the viscous friction coefficient or viscous friction constant. The dimension of b is [torque/angular velocity] = [Nms/rad] in SI units. A damper is an element that provides resistance in mechanical motion, and, as such, its effect on the dynamic behavior of a mechanical system is similar to that of an electrical resistor on the dynamic behavior of an electrical system. Consequently, a damper is often referred to as a mechanical resistance element and the viscous friction coefficient as the mechanical resistance. Practical Eamples. Pictures of various eamples of realworld dampers are found below. Practical damper versus ideal damper All practical dampers produce inertia and spring effects. An ideal damper is massless and springless, dissipates all energy, and obeys the linear forcevelocity law (or linear torqueangular velocity law). Nonlinear friction. Friction that obeys a linear law is called linear friction, whereas friction that does not is described as nonlinear. Eamples of nonlinear friction include static friction, sliding friction, and squarelaw friction. Square lawfriction occurs when a solid body moves in a fluid medium. Figure 35 shows a characteristic curve for squarelaw friction. Figure 35 Characteristic curve for squarelaw friction. 6/34
7 3.3 MATHEMATICAL MODELING OF SIMPLE MECHANICAL SYSTEMS A mathematical model of any mechanical system can be developed by applying Newton s laws to the system. Rigid body. When any real body is accelerated, internal elastic deflections are always present. If these internal deflections are negligibly small relative to the gross motion of the entire body, the body is called rigid body. Thus, a rigid body does not deform. Newton s laws. Newton s first law: (Conservation of Momentum) The total momentum of a mechanical system is constant in the absence of eternal forces. Momentum is the product of mass m and velocity v, or mv, for translational or linear motion. For rotational motion, momentum is the product of moment of inertia J and angular velocity ω, or Jω, and is called angular momentum. Newton s second law: Translational motion: If a force is acting on rigid body through the center of mass in a given direction, the acceleration of the rigid body in the same direction is directly proportional to the force acting on it and is inversely proportional to the mass of the body. That is, or force acceleration = mass ( mass)( acceleration ) = force Suppose that forces are acting on a body of mass m. If F is the sum of all forces acting on a mass m through the center of mass in a given direction, then F = ma (37) where a is the resulting absolute acceleration in that direction. The line of action of the force acting on a body must pass through the center of mass of the body. Otherwise, rotational motion will also be involved. Rotational motion. For a rigid body in pure rotation about a fied ais, Newton s second law states that 7/34
8 or ( moment of inertia)( angular acceleration ) = torque T = Jα (38) where T is the sum of all torques acting about a given ais, J is the moment of inertia of a body about that ais, and α is the angular acceleration of the body. Newton s third law. It is concerned with action and reaction and states that every action is always opposed by an equal reaction. Torque or moment of force. Torque, or moment of force, is defined as any cause that tends to produce a change in the rotational motion of a body in which it acts. Torque is the product of a force and the perpendicular distance from a point of rotation to the line of action of the force. [ ] [ ] Torque = force distance = Nm in SI units Moments of inertia. The moment of inertia J of a rigid body about an ais is defined by J = rdm (39) Where dm is an element of mass, r is the distance from ais to dm, and integration is performed over the body. In considering moments of inertia, we assume that the rotating body is perfectly rigid. Physically, the moment of inertia of a body is a measure of the resistance of the body to angular acceleration. Figure 36 Moment of inertia Parallel ais theorem. Sometimes it is necessary to calculate the moment of inertia of a homogeneous rigid body about an ais other than its geometrical ais. 8/34
9 Figure 37 Homogeneous cylinder rolling on a flat surface As an eample to that, consider the system shown in Figure 37, where a cylinder of mass m and a radius R rolls on a flat surface. The moment of inertia of the cylinder is about ais CC is Jc 1 = mr The moment of inertia J of the cylinder about ais is 1 3 J = Jc + mr = mr + mr = mr Forced response and natural response. The behavior determined by a forcing function is called a forced response, and that due initial conditions is called natural response. The period between initiation of a response and the ending is referred to as the transient period. After the response has become negligibly small, conditions are said to have reached a steady state. Figure 38 Transient and steady state response Parallel and Series Springs Elements. In many applications, multiple spring elements are used, and in such cases we must obtain the equivalent spring constant of the combined elements. 9/34
10 Parallel Springs. For the springs in parallel, Figure 39, the equivalent spring constant k eq is obtained from the relation k 1 k e q k F F Figure 39 Parallel spring elements where F = k + k = k + k = k 1 1 eq keq = k1+ k for parallel springs (39) This formula can be etended to n springs connected sidebyside as follows: n k eq i= 1 i ( for parallel springs) = k (310) Series Springs. For the springs in series, Figure 310, the force in each spring is the same. Thus y k k1 F k eq F Figure 310 Series spring elements 1 F = k y, F = k y Eliminating from these two equations yields or F F = k k1 k k k1+ k F = k F k= F+ F = F k1 k1 k1 10/34
