Quiz2 - EGM 3400/3401 Quiz #2 Fall 2010 4 October 2010 1...

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Unformatted text preview: EGM 3400/3401 Quiz #2 Fall 2010 4 October 2010 1 Materials Allowed During Examination • One 8.5 inch by 11 inch sheet of paper with anything you would like to write on it provided that everything written on the sheet is in your own original handwriting. • Pens, pencils, and erasers. Books, calculators, or other materials are not permitted for use during the exam. Guidelines for Solutions As I have stressed in class, communication is an extremely important part of demonstrating that you understand the material. To this end, the following guidelines are in effect for all problems on the examination: • Your handwriting must be neat. I will not try to decipher sloppy handwriting and will assume that something is incorrect if I am unable to read your handwriting. • You must be crystal clear with every step of your solution. In other words, any step in a derivation or statement you write must be unambiguous (i.e., have one and only one meaning). If it is ambiguous as to what you mean in a step, then I will assume the step is incorrect. In short, please write your solutions in a orderly fashion so that somebody else can make sense of what you are doing and saying. Point Distribution The exam consists of two questions and the value of each question is clearly indicated. Unless otherwise stated, full credit will be given for a proper application of a relevant concept (e.g., proper description of kinematics and kinetics, understanding how to compute a transfer function). Contrariwise, no credit will be given for a concept applied incorrectly, even if the final answer is correct. University of Florida Honor Code On your exam you must state and sign the University of Florida honor pledge as follows: I pledge on my honor that I did not violate the University of Florida honor code during any portion of this exam. Signature: Date: 2 Question 1: 40 Points For each question below, answer “True” or “False” and provide a justification for your answer. Every correct answer of either “True” or “False” will receive 1 point while every correct explanation with receive 7 points. (a) Let O and P be points in three-dimensional Euclidean space. Furthermore, let A and B be distinct A reference frames. Finally, assume that O is fixed in reference frame A. Then drP/O /dt is the velocity of point P as viewed by an observer in reference frame B . (b) A point in three-dimensional Euclidean space can rotate. (c) If a vector b is fixed in a reference frame A, then b must have constant magnitude. (d) A reference frame is a collection of at least three noncollinear points. (e) If the magnitude of a vector changes with time, then there cannot exist a reference frame in which the vector is fixed. 3 Question 2: 60 Points A point P moves along a track in the form of a linear spiral. The equation for the spiral is given as r = bθ where b is a constant. Assuming that the θ is the angle between the horizontal direction (that is, the rightward direction) and the direction from O to P , determine (a) the velocity and acceleration as viewed by an observer fixed to the track and (b) the intrinsic basis {et , en , eb } relative to the track. P r θ O Note: No shortcuts are allowed. In other words, it is disallowed to determine any of the intrinsic basis vectors by using the fact that the particle is moving in a fixed plane. 4 Solution to Quiz 2 Solution to Question 1 (a) False. The point from which all distances are measured must be fixed in the reference frame in which the rate of change is desired. Point O is fixed in A and, thus, cannot be used as a point to measure distance in reference frame B . (b) False. A point is a zero-dimensional object and, thus, cannot rotate. (c) True. Let b = be where b = ￿b￿ and e is a unit vector. Then A A db de ˙ = be + b dt dt A ˙ In order for b to be fixed in A, both b and de/dt must be zero. Thus, b must have constant magnitude. (d) False. A reference frame is a collection of at least three noncolinear points whose distance from one another is constant. (e) True. See the solution to (c). 5 Solution to Question 2 Let T be the track. Then choose the following coordinate system fixed to T : Origin at O = = = Ex Ez Ey To the right Out of Page Ez × Ex . Next, let A be plane defined by OP and Ez . Then choose the following coordinate system fixed to A: Origin at O = = = er ez eθ Along OP Ez ez × er . The position of the particle is then given as rP/O ≡ r = rer = bθer (1) The velocity of point P as viewed by an observer fixed to the track is then given as T v≡ T Now we have dr Adr T A = + ω ×r dt dt T Furthermore, A dr dt TA ω ×r ˙ ω A = θ ez = ˙ bθer = (2) (3) ˙ ˙ θez × bθer = bθθeθ (4) Therefore, the velocity of point P as viewed by an observer fixed to the track is given as T ˙ ˙ ˙ ˙ v = bθer + bθθeθ = bθ(er + θeθ ) (5) Next, the acceleration of point P as viewed by an observer fixed to the track is given as T Now we have a≡ d ￿T ￿ A d ￿T ￿ T A T v= v+ ω × v dt dt A T Therefore, T d ￿T ￿ v dt ωA × T v T = ¨ ¨ ˙ bθer + b(θθ + θ2 )eθ = = ˙ ˙ ˙ θez × (bθer + bθθeθ ) ˙2 θer + bθ2 eθ ˙ − bθ (6) ¨ ˙ ¨ ˙ a = (bθ − bθ2 θ)er + (bθθ + 2bθ2 )eθ (7) (8) (b) Intrinsic Basis Relative to the Track The intrinsic basis relative to the track is obtained as follows. First, the tangent vector relative to the track is given as T v (9) et = T ￿ v￿ 6 Substituting T v from Eq. (5), we have the tangent vector as et = ˙ ˙ bθ(er + θeθ ) er + θeθ =√ = (1 + θ2 )−1/2 (er + θeθ ) ˙(er + θeθ )￿ ˙ 1 + θ2 ￿b θ (10) Next, the principal unit normal relative to the track is given as en = Now T Furthermore, A det dt = = = = = T ω A × et = = T det /dt T ￿ det /dt￿ A det d et T A = + ω × et dt dt 1 ˙ ˙ − (1 + θ2 )−3/2 2θθ(er + θeθ ) + (1 + θ2 )−1/2 θeθ 2 ˙ ˙ −(1 + θ2 )−3/2 θθ(er + θeθ ) + (1 + θ2 )−3/2 (1 + θ2 )θeθ ￿ ￿ ˙ (1 + θ2 )−3/2 θ −θ(er + θeθ ) + (1 + θ2 )eθ ￿ ￿ ˙ (1 + θ2 )−3/2 θ −θer − θ2 eθ + eθ + θ2 eθ (11) (12) (13) ˙ (1 + θ2 )−3/2 θ [−θer + eθ ] ˙ θez × (1 + θ2 )−1/2 (er + θeθ ) ˙ (1 + θ2 )−3/2 θ(1 + θ)2 (−θer + eθ ) Adding the expression for each of the two terms, we obtain T ￿ ￿ d et ˙ ˙ = (1 + θ2 )−3/2 θ −θer + eθ + (1 + θ)2 (−θer + eθ ) = (1 + θ2 )−3/2 θ(−θer + eθ )(2 + θ2 ) dt Therefore, −θer + eθ en = √ 1 + θ2 (14) (15) Finally, the principal unit bi-normal vector is given as er + θeθ −θer + eθ eb = et × en = √ ×√ = ez 2 1+θ 1 + θ2 7 (16) ...
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This note was uploaded on 09/30/2011 for the course EGM 3401 taught by Professor Carlcrane during the Summer '08 term at University of Florida.

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