EN40: Dynamics and Vibrations. Final Examination Monday May : 2pm-5pm

Transcription

1 EN40: Dynaics and Vibrations Final Exaination Monday May : p-5p School of Engineering Brown University NAME: General Instructions No collaboration of any kind is peritted on this exaination. You ay bring double sided pages of reference notes. No other aterial ay be consulted Write all your solutions in the space provided. No sheets should be added to the exa. Make diagras and sketches as clear as possible, and show all your derivations clearly. Incoplete solutions will receive only partial credit, even if the answer is correct. If you find you are unable to coplete part of a question, proceed to the next part. Please initial the stateent below to show that you have read it `By affixing y nae to this paper, I affir that I have executed the exaination in accordance with the Acadeic Honor Code of Brown University. PLEASE WRITE YOUR NAME ABOVE ALSO! 1-8 [0 points] 9 [7 POINTS] 10 [13 POINTS] 11 [10 POINTS] 1 [10 POINTS] TOTAL [60 POINTS]

2 FOR PROBLEMS 1-8 WRITE YOUR ANSWER IN THE SPACE PROVIDED. ONLY THE ANSWER APPEARING IN THE SPACE PROVIDED WILL BE GRADED. ILLEGIBLE ANSWERS WILL NOT RECEIVE CREDIT. 1. The figure shows the path of a predator in the predator-prey copetition, which travels along the dashed line fro A to B. At the instant shown, the predator travels at constant speed with velocity v. It is subjected to a viscous drag force F D = cv and the rando force is zero. On the figures below, draw arrows showing the directions of (a) the viscous drag force; (b) the propulsive force; and (c) the resultant force. Drag force Propulsive force Resultant force B B B A A A ( POINTS). The figure shows a power curve (with power in kw and engine speed in rad/s) for an internal cobustion engine. It powers a vehicle with ass 1000kg that runs at constant speed up a 5% grade (i.e. sin( θ ) 1/ 0 ). The engine runs at rad/s. If transission losses and air resistance can be neglected, the speed of the vehicle is (a) 10 /s (b) 15 /s (c) 0 /s (d) 5 /s (e) none of the above Power (kw) Engine speed (rad/sec) The power at rad/s is 10kW. The power is related to speed by P = gv sinθ v 1/ 0 = v = 0 / s ANSWER C ( POINTS) 3. The vehicle described in the preceding proble has a wheel radius of 0.5. The transission has gear ratio (otor angular speed/axle angular speed) (a) 400 (b) 50 (c) 1/50 (d) 1/400 (e) None of the above The axle angular speed is related to the vehicle speed by v= warw 0 = wa 0.5 wa = 40 w / wa = / 40 = 50 ANSWER B ( POINTS)

3 4. The figure shows the variation of ipact force with tie during the straight-line collision between two identical spheres with 0.5kg ass. Before the collision, one sphere is stationary; the other has speed 8/s. A R 8 /s B R A B R R Collision 4.1 The ipulse of the force is 6 5 (a) 1 Ns (b) 1.5 Ns (c) Ns (d) 3 Ns (e) None of the above Force (kn) The ipulse of the force is I = = 1.5 Ns Tie (illiseconds) ANSWER B ( POINTS) 4. The velocities of the spheres after ipact are (a) VA1 = / s VB1 = 6 / s (b) VA1 = 0 / s VB1 = 8 / s (c) VA1 = 4 / s VB1 = 4 / s (d) VA1 = 1 / s VB1 = 7 / s (e) None of the above After ipact, the two spheres have speeds V = V I / = / 0.5 = / s A1 A0 VB1 = I / = 1.5 / 0.5 = 6 / s ANSWER A ( POINTS) 4.3 The restitution coefficient for the collision is (a) e=0 (b) e=0.5 (c) e=0.5 (d) e=0.75 (e) e=1 The restitution coefficient follows as VB1 VA1 6 e = = = 0.5 VA0 8 ANSWER C ( POINTS) 3

