Question

The motion of pin Pis guided by slots cut in rods AE and BD. Knowing that the rods rotate with the constant angular velocities WA=4 rad/s clockwise and W8= 4.8 rad/s counterclockwise, determine the velocity of pin Pfor the position shown — 250 mm 1 The velocity of pin Pfor the position shown is mm/s

Answer 1

Solution | Page No. 1 250 mm al ol length (dBP): tan 30º = dep | CAB as ten 3 = To length dap) , depa das tan 30° cap =(0, 45o) lơn 30° = 0-144 m dep= 0.144m 1-0 Co3 36 = CAB = 050 , CAP= 0 350 dAP dap Cos 30° P= 0. 3786 m -9) At paint P equation of velo city on the body having motion relative to a rotating frome. Vp = Vpi + UPLA - I Consider the point in the vlod BD. » Up' = WBOX OBP

Page No.2 | = Vp' = (4.8)x (0.1443) = 0.693 m/s → VPYF = (01) Positive y-direction B = 4, 1 Hence from ② equation, we get Up = 0.693 m/s +0,9 - ☺ Consider the point pe in the rod AE Vp'= WAE Xd AP X Cos 30° Vp's 4X 0.9886 X (08 30° Up'= 1 m/s alsog VPLE = (Ulz) Positive y-disiection = Uz Cos 60° = 0.5427 Hence forom equation * wet Vp = imlo x + 0.5024 - 5 Equate equation and ☺ foge vp, and cluesolve into Vertical Com bon ent8 U, = -(1+0.542) - © Equate equation D and 6 fou up and resolve into Holi? Ontal components :-

Page No-3 | → 0-693 = - (0.9886) (4) Sin 30° + Uz singoo 0.693 = -0.5772 +42 Singo ų Sinbo = 0.693 +0.5772 = 1.2702 42 = 1.9702 - 1.2702 – 1.4667 mis I sin bo° 0.8660 1 49 = 1.4667 mls / Substitute 42 = 14667m/s in equation ©, we get U = -(H+(0.5) (1.4667)) = 017333 m/s 14, = 1.7334m18! Vp= 0.693 m/ → + 1.7334 m) 8 1 Callulate magnitude of vector Vp Maguitude of yp: V 10.693)2 + (1 73342 Magnitude of up= 10 74 78 mis B4 7Magnitude of up = 1747 mm/s 04 1747.2 mm). Calculate the direction of vector up Dissection of up tam ( 1.7334) Digue ction of vp = 68.doº ( 0.693

- Answers 9) Magnitude of Vp = 1747 mm/s ou 1747.8 mms b) Direction of Vp = 68. 20°