11 or where k + k k k 1 1 = F F = = kk k1+ k + k1 k k eq = + for series springs k k k eq 1 (311) which can be etended to the case of n springs connected endtoend as follows n 1 1 = for series springs k k eq i= 1 i (31) Free Vibration without damping. Consider the mass spring system shown in Figure The equation of motion can or where be given by 1 m && + k= 0 k && + = && + = m 0 ωn 0 ω = n k m K is the natural frequency of the system and is epressed in rad/s. Taking LT of both sides of the above equation where ( 0) = o and & ( 0) = & o gives & [&&] sx s s ω X s = n L () t M Figure 311 Mass Spring System rearrange to get 1 The weight W = mg of the mass m is equal to the static deflection of the spring. Therefore, they will cancel each other, (See Appendi A). 11/34
12 so + & o X( s) =, Remember poles are s=± jω s + ω 1443 n n & o ωn s X() s = + o ω s + ω s + ω and the response t is given by It is clear that the response n n n & t nt o nt ω o () = sin( ω ) + cos( ω ) n comple conjugates t consists of a sine and cosine terms and depends on the values of the initial conditions o and & o. Periodic motion such that described by the above equation is called simple harmonic motion. ( t) slope = & o Im s plane o ω n t Re ω n π Period = T = ω n Figure 31 Free response of a simple harmonic motion and pole location on the splane & 0 = & = 0, if o cos( ω ) t t = o n The period T is the time required for a periodic motion to repeat itself. In the present case, Period π T = seconds ω The frequency f of a periodic motion is the number of cycles per second (cps), and the standard unit of frequency is the Hertz (Hz); that is 1 Hz is 1 cps. In the present case, n 1/34
13 Frequency f 1 = = T ω π Hz The undamped natural frequency ω n is the frequency in the free vibration of a system having no damping. If the natural frequency is measured in Hz or in cps, it is denoted by f n. If it is measured in rad/sec, it is denoted by ω n. In the present system, k ωn = π fn = rad/sec m Rotational System. Rotor mounted in bearings is shown in figure 313 below. The moment of inertia of the rotor about the ais of rotation is J. Friction in the bearings is viscous friction and that no eternal torque is applied to the rotor. Tt () J ω b J ω bω Figure 313 Rotor mounted in bearings and its FBD. Apply Newton s second law for a system in rotation form or M = J&& θ = J & ω J & ω + bω = 0 & ω+ b/ J ω = 0 1 & ω+ ω = 0 ( J/ b) Define the time constant τ = ( J / b), the previous equation can be written in the 1 & ω + ω = 0, ω( 0) = ω o τ which represents the equation of motion as well as the mathematical model of the system shown. It represents a first order system. To find the response ω ( t) LT of both sides of the previous equation., take 13/34
14 1 sω( s) ω ( 0) + Ω ( s) = τ { L[ ω] L[ ω & ] 1 s+ Ω ( s) = ω τ + 1 τ + 1τ = 0 o ( s) where the denominator equation above equation will give the epression of ω ( t) ωo Ω = s + ( 1 τ ) s is known as the characteristic polynomial and the s is called the characteristic equation. Taking inverse LT of the b/ J t 1/τ t αt o o o ω t = ω e = ω e = ω e It is clear that the angular velocity decreases eponentially as shown in the figure ( t / τ ) below. Since lim e = 0; then for such decaying system, it is convenient to t depict the response in terms of a time constant. 1.5 Free response of a rotor bearing system α=0 ω o ω(t) α=0. α=0.5 α= α=1 α= α=5 α= time (t) Figure 314 Graph of ω α t o e for ranges of α. 14/34
15 A time constant is that value of time that makes the eponent equal to 1. For this system, time constant τ = J/ b. When t = τ, the eponent factor is ( t / τ) ( τ / τ) 1 This means that when time constant=, the time τ response is reduced to 36.8 % of its final value. We also have τ = J / b = time constant ωτ = 0.37 ωo ω 4τ = 0.0ω e = e = e = = 36.8 % o Figure 315 Curve of angular velocity ω versus time t for the rotor shown in Figure SpringMassDamper System. Consider the simple mechanical system shown involving viscous damping. Obtain the mathematical model of the system shown. b k Fb = b& Fk = k m m ( t) FreeBody Diagram(FBD) Figure 316 Mass Spring Damper System and the FBD. i) The FBD is shown in the figure ii) Apply Newton s second law of motion to a system in translation: Rearranging F = m&& b& k = m&& 15/34
16 m&& + b& + k= 0 The above equation represents the mathematical model as well as the free vibration for a second order model. If m= 01. kg, b= 04. N/ms, and k = 4N/m, the above differential equation becomes 0.1 && + 0.4& + 4 = 0 && + 4& + 40 = 0 To obtain the free response ( t ), assume ( 0) = o and transform of both sides of the given equation & () () & Take Laplace 0 = 0. sx s s sx s X s = L && t L & t L t Substitute in the transformed equation ( 0) obtain or solving for which can be written as () = o and ( 0) = 0, [ ] s + 4s+ 40 X s = so + 4 o X s yields ( so + 4o) ( s + 4) X( s) = = s + 4s s + 4s+ 40 G s Characteristic polynomial ( s + 4) X s = = o 1443 s + 4s + 40 Characteristic polynomial & and rearrange, we where G( s ) is referred to as the transfer function that gives the relationship between the input and the output X( s ). G( s ) can be shown graphically as: o o o ( s + 4) s s + 40 Characteristic polynomial Transfer function Xs Figure 317 Transfer function between input and output. iii) It is clear that the characteristic equation of the system is s + 4s + 40= 0 and has comple conjugates roots. 16/34
17 ( s+ ) s + 4s+ 40= s s = s = 0 The roots of the above equation are therefore comple conjugate poles given by s1 = + j6 and s = j6 iv) The epression of X( s ) can be written now as: s + 4s+ 40 ( s + ) 1 6 o ( s+ ) + 3 ( s+ ) + s+ 4 s+ + s+ X( s) = = = + s+ + 6 s+ + 6 s+ + 6 o o o o = v) Solving for ( t 1 ) = L X ( s ) yields () t = cos + Or 10 t o t () = e o ( sin 6t ) t e t T d = π ω d 3 o 1 t t o e 6t e sin 6t 3 s1 = + j6 t 1 t t = o e cost 6 e sint t Im j6 j6 s = j6 s plane Re Figure 318 Free Vibration of the massspringdamper system described by && + 4& + 40 = 0 with initial conditions ( 0) = o and & 0 = 0,. (See Appendi B). 17/34
