Question

Using a forked rod, a smooth 3-lb particle is forced to move along around the horizontal path in the shape of a limaçon, r = (5+sin o) ft. If 0 = -12 rad, where t is in seconds, determine the force of the rod and the normal force of the slot on the particle at the instant t= 3sec. The fork and path contact the particle on only one side.

Answer 1

Solution : Given; mass me 3-4b go (5+ Sino) ft o dt? xad {t is in seconds) at t= 3 sec, force of the rod (f) and normal force (N) 4 slot on particle = ? . = (5+ Sino) j = Caso (o) - @ = - Sino * (0)2 + Cas@x ë) - © Now, 9 - 1+2 0 = At = 3 seci 2: 9 = 1.125 vad, -1 sippe Ē - (E)Ť = 0 Ő tradis? Now put value of 0.0 & ☺ at t-3 sec in equation 0.0 4 © 85 (5+ Simo) - 5+ sin (4:125X180) = 5-90ft ģe cas (1.125X180) *3 = 0.323 ft/s

8 = [ şim(1.125x180) ()*]+[cas (1125x180)*(+)] = -0.507 +0:107 = -0.4.ft/s? tangent LLLLL G1-125xed radial acceleration 4 particle ax=8-ro? = -0.4 - 5.90 x(3)? = - 3.718 $7152 togential acceleration of particle Aug = së + 286 = 5.90 () + 210-323) X(X) = 1.95 ftis ? angle between normal md tangent tomd - 8 =15+ sin or dildo caso late = 1.125xad tema = 13.68 >> a = 85.82 190- Nowo write force equation on particle i Efr=m.ar; -n cas (902)-miar -N con 4.18' = 3224-3.718) = 85.82 LX

IN= 0·3.47 3-16 TE fo = m.agi F-N Sin 4.18 = 3x1.95 32.2 F = 3X1.95 + CO-347) Sin 4-18 1. 32.2 P = 0.206 - eb