Please see the attachments.
In solving use the coriolis accel concept and i j k vector coordinates.

15.148 and 15.149 Pin P is attached to the wheel shown and slides in a slot cut in bar BD. The wheel rolls to the right without slipping with a constant angular velocity of 20 rad/s. Knowing that x = 480 mm when theta = 0, determine the angular velocity of the bar and the relative velocity of pin P with respect to the rod for the given data.

15.148 a) theta = 0, b) theta = 90 degrees
15.149 theta = 30 degrees

r = 140 mm
R = 200 mm
velocity of center of wheel (vc) = w*R = 20*200 = 4000 mm/s = 4 m/s

a.)
theta = 0 degree
velocity of P (vp)= vc + w*r = 4 + 20*0.140 = 6.8 m/s
AB = x = 480 mm = 0.48 m
AP = 140 mm = 0.14 m
Hence,
BP = sqrt(0.48^2 + 0.14^2) = 0.5 m
Let angle PBA = phi
Hence, velocity omponent of P, perpendicular to BD:
v_perp = vp*sin(phi) = 6.8*(AP/BP) = 6.8*0.14/0.5 = 1.904 m/s

Hence, angular velocity of BD:
wbd = v_perp/BP = 1.904/0.5 = 3.808 == 3.81 m/s --Answer

velocity of P with respect to rod:
v_parallel = 6.8*cos(phi) = ...

Solution Summary

This solution solves equations for velocity step-by-step to find the final velocity of points on the system.

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