18 10 PoleZero Map 5 Imaginary Ais Real Ais 1.5 Free vibration of massspringdamper system o (t) t (sec) Figure 318 Free Vibration of the massspringdamper system described by && + 4& + 40 = 0 with initial conditions ( 0) = o and & 0 = 0,. 3.4 WORK ENERGY, AND POWER Work. The work done in a mathematical system is the product of a force and a distance (or a torque and the angular displacement) through which the force is eerted with both force and distance measured in the same direction. d F d θ F W = F d W = F dcosθ Figure 319 Work done by a force The units of work in SI units are : [ work] = [ force distance] = [ Nm] = [ Joule] = [ J]. The work done by a spring is given by: 18/34
19 F 1 W = k d= k { 0 F k l l+ F Figure 319 Work done by a spring. Energy. Energy can be defined as ability to do work. Energy can be found in many different forms and can be converted from one form to another. For instance, an electric motor converts electrical energy into mechanical energy, a battery converts chemical energy into electrical energy, and so forth. According to the law of conservation of energy, energy can be neither created nor destroyed. This means that the increase in the total energy within a system is equal to the net energy input to the system. So if there is no energy input, there is no change in the total energy of the system. Potential Energy. The Energy that a body possesses because of its position is called potential energy. In mechanical systems, only mass and spring can store potential energy. The change in the potential energy stored in a system equals the work required to change the system s configuration. Potential energy is always measured with reference to some chosen level and is relative to that level. h m mg F Figure 30 Potential energy Refer to Figure 30, the potential energy, U of a mass m is given by: U = mg d= mgh 0 19/34
20 For a translational spring, the potential energy U (sometimes called strain energy which is potential energy that is due to elastic deformations) is: 1 U = Fd= kd= k 0 0 If the initial and final values of are 1 and, respectively, then 1 1 Change in potential energy Δ U= Fd= kd= k k1 Similarly, for a torsional spring 1 1 θ 1 1 Change in potential energy Δ U = T dθ = k θ d= k θ k θ θ θ θ 1 1 T T T 1 Kinetic Energy. Only inertia elements can store kinetic energy in mechanical systems. T 1 mv = Kinetic energy= 1 Jθ& (Translation) (Rotation) The change in kinetic energy of the mass is equal to the work done on it by an applied force as the mass accelerates or decelerates. Thus, the change in kinetic energy T of a mass m moving in a straight line is Change in kinetic energy d Δ T =Δ W = Fd= F dt dt = Fvdt= mvv & dt= mvdv = 1 mv 1 mv t 1 t1 t t v t t v 1 The change in kinetic energy of a moment of inertia in pure rotation at angular velocity θ & is 1 1 Change in kinetic energy Δ T = J & θ J & θ 1 0/34
21 Dissipated Energy. Consider the damper shown in Figure 31 in which one end is fied and the other end is moved from 1 to. The dissipated energy Δ W in the damper is equal to the net work done on it: t t d Δ W = Fd= b & d= b dt = b dt { & dt & F t t b Figure 30 Damper. Power. Power is the time rate of doing work. That is, Power = P = dw dt where dw denotes work done during time interval dt. In SI units, the work done is measured in Newtonmeters and the time in seconds. The unit of power is : Nm Joule Power = = = Watt = W s s [ ] [ ]. Passsive Elements. Nonenergy producing element. They can only store energy, not generate it such as springs and masses. Active Elements. forces and torques. Energy producing elements such as eternal Energy Method for Deriving Equations of Motion. Equations of motion are derive from the fact that the total energy of a system remains the same if no energy enters or leaves the system. Conservative Systems. Systems that do not involve friction (damping) are called conservative systems. 1/34
22 ( T U) { W Δ + = Δ 1443 Change in the total energy Net work done on the system by eternal forces If no eternal energy enters the system ( W 0 forces) then Δ =, no work done by eternal or ( T U) 0 Δ + = ( T + U) = constant Conservation of energy only for conservative systems (No friction or damping) An Energy Method for Determining Natural Frequencies. The natural frequency of a conservative system can be obtained from a consideration of the kinetic energy and the potential energy of the system. Let us assume that we choose the datum line so that the potential energy at the equilibrium state is zero. Then in such a conservative system, the maimum kinetic energy equals the maimum potential energy, or Solved Problems. Eample 35 Page 80 (Tetbook) T ma = U ma Consider the system shown in the Figure shown. The displacement is measured from the equilibrium position. The Kinetic energy is: The potential energy is: The total energy of the system is The change in the total energy is d dt 1 T = m& 1 U = k Since & is not zero then we should have 1 1 T + U = m& + k ( && k) T U d d m k + = & + = t 1 1 = m &&& + k& = 0 = & m+ = 0 m&& + k = 0 m k /34
23 or where k && + = && + = m 0 ωn 0 ω = n k m is the natural frequency of the system and is epressed in rad/s. Another way of finding the natural frequency of the system is to assume a displacement of the form = Asinωnt Where A is the amplitude of vibration. Consequently, 1 1 & 1 1 ( cosω ) T = m = m A ω nt ( sinω ) U = k = k A nt Hence the maimum values of T and U are given by Since Tma = Uma, we have From which 1 1 Tma = m A ω, U = k A n ma 1 1 maω = ka n ω = n k m 3/34
24 Problem 4/34
25 Problem 5/34
26 6/34
27 7/34
28 8/34
29 9/34
30 TABLE 1. Summary of elements involved in linear mechanical systems Element Translation Rotation Inertia F 4 F 3 m F = m a F F 1 T J T = J α θ Spring F 1 k F T θ 1 k θ T F = k ) = k ( 1 T = k θ θ ) = kθ ( 1 Damper F & & 1 b F T & θ 1 b θ & T F = b & & ) = b& ( 1 T = b & θ & θ ) = b & θ ( 1 PROCEDURE The motion of mechanical elements can be described in various dimensions as translational, rotational, or combination of both. The equations governing the motion of mechanical systems are often formulated from Newton s law of motion. 1. Construct a model for the system containing interconnecting elements.. Draw the freebody diagram. 3. Write equations of motion of all forces acting on the free body diagram. For translational motion, the equation of motion is Equation (1), and for rotational motion, Equation () is used. 30/34
