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Introduction to Dynamics
 

Introduction to Dynamics

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The basic concepts associated with some elements of dynamic engineering systems and calculations are discussed. The theory associated with Linear and angular motion, energy and simple harmonic motion ...

The basic concepts associated with some elements of dynamic engineering systems and calculations are discussed. The theory associated with Linear and angular motion, energy and simple harmonic motion are outlined with sample calculations for each topic being given.

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    Introduction to Dynamics Introduction to Dynamics Document Transcript

    • Introduction To Dynamics Edexcel HN Unit: Engineering Science (NQF L4) Author: Leicester College Date created: Date revised: 2009 Abstract; The basic concepts associated with some elements of dynamic engineering systems and calculations are discussed. The theory associated with Linear and angular motion, energy and simple harmonic motion are outlined with sample calculations for each topic being given. Contents Introduction To Dynamics.................................................................................................................................1 Dynamic Systems..............................................................................................................................................2 Angular motion - Equations of Angular motion...........................................................................................3 Energy and Rotation..........................................................................................................................................5 Simple Harmonic Motion – Pendulum..............................................................................................................7 Credits................................................................................................................................................................9 These files support the Edexcel HN unit – Engineering Science (Mechanical) File name Unit Key words outcome Stress 1.1 Stress, strain, statics, young’s modulus introduction BM, shear force 1.1 Shear force, bending moment, stress diagrams Selecting beams 1.2 Beams, columns, struts, slenderness ratio Torsion 1.3 Torsion, stiffness, twisting introduction Dynamics 2.1/2.2 Linear motion, angular motion, energy, kinetic, potential, rotation introduction For further information regarding unit outcomes go to Edexcel.org.uk/ HN/ Engineering / Specifications © Leicester College 2009 This work is licensed under a HYPERLINK "http://creativecommons.org/licenses/by/2.0/uk/" n _parentCreative Commons Attribution 2.0 License.
    • Introduction to Dynamics Dynamic Systems These notes give some sample examples of Dynamic Mechanics. (Outcome 2) Linear Motion Equations of Linear Motion v = u + at s = (u + v ) t Where u = initial velocity (m/s) 2 v = final velocity (m/s) 2 2 v = u + 2as s = distance (m) s = ut + ½ at2 t = time (s) a = acceleration (m/s2) Example 1 A vehicle starting from rest accelerates at a rate of 3m/s2 for 8 seconds. It then continues at this velocity for 20 seconds before coming to rest again in 5 seconds. Determine the maximum velocity, rate of deceleration and total distance travelled. First – Draw a v/t graph of this motion; V (m/s) 3m/s2 1 2 3 8 28 33 t(s) We know u = 0, t1 = 8s, t2 = 20s, t3 = 5s and a = 3m/s2 To find v; v = u + at1 = 0 + 3 x 8 = 24 m/s To find s1 s = (u + v ) t1 = (0 + 24) 8 = 96m 2 2 To find s2 s = (u + v ) t2 = (24 + 24) 20 2 2 = 480m To find s3 s = (u + v ) t3 = (24 + 0) 5 = 60m 2 2 Stotal = s1 + s2 + s3 = 636m Page PAGE 9 of NUMPAGES *Arabic 9
