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A disk with a diameter of 0.05 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 12 g's (i.e., 12 times the gravitational acceleration g)? f =

User Thukaram
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1 Answer

6 votes

Answer:

f= 7.8Hz

Step-by-step explanation:

a= 12g= 120m/s²

a= ω²r

120= ω²×0.05

ω= 48.99 rad/s

2πf= 48.99

f= 7.8Hz

User Eugene Trofimenko
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