31 APPENDIX A: Static Equilibrium k L o k k δ st m m kδ st m m k ( + δ st ) mg mg At static equilibrium, we have: mg = Kδ st Let the mass be displaced a distance downward from its equilibrium, the spring force becomes f = K( + δ st ). Apply Newton s second law for the massspring system, one can write: or or Therefore ( δ ) F = m && K + + mg = m && st K Kδ + mg = m && st 1443 = 0 Static Equilibrium m && + K = 0 31/34
32 The above equation represents the mathematical model for the massspring system. It is a second order ordinary differential equation with constant coefficients. It is clear that the weight mg cancels with the static deflection Kδ of the system. Therefore, it does not appear in the equation of motion. st 3/34
33 APPENDIX B: The epression of acos( ωt) + bsin( ωt ) can be written in terms of sin( ωt ), that is cos ωt or a b a cos ( ωt ) + b sin ( ωt ) = a + b cos ( ωt ) + sin ( ωt ) a + b a + b Define ψ such that cosψ = sin ψ = a a + b 1 b ψ tan b = a a + b Therefore a cos ( ωt ) + b sin ( ωt ) = a + b cosψ cos ( ωt ) + sin ψ sin ( ωt ) Using the identity ( ) = + cos ωt ψ cos ωt cosψ sin ωt sinψ Therfore, or Where ( ω ) + ( ω ) = + ( ω ψ) acos t bsin t a b cos t ( ω ) + ( ω ) = + ( ω + ϕ) a cos t b sin t a b sin t sin ϕ = cos ϕ = a + ϕ tan b = b a + b a b 1 a 33/34
34 Useful Sites Torsion Mass Spring System Response of First and second Order Systems 34/34
Hooke s Law. Spring. Simple Harmonic Motion. Energy. 12/9/09 Physics 201, UWMadison 1
Hooke s Law Spring Simple Harmonic Motion Energy 12/9/09 Physics 201, UWMadison 1 relaxed position F X = kx > 0 F X = 0 x apple 0 x=0 x > 0 x=0 F X =  kx < 0 x 12/9/09 Physics 201, UWMadison 2 We know
More informationChapter 24 Physical Pendulum
Chapter 4 Physical Pendulum 4.1 Introduction... 1 4.1.1 Simple Pendulum: Torque Approach... 1 4. Physical Pendulum... 4.3 Worked Examples... 4 Example 4.1 Oscillating Rod... 4 Example 4.3 Torsional Oscillator...
More informationHW 7 Q 14,20,20,23 P 3,4,8,6,8. Chapter 7. Rotational Motion of the Object. Dr. Armen Kocharian
HW 7 Q 14,20,20,23 P 3,4,8,6,8 Chapter 7 Rotational Motion of the Object Dr. Armen Kocharian Axis of Rotation The radian is a unit of angular measure The radian can be defined as the arc length s along
More informationSlide 10.1. Basic system Models
Slide 10.1 Basic system Models Objectives: Devise Models from basic building blocks of mechanical, electrical, fluid and thermal systems Recognize analogies between mechanical, electrical, fluid and thermal
More informationA B = AB sin(θ) = A B = AB (2) For two vectors A and B the cross product A B is a vector. The magnitude of the cross product
1 Dot Product and Cross Products For two vectors, the dot product is a number A B = AB cos(θ) = A B = AB (1) For two vectors A and B the cross product A B is a vector. The magnitude of the cross product
More information1.10 Using Figure 1.6, verify that equation (1.10) satisfies the initial velocity condition. t + ") # x (t) = A! n. t + ") # v(0) = A!
1.1 Using Figure 1.6, verify that equation (1.1) satisfies the initial velocity condition. Solution: Following the lead given in Example 1.1., write down the general expression of the velocity by differentiating
More informationSIMPLE HARMONIC MOTION: SHIFTED ORIGIN AND PHASE
MISN026 SIMPLE HARMONIC MOTION: SHIFTED ORIGIN AND PHASE SIMPLE HARMONIC MOTION: SHIFTED ORIGIN AND PHASE by Kirby Morgan 1. Dynamics of Harmonic Motion a. Force Varies in Magnitude and Direction................
More informationChapter 07: Kinetic Energy and Work
Chapter 07: Kinetic Energy and Work Conservation of Energy is one of Nature s fundamental laws that is not violated. Energy can take on different forms in a given system. This chapter we will discuss work
More informationRotational Dynamics. Luis Anchordoqui
Rotational Dynamics Angular Quantities In purely rotational motion, all points on the object move in circles around the axis of rotation ( O ). The radius of the circle is r. All points on a straight line
More informationRotation, Angular Momentum
This test covers rotational motion, rotational kinematics, rotational energy, moments of inertia, torque, crossproducts, angular momentum and conservation of angular momentum, with some problems requiring
More informationAngular velocity and angular acceleration CHAPTER 9 ROTATION. Angular velocity and angular acceleration. ! equations of rotational motion
Angular velocity and angular acceleration CHAPTER 9 ROTATION! r i ds i dθ θ i Angular velocity and angular acceleration! equations of rotational motion Torque and Moment of Inertia! Newton s nd Law for
More informationSolution Derivations for Capa #10
Solution Derivations for Capa #10 1) The flywheel of a steam engine runs with a constant angular speed of 172 rev/min. When steam is shut off, the friction of the bearings and the air brings the wheel
More informationModeling Mechanical Systems
chp3 1 Modeling Mechanical Systems Dr. Nhut Ho ME584 chp3 2 Agenda Idealized Modeling Elements Modeling Method and Examples Lagrange s Equation Case study: Feasibility Study of a Mobile Robot Design Matlab
More informationSimple Harmonic Motion Concepts
Simple Harmonic Motion Concepts INTRODUCTION Have you ever wondered why a grandfather clock keeps accurate time? The motion of the pendulum is a particular kind of repetitive or periodic motion called
More informationChapter 8 Rotational Motion
Chapter 8 Rotational Motion Textbook (Giancoli, 6 th edition): Assignment 9 Due on Thursday, November 26. 1. On page 131 of Giancoli, problem 18. 2. On page 220 of Giancoli, problem 24. 3. On page 221
More informationChapter 13, example problems: x (cm) 10.0
Chapter 13, example problems: (13.04) Reading Fig. 1330 (reproduced on the right): (a) Frequency f = 1/ T = 1/ (16s) = 0.0625 Hz. (since the figure shows that T/2 is 8 s.) (b) The amplitude is 10 cm.
