Answer to Question #82050 in Physics for lee

Question

Let’s assume you wanted to build a very complicated record turntable with the surface divided into a bunch of rings that turned slower and slower as you moved out from the centre. The rule for the speed would be Kepler’s law that T2 is proportional to r3, where r is the radius of the ring and T is the period. If the ring that would be under the last rack on a record album (closest to the label) turns at 33 1/3 rpm, what would be the speed (in rpm) of the outside track? The diameter of a vinyl record album is 12 in and the label in the centre has a diameter of 4 in.

Answer 1

(T_2/T_1 )^2=(r_2/r_1 )^3

T_2/T_1 =ω_1/ω_2

(r_2/r_1 )=d_2/d_1 .

Thus,

(ω_1/ω_2 )^2=(d_2/d_1 )^3

ω_2=ω_1 (d_1/d_2 )^(3/2)=(33(3)+1)/3 (4/12)^(3/2)=6.415 rpm.