    • Introduction to Dynamics To find deceleration; v = u + at where v = 0 a = - u/t = -24/5 = 4.8 m/s2 Angular motion - Equations of Angular motion θ = (ω1 + ω2 ) Where θ = Angular displacement(rads) 2 ω1= Initial Ang. Velocity (rad/s) ω2 = ω1 + άt ω2 = Final Ang. velocity (rad/s) ά = Angular acceleration (rad/s2) ω22= ω12 + 2άθ θ = ω1t + ½ άt2 Conversion of angular to linear to angular motion v = ωr and a = άr Example An electric motor starting from rest is accelerated to 500rpm in 3 seconds. It continues at this velocity for 20 seconds before coming to rest in 4 seconds. Determine the rate of acceleration and deceleration, the maximum angular velocity and the total number of revolutions during this motion. 500rpm to Rad/sec - 500 x 2π / 60 = 52.36 rad/s First – Draw a v/t graph of this motion; ω (rad/s) 1 2 3 3 23 27 t(sec) Angular acceleration - ω2 = ω1 + άt and ω1 = 0 So ά = ω2 / t = 52.36 / 3 = 17.45 rad/s2 Revs in period 1 - θ = (ω1 + ω2 )t = (0 + 52.36) 3 2 2 = 78.54 Rads Page PAGE 9 of NUMPAGES *Arabic 9
    • Introduction to Dynamics Revs in period 2 - θ = (ω1 + ω2 )t = (52.36 + 52.36) 20 2 2 = 1047.2 Rads Revs in period 3 - θ = (ω1 + ω2 )t = (52.35 + 0) 4 2 2 = 104.7 Rads Total radians = 1230.46 Convert to revs – 1 / 2π = 195.83 Revolutions Deceleration As acceleration; ω2 = 0 and t = 4s ά = ω1 / t = 52.36 / 4 = 13.09 rad/s2 Page PAGE 9 of NUMPAGES *Arabic 9
    • Introduction to Dynamics Energy and Rotation Principle of conservation of energy – Energy cannot be created or destroyed but only converted from one for to another. There are two types of MECHANICAL energy. For linear motion they are; • Potential energy – Pe = mgh (The energy of position) • Kinetic energy - Ke = ½ mv2 (the energy of motion) When an object rotates there is no PE and Ke = ½ I ω2 Where I is the ‘moment of Inertia’ – this is a measure of an object’s resistance to change of motion or state. How we determine the value of I depends on the physical shape of the object; For a solid disk or cylinder I = mr2/2 For a concentrated mass I = mr2. Where m = mass (kg) and r = radius (m) In order to change the state of uniform motion or rest of a rotating object about a central point we need to apply a TORQUE. T = Fr = I ά Example1 A cylinder, 150mm diameter, with a mass of 8.5 kg, rolls along a flat surface at a constant velocity of 1.6 m/s. Determine the total KE of the cylinder (ie both linear and rotational KE) In this case I = mr2/2 = (8.5 x 0.0752)/2 = 0.0239 kgm2 Ke linear = ½ mv2 = 0.5 x 8.5 x 1.62 = 10.88J From previous examples v = ω r therefore ω = v/r = 1.6/0.075 = 21.33 rads/sec Ke rotational = ½ I ω2 = ½ x 0.0239 x 21.332 = 5.438J Total KE = Ke Linear + Ke rotational = 10.88 + 5.438 = 16.32J Example 2 A motor vehicle starting from rest free wheels down a slope of incline 1 in 8. Neglecting resistance to motion determine its velocity after it has traveled 200m down the slope. Pe lost = Ke gained (From conservation of energy) Page PAGE 9 of NUMPAGES *Arabic 9
    • Introduction to Dynamics Mgh = ½ mv2 The slope is 1/8 = 0.125. Becomes gh = ½ v2 Therefore h = 0.125 x 200 = 25m So v2 = 2gh and v = √(2 x 9.81 x 25) = 490J Example 3 A solid cylinder 0.1m diameter and 0.05m long is allowed to roll down a 15 degree slope. What is its linear velocity after it has traveled 0.8 m down the slope. Assume a density of cylinder material of 6800 kg/m3. Solid cylinder 150 Volume of cylinder = π r2l = π x 0.052 x 0.05 = 3.927 x 10-4 m3 Mass = density x volume = 6800 x 3.927 x 10-4 = 2.67 kg Pe = mgh where h = 0.8 x sin 15 = 0.207m = 2.67 x 9.81 x 0.207 = 5.422J Following the principle of conservation of energy; Pe lost = Ke gained = Ke linear + Ke rotational Ke linear = ½ mv2 = 0.5 x 2.67 x v2 = 1.335v2 Ke rotational = ½ I ω2 where I = ½ mr2 = ½ x 2.67 x 0.052 = 3.3375 x 10-3 And ω = (v/r)2 So Ke rotational = ½ I(v/r)2 = ½ x 3.3375 x 10-3 x v2/2.5x10-3 = 0.6667v2 According to the principle of conservation of energy: Pe = Ke lin. + Ke rot 5.422 = (1.335 + 0.6675)v2 v = √(5.422/2.0045) = 1.648 m/s Page PAGE 9 of NUMPAGES *Arabic 9
