Quiz1 - EGM 3400/3401 Quiz #1 Fall 2010 15 September 2010 1...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EGM 3400/3401 Quiz #1 Fall 2010 15 September 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) A vector can be expressed in any system of coordinates. (b) The rate of change of a vector depends upon the coordinate system in which the vector is expressed. A (c) Let b be a vector, A and B be an arbitrary reference frames where A ￿= B , and db/dt be the rate of A change of b as viewed by an observer in reference frame A. Then observations of the vector db/dt will be different in references frames A and B . (d) A coordinate system can be fixed simultaneously fixed in two different reference frames A and B (that is, A = B ). ￿ (e) Let A and B be distinct reference frames that rotate relative to one another. Also, let A ω B be the angular velocity of reference frame B as viewed by an observer in reference frame A. Then the rate of change of AB ω in reference frame A is not equal to the rate of change of A ω B in reference frame B . 3 Question 2: 60 Points A collar, located at point P , slides along a slender uniform rigid rod. The rod is mounted to an annulus of radius R such that its ends are constrained to slide in a slot along the circumference of the annulus. The annulus rotates with constant angular velocity Ω relative to the ground about an axis from A to B (where A and B are the locations of the left and right support points, respectively, of the annulus). Knowing that r describes the distance from the center of the rod to the collar and that the angle θ describes the orientation of the rod relative to the direction of Ω, determine the velocity and acceleration of point P as viewed by an observer fixed to (a) the annulus and (b) the ground. Q D R r A P θ O B Ω C Hint: For convenience, Q is defined to be a point that is fixed to the annulus and is where the point D on one end of the rod would be if θ were equal to π /2. 4 Solution to Quiz 1 Solution to Question 1 (a) True. A vector is a coordinate-free quantity and, thus, can be expressed in any system of coordinates. (b) False. Because a vector is coordinate-free, its rate of change has nothing to do with the coordinate system in which the vector is expressed. (c) False. A db/dt is a vector and, thus, is observed the same in all reference frames. (d) False. By definition, a coordinate system can be fixed in only one distinct reference frame. (e) False. Applying the transport theorem to A ω B , we have A d ￿A B ￿ B d ￿ A B ￿ A B A B ω= ω +ω×ω dt dt Because A ω B × A ω B = 0, we have A d ￿A B ￿ B d ￿ A B ￿ ω= ω. dt dt 5 Solution to Question 2 Let G be the ground. Then choose the following coordinate system fixed to G : Ex Ey Ez Origin at O = = = Along OB Along OQ when t = 0 Ex × Ey . Next, let A be the annulus. Then choose the following coordinate system fixed to A: ex ey ez Origin at O = = = Finally, let B be the plane defined by the direction of OP coordinate system fixed to B : Origin at O ur = uz = uθ = Along OB Along OQ ex × ey . and the direction of ez . Then choose the following Along OP ez uz × ur . Fig. 1 shows the bases {ex , ey , ez } and {ur , uθ , uz }. Given the alignment of the coordinate systems, we have ey uθ ur θ θ u z ,ez ex Figure 1: Diagram Showing Bases {ex , ey , ez } and {ur , uθ , uz } for Quiz 1 Problem. the following relationships for ex and ey in terms of the basis {ur , uθ , uz }: ex ey = = cos θ ur − sin θ uθ sin θ ur + cos θ uθ (1) Given the choice of reference frames and coordinate systems, the position of the collar relative to the ground is given as rP/O ≡ r = rur (2) Now, we can proceed to solve parts (a) and (b) of this problem. (a) Velocity & Acceleration of P As Viewed By an Observer Fixed to the Annulus Because the basis {ur , uθ , uz } is fixed in reference frame B and we want the first and second rates of change of r in reference frame A (that is, the annulus), we apply the transport theorem from reference frame B to reference frame A. First, the velocity of point P in reference frame A is given as A v≡ A dr Bdr A B = + ω ×r dt dt 6 (3) Now we have A Furthermore, B dr dt AB ω ×r Therefore, A Next, we have A Now we have Therefore, a≡ B d ￿A ￿ v dt AB ω × Av ˙ ω B = θuz = rur ˙ = ˙ ˙ θuz × rur = rθuθ ˙ v = rur + rθuθ ˙ A d ￿A ￿ B d ￿A ￿ A B A v= v+ ω× v dt dt = ¨ rur + (rθ + rθ)uθ ¨ ˙˙ = A (4) (5) (6) (7) ˙ ˙ ˙ ¨ θuz × (rur + rθuθ ) = −rθ2 ur + rθuθ ˙ ˙ ¨ a = (¨ − rθ2 )ur + (2rθ + rθ)uθ r ˙˙ (8) (9) (b) Velocity & Acceleration of P As Viewed By an Observer Fixed to the Ground Again, it is seen that r is expressed in the basis {ur , uθ , uz } and {ur , uθ , uz } is fixed in B . Furthermore, we want the rates of change in reference frame G . Thus, in this case, we need to apply the transport theorem from reference frame B to reference frame G . First, the velocity of point P in reference frame G is given as G First, we have v≡ G dr Bdr G B = + ω ×r dt dt (10) G (11) G Now we know that ωB = G ωA + A ωB ωA ωB Ω = Ωex ˙ θ uz (12) A = = In order to express all quantities in the basis {ur , uθ , uz }, we need to express G ω A in terms of {ur , uθ , uz }. Using Eq. (1), we have GA ω = Ω = Ωex = Ω cos θ ur − Ω sin θ uθ (13) Therefore, We then obtain G B dr dt GB ω ×r Therefore, ˙ ω B = Ω cos θ ur − Ω sin θ uθ + θuz = rur ˙ = = ˙ (Ω cos θ ur − Ω sin θ uθ + θuz ) × rur ˙uθ + rΩ sin θ uz rθ G Next, G a≡ ˙ v = rur + rθuθ + rΩ sin θ uz ˙ G d ￿G ￿ B d ￿G ￿ G B G v= v+ω×v dt dt 7 (14) (15) (16) (17) Now we have B d ￿G ￿ v dt GB ω × Gv = = = ¨ ˙ rur + (rθ + rθ)uθ + (r sin θ + rθ cos θ )Ωuz ¨ ˙˙ ˙ ˙ ˙ (Ω cos θ ur − Ω sin θ uθ + θuz ) × (rur + rθuθ + rΩ sin θ uz ) ˙ 2 ˙ ˙ rΩθ cos θ uz − rΩ cos θ sin θ uθ + rΩ sin θ uz − rΩ2 sin2 θur + rθuθ − rθ2 ur ˙ ˙˙ (18) Adding the two parts of Eq. (18) and simplifying, we obtain G ˙ ¨ ˙ a = (¨ − rθ2 − rΩ2 sin2 θ)ur + (2rθ + rθ − rΩ2 cos θ sin θ )uθ + 2(rΩθ cos θ + rΩ sin θ )uz r ˙˙ ˙ 8 (19) ...
View Full Document

Ask a homework question - tutors are online