4 5. The spring-ass syste shown in the figure is critically daped. If one spring is reoved, the daping factor is (a) ζ = 1 (b) ζ = (c) ζ = (d) ζ = 1/ k c k (e) ζ = 1/ The original syste has daping factor After the odification, the daping factor changes to ζ = 1/ k and for critically daped systes ζ = 1 ζ = 1/ k. Therefore ζ =. ANSWER C ( POINTS) 6. In the figure shown, the ring gear is stationary and the sun gear rotates in the counter-clockwise direction with angular speed ω. The angular speed of the planet gear is (a) ωp = ωr/ r counterclockwise (b) ωp = ωr/ r clockwise (c) ωp = ωr/ ( r) counterclockwise (d) ωp = ωr/ ( r) clockwise (e) None of the above w Sun gear R r Planet gear Ring gear The planet gear is stationary where it touches the ring gear, and oves with the sae speed as the sun gear where it touches the sun gear. By inspection its rotation will be in the opposite direction to the sun gear. Therefore ωp = ωr/ ( r) clockwise ANSWER D ( POINTS) 4

5 7. The dashpot shown in the figure has dashpot coefficient c = 100 Ns /. The end of the dashpot is subjected to a force that causes its length to vary as indicated in the graph. The total work done by the force after sec is c=100 Ns/ F (a) 0 (b) 1.5 J (c) 5 J (d) 50 J (e) None of the above Length Change () The extension rate is 0.5/s. The agnitude of the force is therefore F = = 50N and the power dl developed is F = 5W. The total work done is therefore 5 50J dt = Tie (seconds) ANSWER D ( POINTS) 8. The ass oent of inertia about the center of ass of a slender rod with ass length L is IG = L /1. Two slender rods with ass are rigidly connected to create a right-angle section as shown in the figure. The ass oent of inertia of the right angle section about the k axis through point O is O j L i (a) IO = L /3 (b) IO = 17 L /1 (c) IO = 5 L /3 (d) IO = L /6 (e) None of the above L 1 L 1 L 5 L L L = L ANSWER C ( POINTS) 5

6 9. The figure shows the forces acting on a baseball bat at the instant when it strikes the baseball (in plan view). The reaction forces R, R are exerted by the batter s x y hands; the force F is exerted by the ball. The bat has 4 ass and ass oent of inertia IO = L about 3 the point O (NOT the COM) where L is the distance of the center of ass of the bat fro the handle. At the instant shown, the bat rotates in the horizontal plane about the batter s grip (at O) with instantaneous angular speed ω and acceleration α. L O w COM d j i O R y R x F 9.1 Using the rigid body kineatics forulas, find a forula for the acceleration of the center of ass of the bat in ters of L, ω and α, expressing your answer in {i,j} coponents. You do not need to use Newton s laws to answer this part. αg= αo+ αk rgo / + ωk ωk rgo / = 0+ αli+ ω Lj [ POINTS] 9. Using Newton s laws, the equation of rotational otion, and the solution to 10.1, find forulas for the reaction forces R, R in ters of, L, d ω and the ipact force F. x y Newton s laws and rotational otion around O gives ( Rx F) i+ Ryj= αli+ ω Lj I0αk = Fdk 4 I0 = L 3 3dL Rx = F 1 R 4 y = ω L L 9.3 Find a forula for the distance d that will iniize the agnitude of the reaction force at O. The agnitude is iniized if R = 0 d = 4 L/3 x [ POINTS] 6

7 10. Ruble strips are roughened regions on a road that warn drivers when their vehicle approaches the side of a highway. The figure shows an idealization of one wheel of a vehicle driving over a ruble strip with aplitude H and wavelength L. The ass represents the wheel; the spring with stiffness k T odels the deforation of the tire, and the cobined spring and daper ks, c S represent the car s suspension. The vehicle travels with steady horizontal speed V. As a result, the contact point of the tire with the road at A experiences a vertical haronic otion ht ( ) = Hsinωt, where ω = πv / L. The tire vibrates vertically with a displaceent fro its static equilibriu position y(t). Vertical otion of the car body can be neglected. c S A k T k S V y(t) h(t) L Car body Suspension Wheel Tire H 10.1 Draw a free body diagra showing all the tie dependent forces acting on the wheel. Gravitational forces ay be neglected throughout this proble since they do not vary with tie. c S dy dt k S y(t) k T (y-h) 10. Hence, show that the equation of otion for the tire displaceent y(t) is d y dy + c ( ) sin S + kt + ks y = kth ωt dt dt Newton s law gives expression stated. d y dy c ( ) S kt y h ksy dt dt =. Substituting for h and re-arranging gives the 7

