Solution
coordinate system-Since time-parametric equation of the are given,it is not necessary to relate r to theta.
Velocity and acceleration- determining the time derivatives and evaluating them when t = 1s, we have
r = 100t2 |t=1s = 100 mm theta = t3 |t=1s = 1 rad = 57.30
r' = 200t |t=1s = 200 mm/s theta = 3 t3 |t=1s = 3 rad/s
r'' = 200 |t=1s = 200 mm/s2 theta = 6 t |t=1s = 6 rad/s2
v = rur + r thetha u theta = { 200 ur + 300utheta } mm/s
the magnitude of v is
v = sqrt 2002 + 3002 = 361 mm/s
sigma = tan-1 (300/200) = 56.30 sigma + 57.30 = 1140
also
a = [200-100(3)2]ur + [100(6) + 2(200) 3]utheta
the magnitude of a is
a = sqrt 7002 + 18002 = 1930 mm/s2
delta = tan-1 (1800/700)=68.70 (1800-delta) +57.30 = 1690
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