Question

Equations of Motion: Cylindrical Coordinates Learning Goal: To set up and analyze equations of motion in a cylindrical coordinate system. The mechanism shown in the figure below rotates about the vertical axis. The collar has mass m=3.05 kg. The spring has an unstretched length of 310 mm and the spring constant is k=160 N/m. The distance d=200 mm , and the collar is required to stay a fixed distance r=480 mm from the vertical axis. (Figure 1)

Part A - The angular velocity for a smooth shaft If there is no friction between shaft AB and the collar, what angular velocity ?? must the mechanism have to keep the collar at r=480 mm from the vertical axis? Express your answer to three significant figures. ??= 4.60 rad/s SubmitHintsMy AnswersGive UpReview Part Correct Part B - The minimum required angular velocity when there is friction Consider the same mechanism again, with m=3.05 kg, d=200 mm, k=160 N/m, only now, instead of being smooth, the collar and shaft have a maximum coefficient of friction of ?s=0.67. What is the minimum angular velocity required to keep the collar at a constant distance r=480 mm from the axis of rotation? Express your answer to three significant figures. ??= 1.21 rad/s SubmitHintsMy AnswersGive UpReview Part Correct Notice that this angular velocity is less than that found in Part A, as expected. Part C - The maximum allowable angular velocity when there is friction Consider the same mechanism again, with m=3.05 kg, d=200 mm, k=160 N/m, and ?s=0.67. What is the maximum angular velocity allowable if the collar is to remain at a constant distance r=480 mm from the axis of rotation? Express your answer to three significant figures.

??=

rad/s

Answer 1

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

= sqrt(200^2 + 480^2) = 520 mm

x = 520 - 310 = 210 mm = 0.210 m

angle of spring will shaft, @ = tan^-1(d/r)

@ = tan^-1 ( 200/480) = 22.62 degrees

balancing forces on collar,

kx cos@ = mw^2 r

160 x 0.210 x cos22.62 = 3.05 x w^2 x 0.480

w = 4.60 rad/s ...........Ans

B) friction = uN

and N = kxsin@ + mg

N = 160x0.210xsin22.62 + 3.05x9.81 = 42.84 N

f = 0.67 x 42.84 = 28.70 N

now, On collar

kx cos@ - f = mw^2 r

160 x0.210 x cos22.62 - 28.70 = 3.05 x w^2 x 0.480

w = 1.25 rad/s

C) for maximum,

kx cos@ + f = mw^2 r

160 x0.210 x cos22.62 + 28.70 = 3.05 x w^2 x 0.480

w = 6.39 rad/s