More informationPhysics 211 Week 12. Simple Harmonic Motion: Equation of Motion
Physics 11 Week 1 Simple Harmonic Motion: Equation of Motion A mass M rests on a frictionless table and is connected to a spring of spring constant k. The other end of the spring is fixed to a vertical
More informationAdvanced Higher Physics: MECHANICS. Simple Harmonic Motion
Advanced Higher Physics: MECHANICS Simple Harmonic Motion At the end of this section, you should be able to: Describe examples of simple harmonic motion (SHM). State that in SHM the unbalanced force is
More informationPeriodic Motion or Oscillations. Physics 232 Lecture 01 1
Periodic Motion or Oscillations Physics 3 Lecture 01 1 Periodic Motion Periodic Motion is motion that repeats about a point of stable equilibrium Stable Equilibrium Unstable Equilibrium A necessary requirement
More informationHomework #7 Solutions
MAT 0 Spring 201 Problems Homework #7 Solutions Section.: 4, 18, 22, 24, 4, 40 Section.4: 4, abc, 16, 18, 22. Omit the graphing part on problems 16 and 18...4. Find the general solution to the differential
More informationp = F net t (2) But, what is the net force acting on the object? Here s a little help in identifying the net force on an object:
Harmonic Oscillator Objective: Describe the position as a function of time of a harmonic oscillator. Apply the momentum principle to a harmonic oscillator. Sketch (and interpret) a graph of position as
More informationResponse to Harmonic Excitation Part 2: Damped Systems
Response to Harmonic Excitation Part 2: Damped Systems Part 1 covered the response of a single degree of freedom system to harmonic excitation without considering the effects of damping. However, almost
More information9.1 Rotational Kinematics: Angular Velocity and Angular Acceleration
Ch 9 Rotation 9.1 Rotational Kinematics: Angular Velocity and Angular Acceleration Q: What is angular velocity? Angular speed? What symbols are used to denote each? What units are used? Q: What is linear
More informationChapter 9 Rotation of Rigid Bodies
Chapter 9 Rotation of Rigid Bodies 1 Angular Velocity and Acceleration θ = s r (angular displacement) The natural units of θ is radians. Angular Velocity 1 rad = 360o 2π = 57.3o Usually we pick the zaxis
More informationOUTCOME 2 KINEMATICS AND DYNAMICS TUTORIAL 2 PLANE MECHANISMS. You should judge your progress by completing the self assessment exercises.
Unit 60: Dynamics of Machines Unit code: H/601/1411 QCF Level:4 Credit value:15 OUTCOME 2 KINEMATICS AND DYNAMICS TUTORIAL 2 PLANE MECHANISMS 2 Be able to determine the kinetic and dynamic parameters of
More informationphysics 111N rotational motion
physics 111N rotational motion rotations of a rigid body! suppose we have a body which rotates about some axis! we can define its orientation at any moment by an angle, θ (any point P will do) θ P physics
More informationPeople s Physics book 3e Ch 251
The Big Idea: In most realistic situations forces and accelerations are not fixed quantities but vary with time or displacement. In these situations algebraic formulas cannot do better than approximate
More informationChapter 1. Oscillations. Oscillations
Oscillations 1. A mass m hanging on a spring with a spring constant k has simple harmonic motion with a period T. If the mass is doubled to 2m, the period of oscillation A) increases by a factor of 2.
More informationRotational Mechanics CHAPTER SOME IMPORTANT OBSERVATIONS
CHAPTER 6 Rotational Mechanics In this chapter, simple singledimensional rotational processes will be treated. Essentially, we will be concerned with wheels rotating around fixed axes. It will become
More informationVersion PREVIEW Practice 8 carroll (11108) 1
Version PREVIEW Practice 8 carroll 11108 1 This printout should have 12 questions. Multiplechoice questions may continue on the net column or page find all choices before answering. Inertia of Solids
More informationPhysics 53. Oscillations. You've got to be very careful if you don't know where you're going, because you might not get there.
Physics 53 Oscillations You've got to be very careful if you don't know where you're going, because you might not get there. Yogi Berra Overview Many natural phenomena exhibit motion in which particles
More informationChapter 17 Planar Kinetics of a Rigid Body: Force and Acceleration
Chapter 17 Planar Kinetics of a Rigid Body: Force and Acceleration 17.1 Moment of Inertia I 2 2 = r dm, 單位 : kg m 或 slug m ft 2 M = Iα resistance to angular acceleration dm = ρdv I = ρ V r 2 dv 172 MOMENT
More informationHooke s Law and Simple Harmonic Motion
Hooke s Law and Simple Harmonic Motion OBJECTIVE to measure the spring constant of the springs using Hooke s Law to explore the static properties of springy objects and springs, connected in series and
More informationRotational inertia (moment of inertia)
Rotational inertia (moment of inertia) Define rotational inertia (moment of inertia) to be I = Σ m i r i 2 or r i : the perpendicular distance between m i and the given rotation axis m 1 m 2 x 1 x 2 Moment
More informationPhysics 9e/Cutnell. correlated to the. College Board AP Physics 1 Course Objectives
Physics 9e/Cutnell correlated to the College Board AP Physics 1 Course Objectives Big Idea 1: Objects and systems have properties such as mass and charge. Systems may have internal structure. Enduring
More informationSimple Harmonic Motion
Simple Harmonic Motion Theory Simple harmonic motion refers to the periodic sinusoidal oscillation of an object or quantity. Simple harmonic motion is eecuted by any quantity obeying the Differential
More informationChapter 10 Rotational Motion. Copyright 2009 Pearson Education, Inc.