    • Introduction to Dynamics Simple Harmonic Motion – Pendulum For a pendulum The maximum linear velocity occurs at A The maximum linear acceleration B occurs at B v= ωr a = ω2r B Where ω = 2f A And f = 1 √(g/l) 2π Example 1 Determine the natural frequency, maximum linear velocity and acceleration of a simple pendulum with a cord length of 0.5m. f = 1 √(g/l) = 1 √(g/l) = 1/6.284 x √(9.81/0.5) 2π 2π = 0.705 Hz. Where ω = 2πf 2πx 0.705 = 4.43 rad/s v= ωr = 4.43 x 0.5 = 2.215 m/s a = ω2r = 4.432 x 0.5 = 9.81 m/s2 Example 2 A simple pendulum has a periodic time of 2s. What is the length of the cord ? Periodic time t = 1/f so f = 0.5Hz f = 1 √(g/l) Transpose for l 2π L= g (2 π f )2 = 0..994m Simple Harmonic Motion – Springs For a spring there are two displacements to consider Initially, an unloaded spring has no displacement. When a mass is placed on the spring it will deflect and remain static. This deflection is due to the stiffness of the spring – k Page PAGE 9 of NUMPAGES *Arabic 9
    • Introduction to Dynamics Where k = f/x = mg/x x is the extension of the spring due to load f m is the mass and g is acceleration due to gravity The second displacement (A – amplitude) is when the spring is extended with the spring loaded and released causing an oscillating motion. This gives us SHM about the point of deflection due to pre-loading of the spring. In this case; f = 1 √(k/m) and k = mg/x so 2π f = 1 √(g/x) Note m cancels 2π The maximum linear velocity, v= ωA and the maximum linear acceleration a = ω2A Example A spring extends by 25mm when a pre load of 1 kg is applied. It is then extended by a further 30mm and released. Determine the natural frequency of oscillation, the maximum velocity and acceleration for this situation. k = f/x = mg/x = 1 x 9.81/ 0.025 = 392.4 N/m f = 1 √(g/x) = 1 √(9.81 /0.025) = 3.15Hz 2π 2π As before ω = 2π f = 2π x 3.15 = 19.81 rad/s maximum linear velocity, v= ωA = 19.81 x 0.030 = 0.594 m/s maximum linear acceleration a = ω2A 19.812 x 0.030 = 11.774 m/s2 Note in this example that x is used to determine frequency and A is used to determine v and a. Page PAGE 9 of NUMPAGES *Arabic 9
    • Introduction to Dynamics Credits This resource was created Leicester College and released as an open educational resource through the Open Engineering Resources project of the Higher Education Academy Engineering Subject Centre. The Open Engineering Resources project was funded by HEFCE and part of the JISC/HE Academy UKOER programme. © 2009 Leicester College This work is licensed under a Creative Commons Attribution 2.0 License. The JISC logo is licensed under the terms of the Creative Commons Attribution-Non-Commercial-No Derivative Works 2.0 UK: England & Wales Licence. All reproductions must comply with the terms of that licence. The HEA logo is owned by the Higher Education Academy Limited may be freely distributed and copied for educational purposes only, provided that appropriate acknowledgement is given to the Higher Education Academy as the copyright holder and original publisher. The Leicester College name and logo is owned by the College and should not be produced without the express permission of the College. Page PAGE 9 of NUMPAGES *Arabic 9