HW6 - VB =(AB)OJAB =(5(12 = 60 in.ls t vB = VB(BE)OJBD 7 P...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
PROBLEM 15.156 Four pins slide in four separate slots cut in a circular plate as shown. When the plate is at rest, each pin has a velocity directed as shown and of the same constant magnitude u. If each pin maintains the same velocity relative to the plate when the plate rotates about 0 with a constant counterclockwise angular velocity OJ, determine the acceleration of each pin. SOLUTION For each pin: Acceleration of the coinciding point P' of the plate. For each pin, ap' = roi towards the center O. Acceleration of thepin relative to the plate. ap/F = 0 For pin~, u2 ap/F =- - r Corio/is acceleration ac' For each pin ac = 20JU with ac in a direction obtained by rotating u through 90° in the sense of 00,i.e. ). Then, a. = [rOJ2 - ] + [2OJu i ] a2 = [rOJ2i]+[2OJu-] 83= [7m2 _ ] + [u: _ ] +[2= _ ] ----
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
SOLUTION D PROBLEM 15.179 The disk shown rotates with a constant clockwise angular velocity of 12 rad/s. At the instant shown, determine (a) the angular velocity and angular acceleration of rod BD, (b) the velocity and acceleration of the point of the rod coinciding with E. Geometry. 5 tanp =10' 10 = 11.1803 in. IAE = cosp p = 26.5650 Velocity analysis.
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: VB = (AB)OJAB = (5)(12) = 60 in.ls t vB' = VB + (BE)OJBD 7 P = [60 t ] + [11.1803OJBD 7 p] VE/BD = [u r\ p], VE = Use V E = VB' + v E/BD and resolve into components. +7 p: =-60sinp + 11.1803OJBD, OJBD = 2.400 rad/s + \ p: = 60cosp-u, u = 53.666m/s vB' = [60 t ]+[(11.1803)(2.400)7 pJ = 53.7in.ls~ 63.40 Acceleration analysis. aB = (AB)OJ~B = (5)(12)2= 720in.ls2-aE, = aB + [(BE)aBD 7 pJ + [(BE)OJ~D \ pJ = [720-] + [11.1803aBD 7 p] + [64.399 \ p] a E/BD = [u r\ p J a E = Corio/is acceleration. 20JBDU = (2)(2.400)(53.666) = [257.60 7 p] PROBLEM 15.179 CONTINUED Use aE = aE' + aE/BD + [2lUBDU ;7' p] and resolve into components. +7 p: =-720cosp + 11.1803aBD + 257.60 aBD = 34.56 rad/s2 + \ p: = -720sinp + 64.399-u, u = -257.59 inls2 aE'=[720-] + [(11.1803)(34.56) 7 p] + [64.399 \ p] =[720-] + [386.39 7' p] + [64.399 \ p] = 365 in./s2 '"' 18.4° Summary: (a) " lUBD = 2.40 rad/s ~, aBD = 34.6 rad/s2 ) ~ (b) VE' = 53.7 in./s ~ 63.4°,---...
View Full Document

This note was uploaded on 08/31/2010 for the course ENGRMAE 80 taught by Professor Staff during the Summer '08 term at UC Irvine.

Page1 / 3

HW6 - VB =(AB)OJAB =(5(12 = 60 in.ls t vB = VB(BE)OJBD 7 P...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online