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 14 g's (i.e., 14 times the gravitational acceleration g)?
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?
3)At this same frequency, what is the period of rotation of this \"halfway point\"?
4)How long does it take a point on the edge of the disk to travel 1 km?
5)Suppose we double the diameter of the disk. We still want the same 14 g acceleration at the outer edge. Let f2 be the number of revolutions per second needed to get that acceleration. What is the ratio R = f2/f, where f is your answer to part (a)? Answer according to the following key:
1 = 0.500
2 = 0.707
3 = 1.000
4 = 1.414
5 = 2.000
6 = none of the above
- NCSLv 71 month agoFavourite answer
1) a = ω²r
14*9.81m/s² = ω² * 0.04m
ω² = 3430 rad/s²
ω = 58.6 rad/s
f = ω / 2π = 9.3 rev/s
not the number you give
(although you're close to the rad/s value)
2) v = ωr' = 58.6rad/s * 0.02m = 11.7 m/s
3) T = 1/f = 0.107 s
for all points on the disk
4) t = d / v = 1000m / (58.6rad/s*0.04m) = 427 s
(just over 7 minutes)
5) Now 14*9.81m/s² = ω² * 0.08m
ω = 41.4 rad/s
R = f₂ / f = ω₂ / ω = 41.4 / 58.6 = 0.707 ◄ #2
Hope this helps!