A disk with a diameter of 0.04 m is spinning with a constant velocity about an a

Question

A disk with a diameter of 0.04 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 13 g's (i.e., 13 times the gravitational acceleration g)?

f = rev/s  

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 = m/s  

3) At this same frequency, what is the period of rotation of this "halfway point"?

T = s  

4) How long does it take a point on the edge of the disk to travel 1 km?

T1000 = s

Explanation / Answer

d = 0.04 m

r = d/2 = 0.04/2 = 0.02 m

1)

a = rw^2

w = sqrt(a/r)

w = sqrt(13g / 0.02)

w = sqrt[(13 x 9.81 / 0.02)

w = 79.85 rad/s

w = 2pif

f = w/2pi = 79.85/2pi

f = 12.71 rev/s

2)

v1 = rw

v1 = 0.02 m x 79.85 rad/s = 1.597 m/s

v1/2 = 0.7985 m/s

3)

T = 2pi/w

T = 2pi / 79.85

T = 0.0787 sec

4)

d = 1000 m

t = d/v

t = 1000 m / 1.597

t = 626.174 s = 10.436 min

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