M Search results – rohan.thomas.98. X Materials Lab #4 Presentation – G X Files X G how to take a screenshot window X X + C O

need some help. the images have all the pertinent information needed.

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M Search results – rohan.thomas.98. XMaterials Lab #4 Presentation – G XFilesXG how to take a screenshot window XX+C Ohttps://utexas.instructure.com/courses/1243874/files/folder/TakeHomeExam-Spring2019?preview=49540896QGRME318TakeHomeExam_Spring2019.pdfDownloadInfoX ClosePage5>of 7-……..-ZOOM+QCODProblem 3 (35 points): Numerical solution of a nonlinear initial value problem (featuring157brake squeal):F brake padMMMp Fx21 y?brake Md contactkd3Disk brakes on an automobile or bike have a stationary brake pad sliding against a moving brakedisk. Braking occurs when brake force F presses the pad against the disk. A large F will stop thecar; a small F can render weird phenomena. As the disk moves with horizontal velocity vi relativeto the pad, the disk pulls on the brake pad via friction. As with many sliders, friction coefficient u= u(v) diminishes with sliding velocity v (see figure) from a static friction coefficient us at v = 0to a kinetic friction coefficient uk at high speed. The negative slope of u vs. v at low speed coupledwith natural frequencies from stiffness and mass gives rise to friction induced vibrations such asbrake noise, squeaky shoes, violin bow on string, or fingernails on a blackboard. For carbon basedbrake materialsu(V) = Uk + a exp(-bv)(1)where uk, a and b must be determined from the data in table 1 below. Applying Newton’s law tothe pad mass Mp along the horizontal and vertical directions givesMy @xildt? = – kp x] + u(vi – dxildt) P(2)My dxaldt? = F – P(3)where x] and x2 are displacements of Mp in the horizontal and vertical directions, and y is thevertical displacement of the brake disk. Stiffness kp holds the pad to the vehicle frame via springforce kp x1. Reactions to disk-pad sliding contact include normal force P perpendicular to thesliding interface and friction force uP tangential to the sliding interface. The sliding velocity v =vi – dxildt for u(v) of equation (1) is the difference between disk speed vi and pad speed dx1/dt.The brake pad’s sliding surface appears flat but is actually curved. When pressed against the disk,the curved surface "squashes" (like a ball) creating a nonlinear relation between force P = C do/2and squash displacement d = x2 – y, d > 0. When d < 0, the bodies separate and P = 0. ThusP -RC(x2-y)., x2 -y>00,x2 -ys0(4)-71O Type here to searchU9?0A (04))10:32 PM5/8/2019