F12–36. Peg P is driven by the forked link OA along the pathdescribed by r = e®, where r is in meters. When 0 = rad, thelink has an angular velocity and angular acceleration ofè = 2 rad/s and ë = 4 rad/s?. Determine the radial andtransverse components of the peg's acceleration at this instant.= e"O 0,06,0

Given, The described path is r = eθ. The angular velocity of the link is θ˙=2 rad/s. The angular…

Answered: F12–36. Peg P is driven by the forked…

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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The hydraulic cylinder gives pin A a constant velocity v = 3.1 m/s along its axis for an interval of motion and, in turn, causes the slotted arm to rotate about O. Determine the values of r, r, and for the instant when 0 = 27°. (Hint: Recognize that all acceleration components are zero when the velocity is constant.) Answers: r = 7 = ÿ = i i 37 310 mm- m/s m/s² rad/s²
The figure shows a rigid link AB of length 2 m which rotates anticlockwise about a fixed point at A at a rate of 2 rad/s, but which is decelerating in its rotation at 1 rad/s². At the instance of time shown in the diagram, with the link AB at 60 degrees to the horizontal, what is the y-component of the absolute acceleration of point B? (The positive sense of the x and y directions are given by the axes in the diagram). O O O O O w = 2 rad/s a = -1 rad/s² 7.93 m/s2 Don't Know -7.93 m/s² 5.93 m/s2 -5.93 m/s² B 60⁰
Write the vector expression for the acceleration a of the mass center G of the simple pendulum in both n-t and x-y coordinates for the instant when 0 = 57° if 0 = 2.31 rad/sec and 0 = 4.475 rad/sec². Answers: a = ( a = ( i i 4.5' 8 en + i+ i i et) ft/sec² j) ft/sec²
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= At the instant shown rod AB has an angular velocity WAB 4 rad/s and an angular acceleration AB 2 rad/s2. Determine the angular velocity and angular acceleration of rod CD at this instant. The collar at C is pin connected to CD and slides freely along AB. = a) using the relative motion analysis and the contact conditions. A 4 rad/s AB= 2 rad/s² WAB 60° 0 = 0.75 m -0.5 m В B D
Please do not answer if you are going to copy the solution form another source. These solutions do not make sense and I need a better understanding of the problem.
I need correct solution of all parts
The disk with radius r = 0.09 m is rotating at a constant angular velocity of ω = 0.9 rad/s (counterclockwise) about the fixed pin support at O. For the instant shown, find the relative acceleration component (aB/A)n, where (aB/A)n = {(ax)i+(ay)j} m/s2. Choose the correct answer: a) ax=-0.292; ay=-0.0729 b) ax=0.786; ay=0.196 c) ax=0.292; ay=0.0729 d) ax=0.0182; ay=0.00349 e) ax=-0.786; ay=-0.196
At the instant shown, the wheel rotates with an angular velocity ofw = 11.6 rad/s and an angular acceleration of a = 5.8 rad/s². (Figure 1) Figure 0.3 m 1.2 m D 45% 0.3 m C Part A Determine the angular acceleration of link BC at the instant shown measured counterclockwise. Express your answer using three significant figures. Enter positive value if the angular acceleration is counterclockwise and negative value if the angular acceleration is clockwise. ■VAΣ¢vec αBC= Submit Request Answer 6 rad/s²
At a given instant the top BB of the ladder has an acceleration aBaBa_B = 2.1 ft/s2ft/s2 and a velocity of vBvBv_B = 4.0 ft/sft/s , both acting downward Determine the acceleration of the bottom AA of the ladder. Determine the ladder's angular acceleration at this instant.
The hydraulic cylinder gives pin A a constant velocity v = 2.0 m/s along its axis for an interval of motion and, in turn, causes the slotted arm to rotate about O. Determine the values of r, r, and 0 for the instant when 0 = 32°. (Hint: Recognize that all acceleration components are zero when the velocity is constant.) Answers: r = 7 = 0 = A i i i 0 34 260 mm- m/s m/s² rad/s²
The vertical rod has a downward velocity v = 1.6 ft/sec and an upward acceleration a = 4.7 ft/sec? when the device is in the position shown. Determine the corresponding angular acceleration of bar AB (positive if counterclockwise, negative if clockwise) and the magnitude of the acceleration of roller Bif R = 19 in. A 1.35R 29° B Answers: a = i -13.97 rad/s2 aB = i ft/sec? -29.85 !!

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