Chapter 10 Rotational Motion Angular Quantities Units of Chapter 10 Vector Nature of Angular Quantities Constant Angular Acceleration Torque Rotational Dynamics; Torque and Rotational Inertia Solving Problems
More informationEQUATIONS OF MOTION: ROTATION ABOUT A FIXED AXIS
EQUATIONS OF MOTION: ROTATION ABOUT A FIXED AXIS Today s Objectives: Students will be able to: 1. Analyze the planar kinetics InClass Activities: of a rigid body undergoing rotational motion. Check Homework
More informationPhysics 41 HW Set 1 Chapter 15
Physics 4 HW Set Chapter 5 Serway 8 th OC:, 4, 7 CQ: 4, 8 P: 4, 5, 8, 8, 0, 9,, 4, 9, 4, 5, 5 Discussion Problems:, 57, 59, 67, 74 OC CQ P: 4, 5, 8, 8, 0, 9,, 4, 9, 4, 5, 5 Discussion Problems:, 57, 59,
More informationTwo massthree spring system. Math 216 Differential Equations. Forces on mass m 1. Forces on mass m 2. Kenneth Harris
Two massthree spring system Math 6 Differential Equations Kenneth Harris kaharri@umich.edu m, m > 0, two masses k, k, k 3 > 0, spring elasticity t), t), displacement of m, m from equilibrium. Positive
More informationGeneral Physics I Can Statements
General Physics I Can Statements Motion (Kinematics) 1. I can describe motion in terms of position (x), displacement (Δx), distance (d), speed (s), velocity (v), acceleration (a), and time (t). A. I can
More informationLecture 17. Last time we saw that the rotational analog of Newton s 2nd Law is
Lecture 17 Rotational Dynamics Rotational Kinetic Energy Stress and Strain and Springs Cutnell+Johnson: 9.49.6, 10.110.2 Rotational Dynamics (some more) Last time we saw that the rotational analog of
More informationSOLID MECHANICS TUTORIAL MECHANISMS KINEMATICS  VELOCITY AND ACCELERATION DIAGRAMS
SOLID MECHANICS TUTORIAL MECHANISMS KINEMATICS  VELOCITY AND ACCELERATION DIAGRAMS This work covers elements of the syllabus for the Engineering Council exams C105 Mechanical and Structural Engineering
More informationSimple Harmonic Motion
Simple Harmonic Motion Objective: In this exercise you will investigate the simple harmonic motion of mass suspended from a helical (coiled) spring. Apparatus: Spring 1 Table Post 1 Short Rod 1 Rightangled
More informationResponse to Harmonic Excitation
Response to Harmonic Excitation Part 1 : Undamped Systems Harmonic excitation refers to a sinusoidal external force of a certain frequency applied to a system. The response of a system to harmonic excitation
More information3 Work, Power and Energy
3 Work, Power and Energy At the end of this section you should be able to: a. describe potential energy as energy due to position and derive potential energy as mgh b. describe kinetic energy as energy
More informationIMPORTANT NOTE ABOUT WEBASSIGN:
Week 8 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution
More informationLecture L222D Rigid Body Dynamics: Work and Energy
J. Peraire, S. Widnall 6.07 Dynamics Fall 008 Version.0 Lecture L  D Rigid Body Dynamics: Work and Energy In this lecture, we will revisit the principle of work and energy introduced in lecture L3 for
More information2.3 Cantilever linear oscillations
.3 Cantilever linear oscillations Study of a cantilever oscillation is a rather science  intensive problem. In many cases the general solution to the cantilever equation of motion can not be obtained
More informationGround Rules. PC1221 Fundamentals of Physics I. Force. Zero Net Force. Lectures 9 and 10 The Laws of Motion. Dr Tay Seng Chuan
PC1221 Fundamentals of Physics I Lectures 9 and 10 he Laws of Motion Dr ay Seng Chuan 1 Ground Rules Switch off your handphone and pager Switch off your laptop computer and keep it No talking while lecture
More informationSolutions 2.4Page 140
Solutions.4Page 4 Problem 3 A mass of 3 kg is attached to the end of a spring that is stretched cm by a force of 5N. It is set in motion with initial position = and initial velocity v = m/s. Find the
More informationSOLID MECHANICS DYNAMICS TUTORIAL INERTIA FORCES IN MECHANISMS
SOLID MECHANICS DYNAMICS TUTORIAL INERTIA FORCES IN MECHANISMS This work covers elements of the syllabus for the Engineering Council Exam D225 Dynamics of Mechanical Systems C103 Engineering Science. This
More informationSHM Simple Harmonic Motion revised June 16, 2012
SHM Simple Harmonic Motion revised June 16, 01 Learning Objectives: During this lab, you will 1. communicate scientific results in writing.. estimate the uncertainty in a quantity that is calculated from
More informationEquivalent Spring Stiffness
Module 7 : Free Undamped Vibration of Single Degree of Freedom Systems; Determination of Natural Frequency ; Equivalent Inertia and Stiffness; Energy Method; Phase Plane Representation. Lecture 13 : Equivalent
More informationCh.8 Rotational Equilibrium and Rotational Dynamics.
Ch.8 Rotational Equilibrium and Rotational Dynamics. Conceptual question # 1, 2, 9 Problems# 1, 3, 7, 9, 11, 13, 17, 19, 25, 28, 33, 39, 43, 47, 51, 55, 61, 79 83 Torque Force causes acceleration. Torque
More informations r or equivalently sr linear velocity vr Rotation its description and what causes it? Consider a disk rotating at constant angular velocity.