8 10.3 Re-write the equation of otion in standard for and find expressions for the natural frequency and daping coefficient ωn, ζ in ters of k, S, kt, c S The standard for is d y cs dy kt H + + y = sinωt ( k ) T + ks dt ( kt + ks) dt ( kt + ks) 1 d y ζ dy + + y = KF 0 sinωt n dt ωn dt Our expression is the sae as the standard forula if we identify ω kt kt + ks c K = F s 0 = H ωn = ζ = kt + ks ( kt + ks) [ POINTS] 10.4 Find an expression for the aplitude of vibration of the wheel, in ters of ωω, n, ζ, kt, ks and H. Fro the forulas given in class, the aplitude is k Y0 = H T k T + k S ω ω 1 + 4ζ ω n ωn 1 [ POINTS] 8

9 11. The ends A and B of an inextensible cable are attached to a spring and ass, as shown in the figure. The cable passes over a pulley with ass, radius R and ass oent of inertia R / Let x denote the extension of the spring (i.e. the difference between the spring s length and its unstretched length). Write down an expression for the total potential energy of the syste in ters of k,x,,g. x A k R B 1 V = kx gx [ POINTS] 11. Assuing that the cable does not slip on the pulley, find an expression for the kinetic energy of the syste, in ters of and dx / dt 1 dx 1 1 dx T = + Iω I = R / ω = dt R dt 13 dx T = dt 11.3 Hence, show that x satisfies the equation of otion 3 d x + kx = g dt Energy is conserved so d 13 dx 1 T + V = C + kx gx = 0 dt dt 3 d x dx + kx g = 0 dt dt This reduces to the expression stated. 9

10 11.4 Find an expression for the natural frequency of vibration of the syste in ters of k and. Rearrange the equation in standard for 3 d x g 1 d y k + ( x ) = 0 + y = 0 ω n = y = x g k dt k ω 3 n dt The natural frequency is therefore ω = n k 3 [ POINTS] 11.5 The syste is released fro rest with x=0. Find an expression for the subsequent axiu velocity of the ass attached to the cable at B. We can solve this using vibrations note that the velocity of the ass is dx / dt so we can solve the EOM for y and hence deduce dx/dt as follows g g dy dx g k y = y = cosωnt = = sinωnt k k dt dt k 3 The axiu value is thus g 3k Alternatively if we didn t learn any vibrations we can note that d x dt = 0 when the velocity is a axiu, and so the EOM tells us that kx = g at the instant of ax velocity. We can get the velocity when x has this value using energy conservation 1 13 T0 + V0 = 0 = T1+ V1 = kx gx + v k k g g g v = x gx = g = 3 3 k k 3k 10

11 1. A thin unifor disk of radius R, ass and ass oent of inertia R / is placed on the ground with a positive velocity v 0 in the horizontal direction, and a counterclockwise rotational velocity (a backspin) ω 0. The contact between the disk and the ground has friction coefficient µ. The disk initially slips on the ground, and for a suitable range of values of ω 0 and v 0 its direction of otion ay reverse. w 0 R v Draw a free body diagra showing the forces acting on the disk just after it hits the ground. g R µ N N 1. Hence, find forulas for the initial acceleration a and angular acceleration α for the disk, in ters of g, R and µ. Note that the contact point is slipping. Newton s law and the rotational equation give 1 N = g µ N = ax R α = µ NR a= g a = g / R 1.3 Find a forula for the tie at which the disk will reverse its direction of otion. [ POINTS] Velocity is reversed where v=0. The constant acceleration straight line otion forula gives v v 0 µ gt t v 0 / µ g = = at the reversal. [ POINTS] 1.4 Find a forula for the tie at which the disk begins to roll on the ground without slip. Hence, show that the disk will reverse its direction only if v0 < ω0 R/ Rolling without sliding starts when v = ωr. We have that 11

12 ω = ω 0 µ gt / R v = v0 µ gt v0 µ gt = ( ω 0R µ gt) t = ( v0 + ω 0R)/3µ g The reversal will only occur if slip continues long enough this eans that ( v0 + ω 0R)/3 µ g > v0 / µ g v0 < ω0r/ 1