Question

The angular position of a point on a rotating wheel is given by 0 = 8.70 + 3.69t2 + 2.11ť, where O is in radians and t is in seconds. At t = 0, what are (a) the point's angular position and (b) its angular velocity? (C) What is its angular velocity at t = 8.43 s? (d) Calculate its angular acceleration at t = 2.70 s. (e) Is its angular acceleration constant? (a) Number 1 SI rad - = (b) Numberî units" Fad units" rad/s nitsraels units"radis12 (c) Number S D D D D u (d) Number ts rad/s^2 (e) Number No D Click if you would like to Show Work for this question: Open Show Work

Answer 1

a At t=0 0 = 8.7 radians b wo do a 28.7 + 3.69 th + 2.11 t) dt - 7.38t + 6.33t c At t=0 w =0 rad/s At t = 8.45 w= (7.38 * 8.43) + (6.33x8.43²) = 512.05 rad's u Reda - 017-38t+ 6-33+) = 7.38 + 12 dt At t=2 .7 x = 41.562 rad/s² No, the angular acceleration is not constant e & a given Win 11.1 rad/ L = a 6.87 rady? we know wf = Wit at t t = wi o be on 6.87 t= 1.62 seconds o wy² – Wi2 - 22 = o- 11.12 - 2x -6.87 = 8.97 radians.