Rotation its description and what causes it? Consider a disk rotating at constant angular velocity. Rotation involves turning. Turning implies change of angle. Turning is about an axis of rotation. All
More informationCh 7 Kinetic Energy and Work. Question: 7 Problems: 3, 7, 11, 17, 23, 27, 35, 37, 41, 43
Ch 7 Kinetic Energy and Work Question: 7 Problems: 3, 7, 11, 17, 23, 27, 35, 37, 41, 43 Technical definition of energy a scalar quantity that is associated with that state of one or more objects The state
More informationChapter 6 Work and Energy
Chapter 6 WORK AND ENERGY PREVIEW Work is the scalar product of the force acting on an object and the displacement through which it acts. When work is done on or by a system, the energy of that system
More informationChapt. 10 Rotation of a Rigid Object About a Fixed Axis
Chapt. otation of a igid Object About a Fixed Axis. Angular Position, Velocity, and Acceleration Translation: motion along a straight line (circular motion?) otation: surround itself, spins rigid body:
More informationBead moving along a thin, rigid, wire.
Bead moving along a thin, rigid, wire. odolfo. osales, Department of Mathematics, Massachusetts Inst. of Technology, Cambridge, Massachusetts, MA 02139 October 17, 2004 Abstract An equation describing
More informationChapter Test. Teacher Notes and Answers Forces and the Laws of Motion. Assessment
Assessment Chapter Test A Teacher Notes and Answers Forces and the Laws of Motion CHAPTER TEST A (GENERAL) 1. c 2. d 3. d 4. c 5. c 6. c 7. c 8. b 9. d 10. d 11. c 12. a 13. d 14. d 15. b 16. d 17. c 18.
More informationRotation, Rolling, Torque, Angular Momentum
Halliday, Resnick & Walker Chapter 10 & 11 Rotation, Rolling, Torque, Angular Momentum Physics 1A PHYS1121 Professor Michael Burton Rotation 101 Rotational Variables! The motion of rotation! The same
More informationExplaining Motion:Forces
Explaining Motion:Forces Chapter Overview (Fall 2002) A. Newton s Laws of Motion B. Free Body Diagrams C. Analyzing the Forces and Resulting Motion D. Fundamental Forces E. Macroscopic Forces F. Application
More informationSIMPLE HARMONIC MOTION
PERIODIC MOTION SIMPLE HARMONIC MOTION If a particle moves such that it repeats its path regularly after equal intervals of time, its motion is said to be periodic. The interval of time required to complete
More informationSample Questions for the AP Physics 1 Exam
Sample Questions for the AP Physics 1 Exam Sample Questions for the AP Physics 1 Exam Multiplechoice Questions Note: To simplify calculations, you may use g 5 10 m/s 2 in all problems. Directions: Each
More information(Most of the material presented in this chapter is taken from Thornton and Marion, Chap. 3)
Chapter 2. Small Oscillations Most of the material presented in this chapter is taken from Thornton and Marion, Chap. 3) 2.1 Introduction If a particle, originally in a position of equilibrium we limit
More information9 ROTATIONAL DYNAMICS
CHAPTER 9 ROTATIONAL DYNAMICS CONCEPTUAL QUESTIONS 1. REASONING AND SOLUTION The magnitude of the torque produced by a force F is given by τ = Fl, where l is the lever arm. When a long pipe is slipped
More informationANALYTICAL METHODS FOR ENGINEERS
UNIT 1: Unit code: QCF Level: 4 Credit value: 15 ANALYTICAL METHODS FOR ENGINEERS A/601/1401 OUTCOME  TRIGONOMETRIC METHODS TUTORIAL 1 SINUSOIDAL FUNCTION Be able to analyse and model engineering situations
More information1 of 10 11/23/2009 6:37 PM
hapter 14 Homework Due: 9:00am on Thursday November 19 2009 Note: To understand how points are awarded read your instructor's Grading Policy. [Return to Standard Assignment View] Good Vibes: Introduction
More informationLesson 04: Newton s laws of motion
www.scimsacademy.com Lesson 04: Newton s laws of motion If you are not familiar with the basics of calculus and vectors, please read our freely available lessons on these topics, before reading this lesson.
More informationDynamics. Today: Linear System Dynamics. EEL5225: Principles of MEMS Transducers (Fall 2003) Instructor: Dr. HuiKai Xie
Today: Instructor: Dr. HuiKai Xie Dynamics Linear System Dynamics Direct integration System functions Superposition Convolution Sinusoidal steady state Reading: Senturia, pp. 78, Chapter 7, pp. 4960
More informationPractice Exam Three Solutions
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01T Fall Term 2004 Practice Exam Three Solutions Problem 1a) (5 points) Collisions and Center of Mass Reference Frame In the lab frame,
More informationMechanics Cycle 1 Chapter 13. Chapter 13
Chapter 13 Torque * How Shall a Force be Applied to Start a Rotation? The Torque Formula Systems in Equilibrium: Statics Glimpse of Newton s rotational second law: Q: Review: What causes linear acceleration
More informationF mg (10.1 kg)(9.80 m/s ) m
Week 9 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution
More informationLinear Centripetal Tangential speed acceleration acceleration A) Rω Rω 2 Rα B) Rω Rα Rω 2 C) Rω 2 Rα Rω D) Rω Rω 2 Rω E) Rω 2 Rα Rω 2 Ans: A
1. Two points, A and B, are on a disk that rotates about an axis. Point A is closer to the axis than point B. Which of the following is not true? A) Point B has the greater speed. B) Point A has the lesser
More informationAP1 Oscillations. 1. Which of the following statements about a springblock oscillator in simple harmonic motion about its equilibrium point is false?
