Question

A digital audio compact disc carries data along a continuous spiral track from the inner circumference of the disc to the outside edge. Each bit occupies 0.60 mm of the track. A CD player turns the disc to carry the track counter-clockwise above a lens at a constant speed of 1.22 m/s. (a) Assuming the acceleration is constant, find the total angular displacement of the disc as it plays from the outside at 2.10 cm to the inside at 6.10 cm. (b) Find the total length of the track.

Answer 1

A full- length recording lasts for 74 min, 33 s ,
t=74 min33 sec =4473 s
ω1=v/R1 =1.22/0.021 = 58.1 rad/s
ω2=v/R2 =1.22/0.061 = 20 rad/s
ε= (ω2- ω1)/t =(20-58.1)/ 4473=-0.0085 rad/s²
φ=εt²/2=0.0085•4473²/2=85032.8 rad
L=v•t=1.22•4473=5457.06 m

Answer 2

I assume that you have inside and outside reversed and I assume that it is the angular acceleration that is constant.

alpha = angular acceleration = dw/dt = constant

r outside = .061 m
r inside = .021 m
I am assuming you mean radius and not diameter. If you copied this question from a book they really messed it up.

.60 mm = .6 * 10*-3 m = 6 * 10^-4 meters/bit

v = w r = 1.22 m/s
v is constant so w =1.22/r

Initial angular velocity wi at outside:
wi = 1.22/.061 = 20 radians/second
Final angular velocity wf at inside:
wf = 1.22/.021 = 58 radians/second

angular acceleration alpha = (58-20)/time

Now to find the time I need how many total bits are on the track. (unless the time was given)

Once you find the angular acceleration, alpha, and knowing the time, integrate to find the total angular displacement

theta = wi t + (1/2) alpha t^2

Answer 3

okay I would just skip it I'm in 7th grade and that looks hard