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Coriolis Acceleration Physics

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Question: Pin P is attached to the collar shown; the motion of the pin is guided by a slot cut in bar BD and by the collar that slides on rod AE. Rod AE rotates with a constant angular velocity of 5 rad/s clockwise and the distance from A to P increases at a constant rate of 2 m/s. Determine at the instant shown (a) the angular acceleration of bar BD, (b) the relative acceleration of pin P with respect to bar BD.

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Solution Summary

This solution consists of a detailed step-by-step response of how to work through the calculations required to complete both parts of this physics-based question. Additionally, a diagram is attached (jpg file) to provide further clarification.

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Note: A diagram is attached.

Let at the instant the AE rod makes an angle theta with the vertical and Bd makes phi with the horizontal, point P be at distance r from A, BP = r', AB = s.

The velocity of point P along the bar AE:
v||AE = u = dr/dt = 2 m/s
The velocity of point P perpendicular to bar AE:
v_perpAE = w*r
Where w = angular velocity of bar AE = 5 rad/s

Hence, velocity of point P perpendicular to bar BD:
v_perpBD = w*r*sin(theta+phi) - u*cos(theta+phi)
Hence, angular velocity of bar BD:
d(phi)/dt = w' = v_perpBD/r'
w' = {w*r*sin(theta+phi) - u*cos(theta+phi)}/r'

From triangle ABP, by sine ...

Solution provided by:
  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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