A standard CD has a diameter of 12.0 cm, and a hole that is centered on the axis of symmetry and has a diameter of 1.50 cm. The CD’s thickness is 1.2 mm, and it's mass is 17 g. If a (very thin) pin extending through the hole of the CD suspends it in a vertical orientation, the CD may oscillate about an axis parallel to the pin, rocking back and forth. How long does it take to complete 8 complete oscillations?
4 answers
sorry, I need a picture.
there isn't a picture
ok, well if I have it right the pin is at the edge of the hole, 0.75 cm above the center of gravity of the CD
for a small angle A the restoring moment about that pin is
-m g (.0075)sin A = -.0075 m g A
= -.0075*.017*9.81 A = -.00125 A
the moment of inertia of the CD about its center is
(1/2) m r^2 - moment of inertia of hole
= (1/2) .017(.06^2 + .0075^2)
= .0085 (.00366) = 3.11*10^-7
use parallel axis to move that up to the pin
I = 3.11*10^-7 + .017(.0075)^2
= 3.11*10^-7 + 9.56*10^-7
= 12.67*10^-6
moment = I d^2 A/dt^2
-.00125 A = 12.67*10^-6 d^2A/dt^2
if A = sin 2 pi t/T
then d^2A/dt^2 = - (2 pi/T)^2 A
so
.00125 = 12.67*10^-6 (2 pi/T)^2
(2 pi/T)^2 = 9.86*10^1 = 98.6
2 pi/T = 9.93
T = .633 seconds
times 8 = 5 seconds
for a small angle A the restoring moment about that pin is
-m g (.0075)sin A = -.0075 m g A
= -.0075*.017*9.81 A = -.00125 A
the moment of inertia of the CD about its center is
(1/2) m r^2 - moment of inertia of hole
= (1/2) .017(.06^2 + .0075^2)
= .0085 (.00366) = 3.11*10^-7
use parallel axis to move that up to the pin
I = 3.11*10^-7 + .017(.0075)^2
= 3.11*10^-7 + 9.56*10^-7
= 12.67*10^-6
moment = I d^2 A/dt^2
-.00125 A = 12.67*10^-6 d^2A/dt^2
if A = sin 2 pi t/T
then d^2A/dt^2 = - (2 pi/T)^2 A
so
.00125 = 12.67*10^-6 (2 pi/T)^2
(2 pi/T)^2 = 9.86*10^1 = 98.6
2 pi/T = 9.93
T = .633 seconds
times 8 = 5 seconds
better get a CD and try it :)
0 answers