Question

Pin Pis 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 3 radis clockwise and the distance from A to Pincreases at a constant rate of 4 fus. Determine at the instant shown the angular acceleration of bar BD and the relativo acceleration of pin P with respect to bar BD. (Round the final answer to one decimal place) E 16 in The angular acceleration of bar BDS The relative acceleration of in P with respect to bar is

Answer 1

Given

$\\ \omega_{AE}=3\ rad/s\ \ CW \\ V_{PAr}=4\ ft/s$

Consder velocity of the mechanism

E V v 기 D PBr PAT P V PAt Y. 16 in PBt 3 AE

$\\ V_{PAt}=\omega_{AE}*16=48\ in/s=4\ ft/s$

By taking component in the tangential and radial direction of BP

VPBt = VP At cos 45° + VP.Ar Cos(90° + 45°) = (VP.At - Vp Ar)/V2 = 0 WBD = VPBt/BP = 0 VpBr = VP At cos 45° + VP Ar Cos(45°) = 5.6569 ft/s

The acceleration of the mechanism

Relative to A, P will have a Coriolis acceleration and a centripetal acceleration

Relative to B, P will have a radial acceleration and a tangential acceleration

E 71 a D PBr a PAcoriolis a PROJ a PBt IW AE

$\\ a_{PAc}=AP*\omega_{AE}^2=144\ in/s^2=12\ ft/s^2 \\ a_{PAcoriolis}=2V_{PAr}\omega_{AE}= 24\ ft/s^2$

Taking the acceleration component in the radial and tangential direction of BP

$\\ a_{PBt}=a_{PAc}\cos45\degree+a_{PAcoriolis}\cos45\degree= 25.456\ ft/s^2 \\ Angular \ acceleration \ of \ BD \\ \alpha_{BD}=a_{PBt}/BP=a_{PBt}\sin45\degree/AP=\frac{25.456\sin45\degree}{16/12}=13.5\ rad/s^2\ \ CW \\ Acceleration \ of\ P\ w.r.t\ BD \\ a_{PBr}=a_{PAc}\cos(90\degree+45\degree)+a_{PAcoriolis}\cos45\degree= 8.4853\ ft/s^2$