Spinning disk revolutions homework

[mattmannmf]

Spinning Disk

A disk with a diameter of 0.06 m is spinning with a constant velocity about an axle perpendicular to the disk and running through its center.

a) How many revolutions per second would it have to rotate in order that the acceleration of the outer edge of the disk be 15 g's (i.e., 15 times the gravitational acceleration g)?

HELP: Use the relation between the centripetal acceleration and angular velocity.
HELP: Make sure you use the correct radius.

b) 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?


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


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


e) Suppose we double the diameter of the disk. We still want the same 15 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

im not sure what to do

 

[rl.bhat]

Use the relation between the centripetal acceleration and angular velocity.
an you find out the formula for the above?

 

[nlightner]

with this information, but here is my response:

I would approach this problem by first understanding the concepts involved. The spinning disk is rotating with a constant velocity about an axle perpendicular to the disk and running through its center. In order to determine the necessary revolutions per second for the disk to have an acceleration of 15 g's at its outer edge, we can use the relationship between centripetal acceleration and angular velocity. This relationship is given by a = ω^2r, where a is the centripetal acceleration, ω is the angular velocity, and r is the radius of the disk. In this case, the radius is given as 0.06 m.

a) To find the necessary revolutions per second, we can rearrange the equation to solve for ω. Plugging in the given acceleration of 15 g's and the radius of 0.06 m, we get ω = √(15*g/0.06) = 63.245 rad/s. To convert this to revolutions per second, we can divide by 2π, giving us f = ω/2π = 10.074 rev/s.

b) At this frequency, the speed of a point halfway between the axis of rotation and the edge of the disk can be calculated as v = ωr/2 = (63.245*0.06)/2 = 1.897 m/s.

c) The period of rotation for this halfway point can be calculated as T = 1/f = 1/10.074 = 0.099 s.

d) To find the time it takes for a point on the edge of the disk to travel 1 km, we can use the formula d = vt, where d is the distance, v is the speed, and t is the time. Plugging in the given distance of 1 km and the calculated speed of 1.897 m/s, we get t = d/v = 1000/1.897 = 527.223 s.

e) If we double the diameter of the disk, the radius will also double, becoming 0.12 m. Using the same formula as before, we can solve for the new angular velocity needed to maintain an acceleration of 15 g's at the outer edge. Plugging in the new radius and the same acceleration, we get ω2 = √(15*g/0.12)