An electric turntable 0.780 m in diameter is rotating about a
fixed axis with an initial angular velocity of 0.300 rev/s . The
angular acceleration is 0.894 rev/s2 .
Part A
Compute the angular velocity after a time of 0.192 s
Part B
Through how many revolutions has the blade turned in this time interval?
Part C
What is the tangential speed of a point on the tip of the blade at time t = 0.192 s ?
Part D
What is the magnitude of the resultant acceleration of a point on the tip of the blade at time t = 0.192 s ?
given that wi = 0.3 rev/sec
alpha = 0.894 rev/s^2
t =0.192 sec
using wf = wi+(alpha*t)
wf = 0.3+(0.894*0.192) = 0.472 rev/sec
B) using theta = (wi*t)+(0.5*alpha*t^2)
theta = (0.3*0.192)+(0.5*0.894*0.192^2) = 0.07407 rev
=0.07407*2*3.142 = 0.465 rad
c) v = r*w = (0.78/2)*(0.472*2*3.142) = 1.15 m/sec
d) ar = v^2/r = 1.15^2/(0.78/2) = 3.4 m/s^2
at = r*alpha = (0.78/2)*0.894*2*3.142 = 2.2 m/s^2
resultant accelaration is a = sqrt(at^2+ar^2) = sqrt(2.2^2+3.4^2) =
4.04 m/s^2
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