Question

131 Consider two observers: O at rest in space (inertial reference frame S), and O' in a noninertial rotating frame, S. S' is rotating relative to S with a time-dependent angular velocity w = 2t î - { +(2+ + 4) Ê. (a) In the rotating frame, a particle has position vector given by r' (t) = (t? + 1)î -- 6th + 4th. Find the true velocity v(t) in reference frame S and the apparent velocity v'(t) in frame S' at time t = 1 s. (b) Find the acceleration in each reference frame at time t = 1 s. (c) Find the magnitude and direction of the centrifugal acceleration in reference frame O' at time t = 1 s. (d) Find the magnitude and direction of the coriolis acceleration in reference frame O' at time t = 1 s.

Answer 1

a) In our problem the position veelon m'in non inertial tofaling frame y (2 241) it - 6+ 3+ 4+3 4 .: velocity in s frame ': dr = 247-603'+12+?" al time t= 1 sec. V=21 - 65+12K? :lvl = 4 +36 + 144 = (1.84 The velocity vins frame ū0+ In our Problem w = at T _{², +(2+4)ű 24 - 2t+4) (t+1) 6t 4+3 -147+12+ + 24t)- 5 (84!224) th (124² + + + + 2) -48-4) at time f= 1 sec. = 321 – 45 -10K ū- (: -65412k):( 32 - 451-104) = (34 ² -10 5 + 2K)

b) da (247-654 12 ) at time t=1sec du = 2 + 24K # dt de = N4+576 = 1580 acceleration in s frame dyda + 2b0xw') + 234*) +(247) at -t² (21+4) 24 -6 +1242 • (-12 + + 12+ +24) 2 -3 ( 24+?- 4+"-84) +(-12+ + 2+3) at time t = 1 sec. (W x 3) = 24 î–12 3 -1o ...

w xv) lt=2 = 32 i-49-08 2 - 16 32 -4 -10 - î ( 16 +24) - 3 (-20 - 192) + 4 (-8 +32) - 34+ 212 3 +24 ---- w = stê - t + (2+4) a dw = 21-2+3 + 2 à 20 h = 21-23 +2 Gî -4 9 -8û id=(22+246) (2,3 -243-208)+(947 2212542298) Putting all the nature = 888 + 18457 +2012 e) and Centrifugal acceleration --x(x*') From eqution 1 = 847 +212] +296 √46676= 216.05

d) Coriolis acceleration, = + 2 (W x D) From equation ④ = 488-245-20k? = 57.27