In a typical CD player, the constant speed of the surface at the point of a lase

Question

In a typical CD player, the constant speed of the surface at the point of a laser-lens system is 1.3 m/s. The innermost first tack in the CD has a radius = 23mm and the outermost final track has a radius = 58mm. The maximum playing time of a standard music disk is 4473seconds. a) I found the angular speed of the inner most track to be 57 rad/s and the angular speed of the outer most track to be 22 rad/s. b) I found that the number of revolutions it took the for the CD to play for 4473seconds to be 2.8 x 10^4 rev. c) I found the angular acceleration of the compact disk over the 4473second time interval to be -0.0076 rad/s^s I DO NOT UNDERSTAND: d) If the mass of the CD is 15grams, calculate the rotational kinetic energy of the CD when the laser-lens system is reading from the innermost track and when it is reading in the outermost track. (Consider the CD is a solid cylinder: where the moment of inertia = 1/2MR^2). e) Repeat the calculation on point d but now considering the CD as a hollow cylinder of interal radius of 10mm and external radius of 60mm. (Consider the CD is a hollow cylinder: where the moment of inertia = 1/2M(R1^2 + R2^2).

Explanation / Answer

you need help with part (d)
First you must find moment of inertia
given that considering as solid cylinder, moment of inertia(I) = 1/2MR^2
= 0.015*0.058^2 / 2 = 25.23*10^-6

Now rotational energy = I 2 / 2

I is moment of inertia and put as 57 and 22 for inner most and outermost part, which give answer as 4.0986*10^-2 J and 1.2211*10^-2 J

e) for part e , just find the new I as 1/2M(R1^2 + R2^2). =1/2*0.015*(0.01^2 + 0.06^2).

all other steps are same.

Hope it helps.

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