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# Aerospace 3 Introduction to Dynamics and Systems ASEN 2003

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This 59 page Class Notes was uploaded by Laila Windler on Friday October 30, 2015. The Class Notes belongs to ASEN 2003 at University of Colorado at Boulder taught by Staff in Fall. Since its upload, it has received 19 views. For similar materials see /class/232176/asen-2003-university-of-colorado-at-boulder in Aerospace Engineering at University of Colorado at Boulder.

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Date Created: 10/30/15

I took calculusbased physics mechanics in A High school B CU Physics 1110 C Elsewhere D None of the above The online homeworks in Physics 1110 prepared me to solve dynamics problems in this class Strongly agree Agree Neutral Disagree Didn t take Physics 1 1 1 O Of the required freshmen Physics courses my least favorite was Chemistry Math Programming Freshman Projects F1909 Remember to post equations Announcements HW 9 due Thursday 1781 1787 17104 1794 HW10 1817 189 1832 188 report result b in radsec Reading pp 366372 I will not rederive these equa ons l ll briefly review moment of inertias for simple shapes and the parallel axis theorem For further reading see the appendix of ch 18 My office hours on Wednesday have been moved from 34 to 121 If Va 20 ms and 00 is 10 radsec clockwise Where is the instantaneous center of rotation L R1111 R1m Figure 17719708P1764 7 Problems 17641765 Copyrrgm 2008 Pearson Prenuce Herr Inc Ax0 B x2 C x1 D x05 Now the bar is in xed rotation about 0 and Va 20 ms What is the velocity of B ms A0 y B 20 71A UB C 40 D 20 E 40 3 7 A B Wheel is piling clockwise at 10 rads VA 5 msec I Is the wheel slipping D 5 300mm l C 7 A A B i Figure 17722717F17 55 r Pmblem 1785 Copyright 2005 Pearson Frenhce Hall Inc A Yes B No C Need more Information about the surface Wheel is rolling clockwise 10 rads VA 5 msec I Is the wheel slipping 45 J 300 mm Figure 17722717F17 55 r Pmblem 1785 Copyright 2005 Pearson Frenhce Hall Inc N A Yes B No C Need more Information about the surface Wheel is rolling wo slipping 005 rads Sand A on 2 radss k What is acceleration at A j y All answers in mss E A A 06i D B 06 f 35 300mm r C i C 06 J A D 06 k E none B Wheel is rolling wo slipping w5 rads lgand A on 2 radss k What is acceleration at E D J Y All answers in mss A 12f 75 D A A i5 300mm B 12139 75j C 74 i l 0121175 3 Bod 75 Topics today Combining acceleration and velocity relations for rigid bodies in planar motion gt gt gt gt 2 Point O is moving to right at constant velocity of 20 ins What is the velocity and acceleration of midpoint G Figure 1735743P17V167 Probiem517187 i7169 Copyright 2008 Pearson Prenlice Haii Inc Did not get this far Given 0 of 1 radsec clockwise for bar OQ length I what is the velocity of point Q I All answers in msec A sin 9 cos 9 B 0036 lsinej C l sin 0 1 cos 0 Figure17722722P17927Problems17921793 D l l Capyrlghl 2008 Pearson Prentice Hall Inc Given clockwise m and counterclockwise oz for bar 00 length I what is the acceleration of point Q All answers in We2 A 40 002 B lwzf la C laf 1602 V A A loci 602 39 Figure 17722722P17927 Problems 17321793 J nnnnnnnnn What is alpha of the second bar gt gt gt gt 2 For a body in fixed axis rotation about point 0 moment of inertia I defined at center of mass moment of inertia o defined at point 0 center of mass velocity Vc angular speed 00 the kinetic energy is 1 2 AOnly 5100 is correct B only MVC2 Ia2 is correct Homework problem fFdf13 dt f13 17dt No work done by friction Massless pulley Pin force sum of tensions Sum of moments at the pin0 4aT4g 20a 209T thure 180639P1816 Problem 18 we Copyngm 2005 Pearson Prenuce Han Inc Now give the pulley mass Pin force tensions weight of pulley0 3M6 Ca R x TBj R x TAj C 15777 For mass B J mBaB mBg TB 4 kg A 20 kg B ngme39 180639P1816 Problem 18 15 Copyngm 2coa Pearson Prenuce Han no For massB mBaBmBgTB For massA A mAaA mA8 TA mAaA mAgTA B 4kglll IIZOkg J C Neither ofthese ngule391806SBP1816 Problem 18 15 Copyngm 2005 Pearson Prsnuce Han Inc For massB mBaBmBg TB Instead of a mass at B now apply a force of 209 downwards ls tension in cable B now quot A Less than before J B B Equal to before ng l F20g Flgule 180639P1816 Problem 18 16 Copyngm 2008 Pearson Prsnllce Hall