1. Which of the following statements about a springblock oscillator in simple harmonic motion about its equilibrium point is false? (A) The displacement is directly related to the acceleration. (B) The
More informationEQUATIONS OF MOTION: ROTATION ABOUT A FIXED AXIS
EQUATIONS OF MOTION: ROTATION ABOUT A FIXED AXIS Today s Objectives: Students will be able to: 1. Analyze the planar kinetics of a rigid body undergoing rotational motion. InClass Activities: Applications
More informationPhys101 Third Major Exam Term 142
Phys0 Third Major Exam Term 4 Q. The angular position of a point on the rim of a rotating wheel of radius R is given by: θ (t) = 6.0 t + 3.0 t.0 t 3, where θ is in radians and t is in seconds. What is
More informationNotes on the Periodically Forced Harmonic Oscillator
Notes on the Periodically orced Harmonic Oscillator Warren Weckesser Math 38  Differential Equations 1 The Periodically orced Harmonic Oscillator. By periodically forced harmonic oscillator, we mean the
More informationPhysics Notes Class 11 CHAPTER 5 LAWS OF MOTION
1 P a g e Inertia Physics Notes Class 11 CHAPTER 5 LAWS OF MOTION The property of an object by virtue of which it cannot change its state of rest or of uniform motion along a straight line its own, is
More informationAR9161 B.Tech. VI Sem. Chemical Engineering Process Dynamics &Control Model Answer
AR9161 B.Tech. VI Sem. Chemical Engineering Process Dynamics &Control Model Answer Ans (1) Section A i. (A) ii. iii. iv. (B) (B) (B) v. (D) vi. vii. viii. ix. (C) (B) (B) (C) x. (A) Section B (1) (i)
More informationForces. using a consistent system of units, such as the metric system, we can define force as:
Forces Force: physical property which causes masses to accelerate (change of speed or direction) a push or pull vector possessing both a magnitude and a direction and adds according to the Parallelogram
More informationPHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013
PHYSICS HOMEWORK SOLUTION #0 April 8, 203 0. Find the net torque on the wheel in the figure below about the axle through O, taking a = 6.0 cm and b = 30.0 cm. A torque that s produced by a force can be
More informationPhysics 1A Lecture 10C
Physics 1A Lecture 10C "If you neglect to recharge a battery, it dies. And if you run full speed ahead without stopping for water, you lose momentum to finish the race. Oprah Winfrey Static Equilibrium
More informationwww.mathsbox.org.uk Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx Acceleration Velocity (v) Displacement x
Mechanics 2 : Revision Notes 1. Kinematics and variable acceleration Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx differentiate a = dv = d2 x dt dt dt 2 Acceleration Velocity
More information23.7. An Application of Fourier Series. Introduction. Prerequisites. Learning Outcomes
An Application of Fourier Series 23.7 Introduction In this Section we look at a typical application of Fourier series. The problem we study is that of a differential equation with a periodic (but nonsinusoidal)
More informationLABORATORY 9. Simple Harmonic Motion
LABORATORY 9 Simple Harmonic Motion Purpose In this experiment we will investigate two examples of simple harmonic motion: the massspring system and the simple pendulum. For the massspring system we
More informationPhysics 201 Homework 8
Physics 201 Homework 8 Feb 27, 2013 1. A ceiling fan is turned on and a net torque of 1.8 Nm is applied to the blades. 8.2 rad/s 2 The blades have a total moment of inertia of 0.22 kgm 2. What is the
More informationForced Oscillations in a Linear System
Forced Oscillations in a Linear System Manual Eugene Butikov Annotation. The manual includes a description of the simulated physical system and a summary of the relevant theoretical material for students
More informationPHY231 Section 2, Form A March 22, 2012. 1. Which one of the following statements concerning kinetic energy is true?
1. Which one of the following statements concerning kinetic energy is true? A) Kinetic energy can be measured in watts. B) Kinetic energy is always equal to the potential energy. C) Kinetic energy is always
More informationPHYS2020: General Physics II Course Lecture Notes Section VII
PHYS2020: General Physics II Course Lecture Notes Section VII Dr. Donald G. Luttermoser East Tennessee State University Edition 4.0 Abstract These class notes are designed for use of the instructor and
More informationSECONDORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS
L SECONDORDER LINEAR HOOGENEOUS DIFFERENTIAL EQUATIONS SECONDORDER LINEAR HOOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS A secondorder linear differential equation is one of the form d
More information= mg [down] =!mg [up]; F! x
Section 4.6: Elastic Potential Energy and Simple Harmonic Motion Mini Investigation: Spring Force, page 193 Answers may vary. Sample answers: A. The relationship between F g and x is linear. B. The slope
More informationUnit  6 Vibrations of Two Degree of Freedom Systems
Unit  6 Vibrations of Two Degree of Freedom Systems Dr. T. Jagadish. Professor for Post Graduation, Department of Mechanical Engineering, Bangalore Institute of Technology, Bangalore Introduction A two
More informationcharge is detonated, causing the smaller glider with mass M, to move off to the right at 5 m/s. What is the
This test covers momentum, impulse, conservation of momentum, elastic collisions, inelastic collisions, perfectly inelastic collisions, 2D collisions, and centerofmass, with some problems requiring
More informationDynamic Response of Measurement Systems
ME231 Measurements Laboratory Spring 1999 Dynamic Response of Measurement Systems Edmundo Corona c The handout Getting Ready to Measure presented a couple of examples where the results of a static calibration
More informationHOOKE'S LAW AND A SIMPLE SPRING DONALD C. PECKHAM PHYSICS 307 FALL 1983 ABSTRACT
HOOKE'S LAW AND A SIMPLE SPRING DONALD C. PECKHAM PHYSICS 307 FALL 983 (Digitized and Revised, Fall 005) ABSTRACT The spring constant of a screendoor spring was determined both statically, by measuring
More informationMechanics 1. Revision Notes
Mechanics 1 Revision Notes July 2012 MECHANICS 1... 2 1. Mathematical Models in Mechanics... 2 Assumptions and approximations often used to simplify the mathematics involved:... 2 2. Vectors in Mechanics....
More informationExamination of Viscous Torque Relationships
Examination of Viscous Torque Relationships Nathan D. Schiffrik Physics Department, The College of Wooster, Wooster, Ohio 4469 April 29, 998 Viscous torque, the resistance that a fluid offers to rotational
More informationChapter 4 HW Solution
Chapter 4 HW Solution Review Questions. 1. Name the performance specification for first order systems. Time constant τ. 2. What does the performance specification for a first order system tell us? How
More information