Equations of Motion: Cylindrical Coordinates 20 of 25 Learning Goal: Part A-The

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Equations of Motion: Cylindrical Coordinates 20 of 25 Learning Goal: Part A-The angular velocity for a smooth shaft To set up and analyze equations of motion ina cyindrical coordinate system The mechanism shown in the figure below rotates about the vertical axis. The colliar has mass m 3.7 kg. The spring has an unstretched length of 575 mm and the spring constant is km 190 N/m. The dstance d 440 mm, and the collar is roqured to stay a fxed distanoe r = 825 mm from the vertical axis. If there is no friction between shat AB and the collar, what anguiar velocty é must the mechanism have to keep the collar at 825 mln from the vertical as? Express your answer to three significant figures View Available Hintis) (Figure 1) rad/s Submit Previous Answers x Incorrect; Try Again; 5 attempts remaining Part B-The minimum required angular velocity when there is friction same mechanism agar, with m-3.7 kg d Consider the smooth, the clar ard shat have a maximum coetscent of t ction of '0.57 what s the mnmm anular velooity requred to keep the collar at a constant distancer 825 mm from the axis of rotation? Express your answer to three significant figures View Available Hints) 440 mm. k-190 N/m. only now metod of being rad/s Submit Part C This question will be shown after you complete previous questions) Figure 1of1 Provide Feedback Next >

Explanation / Answer

length of spring = sqrt(d^2 + r^2)

= sqrt(440^2 + 825^2) = 935 mm

x = 935 - 575 = 360 mm = 0.360 m

angle of spring with shaft, theta = tan^-1(d/r)

Theta = tan^-1 ( 440/825) = 28.07 degrees

balancing forces on collar,

kx cos theta = mw^2 r

190 x 0.36 x cos28.07° = 3.7 x w^2 x 0.825

w = 4.45 rad/s ...........Ans

B) friction = uN

and N = kxsin theta + mg

N = 190 x 0.360 x sin28.07 + 3.7 x 9.81 = 68.48 N

f = 0.57 x 68.48 = 39.04 N

now, On collar

kx cos theta - f = mw^2 r

190 x 0.360 x cos28.07- 39.04 = 3.7 x w^2 x 0.825

w = 2.64 rad/s

Please rate my answer, good luck...

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