Question

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Motor M rotates with an angular velocity of w = a(1 - ebt)Determine the magnitudes of the velocity and acceleration of point P on the fan blade when t = t1. What is the maximum speed at this point? rad ri = 1 in a = 4- r2 = 4 in r3 = 16 in ANSWERS: Vp = 6.30 in Vr = 16.00 SIS Ap = 10.023

Answer 1

Given Motor M rotates with angular velocity wn=all-eo Given a = 40ls - b = -1/ W = 4(1-et) rls radius=s, radius=82 lot let ws (angular velocity of fan). Motor fan relationship between fan ang. vel & fator ang. vel - (Wf) (WM) is given by som = V2 Wp (same velocity ) (Motor) (fan) Given V=ow 8 = 1 in 72-4 in > 1X((1-én)) = 4 wf we = 1-et 8/ .. fan angular velocity (W) = 1-4 6/s.

lan 1-6 angular velocity of fan w angulan acceleration dt angular acceleration of fan - 01 vclocity of point P, V = 8 L 8 = distance from centre to point P 83 = 16 in (briven) We = 1-6 Given to, = 0.55 Wf at tt, is Wp = 1-605 = 0.39348/5 *f at tzt, is & = 0-05 = 0.6065 8/5² => velocity of point P, Vp (at t = 0.5 seconds) = 8z WFT t=0.5 >> Vp = 16 x 0.3934 ~ 6-3 in)s Vo = 6-2955 in)s = 6-3 3/5xin Vp = 6.3 in/s At point p accelcratim has two components. pat Tangentia Tangential component, a, = 0; p = 16 X 06065 ap = 9.704iM) 5 Date normal component, an = 0,6 = 16X0-3934 an = 2-4762 i 15² Total acceleration ap-Jantat - 19.704 +2.47622
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I hope you understand. Comment if you have any doubt.

acceleration ap = 10.0149 iris 10.02 in)s velocity (at any time) of point P, Up = 8 3 Wp = 16 (1-6) inls - At To find man velocity dip = o velocity will be max - -t 16e som et = 16€* dup =0 dt at t= x i velocity will be max when t=0 man velocity up = 16 (1-2 ) = 16 inls. Lot an easy way is ēt is minimum at tz* = 1-ét will be maximum)