Question

Dynamics

Given: The motion of a particle P is defined by the relations r = [kı sin(b) m and 0=(k21°) rad, where t is expressed in seconds. Please note that kı, k2, and b are constants which are greater than zero. For the initial condition, the particle has an angle of o=0° when t = 0 sec. So, when 1 = 1 sec, Find: a) The radial (vr) and transverse (ve) components of velocity of the particle P. b) The magnitude of the total velocity of the particle P. c) The radial (ar) and transverse (ae) components of acceleration of the particle P. d) The magnitude of the total acceleration of the particle P. Solution:

Answer 1

solution Ginsen, g = ki silbe) m sad = Kz 13 dor de 11 a) = dos at do at vo 11 = do z at W = angulors velocity 2 do तर Now de do d [ki simbo)] ki cos(be). b do dar do (kib) cosebe) il de = $(k2+3) = 342.62 (342) 72 de dt Now of - dro de at (kib) cos(be). (3 k2) +2 (3 k, k2 b) +². cosbe (v) VA de.7 Vo = at (3K2) +2. k, sin (be) (3 k, K2) 12 smlbo) VO =

Now t = 1 Sec Vo = (ski kzb) cosebe) Vo = (3kika) smebe) b) A ve total velocity √ ² + v6² Vy Hence, V= [(3 ki ky b)2 tos?(be)] + (3kika) ? simbo) (36 kg) 6² cos?bo) + sinabo) 3 ki kz J62 cos? (6K2) + sim? (bkz) 0 = k2 At t = Isc Oy dur at 11 at dvr do do de [3 ki kab) +2 cos(be) ] . d (k2+3) de (3 ki kz b) +? [-sin(be). b] - [342 kz] 9 kik2b2 ++ [- simbo)] At t= 1 sec 9 K, 26² (- simbo) Skik b? [- si (bk)] Og 6 Bing Arr 11

dre - dig det de 838 & [ 3ki kt? & sim(be)] $(x+3) 3k; kzt? (OS (bo). b (342 K2) 9 ki KE 1² b cos(bo) at t= isec, 6 = K2 g kik b cos (bk) 9 ki kz2b cos (bk) 06 - a = Total acceleration - Jab + ant VER, MS LA SAN )]* [9 kitêb coschers 9 ki kz / b sos (bk) + 6² a = total acceleration race + ars? J[gkik? b cos(bky)]? [9 ki k² 6² 1-sim(bk))]² (gkik?) b2 cos?(bk) + b4 sm?l bk) d) +