A disk with a diameter of 0.08 m is spinning with a constant velocity about an a

Question

A disk with a diameter of 0.08 m is spinning with a constant velocity about an axle perpendicular to the disk and running through its center. 1) How many revolutions per second would it have to rotate in order for the acceleration of the outer edge of the disk to be 16.0 g's (i.e., 16.0 times the gravitational acceleration g)? f = ANSWER IS NOT 624 or 62 please help me 2) For the frequency determined in part (a), what is the speed of a point half way between the axis of rotation and the edge of the disk? v1/2 =

Explanation / Answer

D = 0.08 m; r = D/2 = 0.04 m
a = 16 g = 157 m/s2

(1) a = v2/r; v = r

a = (r)2/r = r 2

= (a/r) = 62.6 This answer is in rad/s; convert it to rps

(62.6 rad/s) x (1 rev/2 rad) = 10 rev/s

[You should do the math carefully on your calculator ratining as many significant figures as you can; I rounded off several of my answers]

(2) For this part, you should use = 62.6 rad/s

r = 0.04/2 = 0.02 m

v = r = (0.02)(62.6)    The answer is in m/s

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