The disk has an initial clockwise rotation speed of 8 rad/sec and at t=0 begins

Question

The disk has an initial clockwise rotation speed of 8 rad/sec and at t=0 begins a counter-clockwise acceleration of (0.6*(t^1.88)). What is the speed of point A at t=3 sec? Point A has a radius of 2 ft. At what time does it come to a momentary stop? What is the total acceleration of point B at t=5 sec? Point B has a radius of 1.5 ft.

Explanation / Answer

Call counterclockwise direction as positive. =>wo = -8 rad/sec alpha = 0.6*(t^1.88) dw/dt = 0.6*(t^1.88) =>dw = 0.6*(t^1.88)dt =>w = 0.6*(t^2.88)*(1/2.88) + C, where C is integration constant. But at t = 0, w = wo = -8 rad/sec =>C = -8 rad/sec =>w = -8 + 0.6t^(2.88)*(1/2.88) At t=3sec, w = -8+(0.6*(3^2.88)*(1/2.88)) = -3.069760961 rad/sec That is 3.069760961 rad/sec, clockwise Speed = v = rw = 3.069760961*2 = 6.139521922 ft/s Now second part: Let it comes to momentary stop at time t =>w = 0 =>-8 + 0.6t^(2.88)*(1/2.88) = 0 =>0.6t^(2.88)*(1/2.88) = 8 =>t^2.88 = (8*2.88)/0.6 =>t = 3.549073622 seconds Now third part: a_t = 1.5*0.6*(t^1.88) = 1.5*0.6*(5^1.88) = 12.36559045 rad/sec^2 at t = 5seconds, w = -8+(0.6*(5^2.88)*(1/2.88)) = 13.46803898 rad/sec v = rw = 13.46803898*1.5 = 20.20205847 ft/s a_r = v^2/r = (20.20205847^2)/1.5 = 272.0821108 ft/s^2 Total acceleration = sqrt((a_t^2)+(a_r^2)) = 272.7136184 ft/s^2

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