PHY101 2nd Assignment Solution

Q# 1 A compact disc stores lectures Physics (PHY101) in a coded pattern of tiny pits 10^-7m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 20.0mm and 50.0mm, respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.5m/s.

(a) What is the angular speed of the CD when the innermost part of the track is scanned the outermost part of the track?

(b) The maximum playing time of a CD is 75.0m. What would be the length of the track on such a maximum duration CD if it were stretched out in a straight line?

(c) What is the average angular acceleration of a maximum duration CD during its 75.0m playing time? Take the direction of rotation of the disc to be positive.

Ans 1. The problem statement, all variables and given/known data
Compact Disc. A compact disc (CD) stores music in a coded pattern of tiny pits 10^-7 m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 mm and 58.0 mm, respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 m/s.
The maximum playing time of a CD is 74.0 min. What would be the length of the track on such a maximum-duration CD if it were stretched out in a straight line?
2. Relevant equations
theta = (w_f + w_0)t
s = r*theta
w = angular velocity
3. The attempt at a solution
This is part 3 of the problem so I already have the angular velocity for the innermost and outermost of the disc--which I assume it is the initial and final angular velocity. Pls correct me if I'm wrong. I've used the eqns above to find theta and the length but I'm coming out with the wrong answer, 5.25km. I took the outer radius - inner radius = r. Can someone pls correct me? I don't know what I'm doing wrong. :frown:

Q # 2

The flywheel of a gasoline engine is required to give up 550J of kinetic energy while its angular velocity decreases from 790 rev/min to 730rev/min. What moment of inertia is required?

Ans

790rpm = 790 x 2pi / 60 rad/s; w1 = 82.73 rad/s
730rpm = 730 x 2pi / 60 rad/s; w2 = 76.45 rad/s
Kineic energy of rotation = Iw2/2 so
550 =(I x 82.73^ /2) - (I x 76.45^2)
1100 = I x (82.73^2 - 76.45^2)
I = 1100 / 999.65
= 1.10 kgm^2