A particle is held and then let go at the edge of a circular shaped hill of radius R =shown below. The angular motion of the particle is governed by the following ODE:+ 0.4 02 - 2 cos 0 + 0.8 sin 0 = 0where is the angle in rad measured from the top (CCW: +), ė5m, as= wis the velocity in rad/s,==a is the angular acceleration in rad/s². Use MATLAB to numerically integrate thesecond order ODE and predict the motion of the particle.(a) Plot and w vs. time(b) How long does it take for the particle to fall off the ring at the bottom?(c) What is the particle speed at the bottom. Hint v = Rw.in deall questionsthe particles inside the tube./2/07/25ParticleR00Reled with

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Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Tp = Fq +°P/Q• (1) Here ip/Q is the "position of point P relative to point Q." Similarly the velocities of the two points are related by õp = bq + Up/Q- (2) The quantity õp/Q is the velocity of point P relative to point Q. I want you to use these ideas to solve the following problems. 1. The figure below shows a view from above of a large boat in the middle of the ocean. So that the crew on the ship can get exercise on long journeys, there is a circular walking/running track on the back deck. CA B- -D Suppose that the radius of the track is R = 6 m, and a person is running on the track at a constant speed of v = 3m/s as measured with a stopwatch by a crew-mate on board the ship. Suppose the runner is running counter-clockwise around the track when viewed from above. Write the velocity vector of the runner in terms of basis (ê1, ê2) as perceived by a crew-mate on the ship. (a) What is the velocity vector when the runner is at point A? (b) What is the velocity vector when the runner is…
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A particle moves along the x axis from x; to xf. Of the following values of the initial and final coordinates, which results in a negative displacement? x = -4m, xf = 2m x₂ = -4m, f =-2 m ○ x₁ = 4m, f = 6 m x₂ = 8 m, f = 4 m x = -4m, f none of these = 4 m
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The kinematic diagram is given as 012=60 degree and the speed of the mechanism is constant and 012=50 rad/sn in the slide mechanism whose limb dimensions are given below. Find the following items. AoA = r2 = 0.2 m A,Bo = r1 = 0.5 m 012 = 60° 12 = 50 rad/s %3D %3D A a) Write the vector closure equations b) Bring the vector closure equations into scalar form c) Find s13,014 d) Find s13,014 e) Find 513,014 S13 014 ri Bo 012 Ao
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0.8 m wo B D E 0.8 m 1.0 m Bar AB rotates about the fixed point A with constant angular velocity wo. The system starts with bar AB horizontal. 1) Use the relative velocity equation to find the velocity of C in terms of the angles 0 and > and their derivatives. 2) Determine the lengths of bars AB and BC so that as bar AB rotates, the collar C moves back and forth between the positions D and E. A design constraint is that bar BC must be longer than bar AB (as shown in sketch). 3) You are given the design constraint that the magnitude of the acceleration of collar C must not exceed 200 m/s². What is the maximum allowable value of wo? 4) Create the following plots for a complete revolution of bar AB using the values for lengths and angular velocity determined above: о о e and > versus time as 2 sub-plots. Position, velocity, and acceleration of C versus time as 3 sub-plots. Velocity and acceleration of C versus its position as 2 sub-plots. - 5) Use your plotted results to describe the…
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The plot below shows the acceleration of a mass placed at the end of a beam as it oscillates: 20 Which we model as: Acceleration (m/s²) 15 10 5 0 -5 -10 -15 -20 0.5 1 Wd 1 1.5 1 2 = wn√1 y(t) = Ae-3wnt sin(wat) y(t) = Ae-Ct sin 1 2.5 Time (s) = 2π T 2π T 1+ 3 2π We will use MATLAB to find the decay constant C and the period T. We can then use these to calculate the damping ratio and the natural frequency @n, using: 3.5 T 4 2 and C = {wn O data zeros peaks decay fit 4.5 Using the information above, derive the expression shown below for the damping ratio 3. 1 3 = 5
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