Inc gt4gt Rolls wo slipping Center of mass M moves distance L What is work done by gravity B Mchos0i C Need to known radius D Mthan0i E None ofthese Find an expression that relates 0L 6 and 00 Moment of inertia of car rear wheels 024 kgmm Front wheels 02 kgmm Tire radius 03 m M1480 kg Car is moving 100 kmhr Car brakes exert constant retarding couple of 650 Nm on each wheel and the tires do not slip what distance does it take to stop Figure 1970945P1927a 7 Problems 19 2719 23 Copyright 2008 Pearson Prentice Hali Inc Hoop mass m rolling wo slipping has a particle of mass m welded to it Time 1 Time 2 A 01 gt02 B 01 02 Massless hoop radius R with particle mass m located r away from center of hoop It is convenient to know b b2 R2 r2 212 005180 6 Massless hoop radius R with particle mass m located r away from center of hoop It is convenient to know b Define potential energy function for gravity using defined coordinate system At 00 what is PE of mass m W909 ng mgRr mgRr none Define potential energy function for gravity using defined coordinate system What is PE of mass m gtlt Dowgt mgrsinE mgrcose ng mgrcosE None of these Sphere is stationary Impulsive external force P acts on the system over a very very very short time Does the angular velocity of the sphere change over this time A Yes B No Angular momentum for a particle depends on where the coordinate system is defined ATrue B False Angular momentum for a rigid body depends on the coordinate system ATrue B False Slender bar b no constraints on its motion hit by sphere a in an elastic collision Does the sphere rotate A Yes G gt NO Bar released from rest No friction on the surfaces The angular acceleration of the bar is A Counterclockwise B Clockwise C Zero Announcements The gravity lab is due at beginning of next Wednesday s ab HW due Tuesday reduced to 188 amp 189 Today Finish problem from last time Dynamics equations for rigid bodies Briefly review moments of inertias Remember to post this gt gt gt gt 2 Point O is moving to right at constant velocity of 20 ins What is the velocity and acceleration of midpoint G 3in LAMA 1013144 Figure i73543P17457 7 Problems 17 167 17169 Copyright 2008 Pearson Prenlice Haii Inc VB 17A DAB XFBA VBwABIEx4f4j4wABj f Ac0 Vc20 Figure 1773549171577 Problems 17167 17169 Cupynght 2003 Pearson Prenluze Hat Inc 73 7C HDBC XFBC VB 20 ch12 x 10f 7 173 20 lowBCj 7ch Figure 1773549171577 Problems 17167 17169 Cupynght 2003 Pearson Prenluze HaH Inc v3 4wABj f v3 20 lowBCj 7003C 400 20 7003C 400143 10 BC Solve wAB 294aBC 118 a a a a 2 613 61A O AB XVBA wAB VBA a3 aABIEx4f4j wABZ4f4j g3 405143 405143 34576 j L4mJlt710in JI FIB C aBCl gtlt 10 7 wBC2 10 7 C73 aBC12x 10f 7 wBC2 10f 7 10aBCj 7aBCf 13912 973 Combine and separate equations aAB 456aBC 432 Use center of mass coordinates and think of a rigid body as being a collection of particles albeit particles that must move together AS a rigid body dH d dR 77 Rgtltmil 2 l 1dr dFI d M 77 Rgtlt 2 dt it m dR d ljEZRixmiwai l A 2 IEmiri For motion about rotation axis L Flur 1 e 6415 r a A murdlnate system wnh me I 3st ahgned wnh y m w mquot m A I y n m Modified from particles gt E F mZz cm Typically resulting in two scalar equations in x and y 2 F x ma 2 y I e About center of mass c 2 gt About fixed axis 0 2 gt About any point 160 10a Moment of Inertias for Planar Motion Assume rotation axis point 0 is perpendicular to this page Definition IO fr2dm Net Force acceleration Net Torque rotation How can you interpret the moment of inertia If forces and moments acting on the system are the same angular acceleration for the right case is bigger So it will roll downhill faster lmR2 12 mR2 D You still need to identify the forces so you can compute the moments A Up B Down C LD D u P1834 Problem 1834 Released from rest and at that instant find a Look at cases where surface isis not rough If surface is rough which way does friction act moowgt UP Down Left Right Friction is zero f surface is smooth which way will point A move 90 gt A i Figure 180661Pi B 47 Problem 1847 Copyrighl 2003 Pearson Prentice Hall Inc 003 Left Right Not enough information If surface is smooth the normal force acting at A is equal and opposite to mg Figure 180661P18 47 Problem 1847 Copyrighl 2003 Pearson Prentice Hall Inc A True B False

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