Test 2 Fall 2009

# Test 2 Fall 2009 - Problem 1(25 Points(a lOpts A bar of...

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Unformatted text preview: Problem 1 (25 Points) (a lOpts) A bar of aluminum has a length of 2 m and a cross-sectional area of 1 x 10‘4 m2. If the macrosc0pic spring stiffness of the bar is equal to 3.1 x 106 N/m, what is the Young’s modulus of aluminum. [AT/1’ A (leq M2) n... y, F/A : kL , (slump/Mm”) i=éan (b lOpts) The density of aluminum is 2.7 g/cm3 and the atomic mass is 27 g/mol. What is the interatomic spring stiffness for aluminum? (c 5pts) If you hit one end of the aluminum bar with a hammer, how long does it take before the other end of the bar experiences this perturbation? If you were unable to complete part (b), express your answer in terms of ksﬂ. (2»Sexlo"°m> Problem 2 (25 Points) A block of mass m hangs from two identical springs each with spring stiffness ks, as show in the accompa- nying ﬁgure. How much does each spring stretch when the block is in static equilibrium? Problem 3 (25 Points) A Ferris wheel is a vertical, circular amuse- ment ride with radius 5 m. The Ferris wheel rotates clockwise at a constant rate, going around once in 10 s. Consider a rider whose mass is 75 kg and sitting in the fourth seat. Note that the ﬁgure is not to scale and the an- gle between seat four and seat ﬁve is actually 6 = 40°. (a 4pts) In the space provided draw a diagram indicating the forces acting on the rider in the fourth seat. Note that the Ferris wheel rotates at a constant rate. \P/ F\ K East,“ :0 g) its,” has “i0 be a3 iohj 0L3 Lima)“ Rem to owl MW cod” 50 1:le I‘S lanai! ‘H‘Mn (mg); 0;? , (b 4pts) What is the parallel component of the rate of change of the riders momentum in the fourth seat? Explain how you know this. <03; :0 L/C \/ '«ComS‘i‘ml/xi' \I (c 4pts) What is the perpendicular component of the rate of change of the riders momentum in the fourth seat? Explain how you know this. of» Maw: maa‘lfrqwzmk dt 1 R R 7-1 H (d 8pts) What is the parallel and perpendicular component of the contact force exerted by the fourth seat on the rider? 1/ leﬂ __FS K T?- ~> a = weeséwnw /&[email protected] T7. : (asrjﬂ‘yem-lkng +171ng =C§E§’ éﬁl=Wm?l (e 5pts) Now consider a rider who is momentarily at the top of the Ferris wheel (i. e. riding in seat one). With what angular speed must the Ferris wheel be turning so that this rider feels weightless? Rwy Leah Nastle NM 5:0 A+ m +0? ' 2 'E +th : M R _ 4 "’1 “BR iM-emsdxsm‘) : 1mg“ (A): V :‘ 7M5”) E 'Vt:"q"“d:\$"? For F5 ) Problem 4 (25 Points) (a 6pts) An electron of mass m is traveling in a straight line in the positive x—direction at a speed vi = 0.990, where c is the speed of light. A constant force brings this electron to a complete stop. What is the change in kinetic ener of the electron? " 2 s - _ gy t; KL+E°‘&/mcl t4: Ewarch % AK: 3"(X-l3mcl Y ‘ thlHKIO/Ty‘el“We” :lf‘iﬁ‘l mo"5 3 (b 6pts) Determine the x—component of the constant force responsible for bringing the electron to a stop if the deceleration took place over a distance of 3219 m. 2 .3 W: AK: EA? ’5 F = _V_U__ : q'qol"o-j .(AL;) AL 62qu = I.SSXI0""M f..l~3.53"0"l°"’ / w Wm (c 6pts) After the electron comes to a stop it is then accelerated up to a speed 11f, at which point the speed remains constant. Once the electron is moving at this ﬁnal speed, it enters a region containing a strong magnetic ﬁeld. The result of the interaction between the electron and the magnetic ﬁeld is that the path of the electron begins to curve. Since the speed remains unaltered, which components of the rate of change of momentum change and why? ' e -‘ f“ " OLE am? 4 IFlM f a 5—. alt dt R 1 1 , \Cﬂi‘ﬁlgt “Mimi 1: O ﬁsh/C ‘ v: (onctémL \ ‘ _ A L . i i a , ‘.'_—.— h C lathej glance, M (0»th \pQ-tehﬂnyv 1?; 4 B is not CAMS‘I‘AWP') chmwjés.' (d 7pts) At some point along this curved trajectory, the osculating (or kissing) circle has a radius R = 0.5 m. If the magnitude of the perpendicular component of the force acting on the electron at this instant is F = 1.62 x 10‘10 N how fast is the electron traveling? at; r ‘Tlm’: o‘PPmecL 75 ian‘E‘ClL. _\ Z t”. : WW2 b/c [ill give: ‘<~J~-I>C‘ 0“ .L K ___ ~ Q,52x(0.'ol\l3(0.§m) I —_> \I if t l -23 ‘ (lt’l Xloq‘ms . m OHxlO log K \ x“ 522 mex‘i’ Pmyﬁ Qt correfi‘ QMSW Comet" ginHOM ‘3“ F:1£:Uy‘{: KW‘VZ: ‘ < 5m Cream '14 cu [u I/ Y P J 6x CAICU (A'Hr 4'th him you‘ H J05" ...
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Test 2 Fall 2009 - Problem 1(25 Points(a lOpts A bar of...

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