CD Rotation Speeds & Acceleration: Analysis of a 12cm Disk

[farrah003]

Unlike the older vinal records which rotated at a constant ω of 33.3 rotations per minute (rpm), compact disks vary their rotation speed during the playing period. A certain CD has a diameter of 12cm and a playing time of 77 minutes. When the music starts, the CD is rotating at 480 rpm. At the end of the music, the CD is rotating at 205 rpm.

a) What is the final rotational velocity of a point on the very outer edge of the disk?
ωf1 = rad/sec

b) What is the final rotational velocity of a point that is half way between the center of the disk and the outer edge?
ωf2 = rad/sec

c) What is the final tangetial speed of a point on the very outer edge of the disk?
vT1 = m/sec

d) What is the final tangential speed of a point half way between the center of the disk and its outer edge?
vT2 = m/sec

e) What is the magnitude of the average angular acceleration of the disk during this playing period?
αAVG = /sec2

 

[anathema]

a) To find the final rotational velocity (ωf1) of a point on the very outer edge of the disk, we can use the equation ωf1 = ωi + αt, where ωi is the initial rotational velocity, α is the angular acceleration, and t is the time. We know that ωi = 480 rpm and t = 77 minutes = 4620 seconds. Since the CD is rotating at a constant rate, we can assume that the angular acceleration is constant. Therefore, we can rearrange the equation to solve for ωf1:

ωf1 = ωi + αt
ωf1 = (480 rpm) + (α)(4620 s)

To find α, we can use the equation α = (ωf - ωi)/t. Substituting in the given values, we get:

α = (205 rpm - 480 rpm)/4620 s
α = -0.058 rpm/s

Now, we can plug this value back into the original equation to find ωf1:

ωf1 = (480 rpm) + (-0.058 rpm/s)(4620 s)
ωf1 = 205 rpm

Therefore, the final rotational velocity of a point on the very outer edge of the disk is ωf1 = 205 rpm.

b) To find the final rotational velocity (ωf2) of a point that is halfway between the center of the disk and the outer edge, we can use the same equation as in part a, but with a different initial rotational velocity. Since the point is halfway between the center and the outer edge, we can assume that the initial rotational velocity is halfway between the initial and final rotational velocities. Therefore, ωi = (480 rpm + 205 rpm)/2 = 342.5 rpm. Substituting this into the equation and solving for ωf2, we get:

ωf2 = (342.5 rpm) + (-0.058 rpm/s)(4620 s)
ωf2 = 152.5 rpm

Therefore, the final rotational velocity of a point halfway between the center and the outer edge is ωf2 = 152.5 rpm.

c) To find the final tangential speed (vT1) of a point on the very outer edge of the disk, we can use the equation vT1 = ωf1