Lecture Notes for Section 16.4 (Absolute Motion Analysis)

EXAMPLE(continued)Solution:By geometry, s A = 2 cos qreferences AqABy differentiating with respect to time,v A = -2 w sin qUsing q = 60° and v A = 8 m/s and solving for w:w = 8/(-2 sin 60°) = - 4.62 rad/s(The negative sign means the rod rotates counterclockwise aspoint A goes to the right.) Differentiating v A and solving for a,a A = -2a sin q – 2w 2 cos q = 0a = - w 2 /tan q = -12.32 rad/s 2

EXAMPLE IIGiven:Crank AB rotates at aconstant velocity ofw = 150 rad/sFind: Velocity of P whenq = 30°Plan: Define x as a function of q and differentiate withrespect to time.Solution:x P = 0.2 cos q + (0.75) 2 – (0.2 sin q) 2v P = -0.2w sin q + (0.5)[(0.75) 2– (0.2sin q) 2 ] -0.5 (-2)(0.2sin q)(0.2cos q)wv P = -0.2w sin q – [0.5(0.2) 2 sin2q w] / (0.75) 2 – (0.2 sin q) 2At q = 30°, w = 150 rad/s and v P = -18.5 ft/s = 18.5 ft/s

CONCEPT QUIZ1. If the position, s, is given as a function of angular position,q, by s = 10 sin 2q, the velocity, v, isA) 20 cos 2q B) 20 sin 2qC) 20 w cos 2q D) 20 w sin 2q2. If s = 10 sin 2q, the acceleration, a, isA) 20 a sin 2q B) 20 a cos 2q - 40 w 2 sin 2qC) 20 a cos 2q D) -40 a sin2 q

GROUP PROBLEM SOLVINGGiven: The w and a of the disk andthe dimensions as shown.Find: The velocity and accelerationof cylinder B in terms of q.Plan: Relate s, the length of cablebetween A and C, to theangular position, q. Thevelocity of cylinder B isequal to the time rate ofchange of s.

GROUP PROBLEM SOLVING(continued)Solution:Law of cosines:s = (3) 2 + (5) 2 – 2(3)(5) cos qv B = (0.5)[34 – 30 cosq] -0.5 (30 sinq)wv B = [15 sin q w]/34 – 30 cos q(15w 2 cosq + 15asinq)a B=+34 -30cosq(-0.5)(15w sinq)(30w sinq)(34 -30cosq) 3/215(w 2 cosq + asinq) 225w 2 sin 2 qa B=-(34-30 cosq) 0.5 (34-30 cosq) 3/2