- #1
Jewelz
- 8
- 0
Homework Statement
A 45 RPM record is a disk with a wide hole in the center. The mass of such record is 45g. It is 7 inches in diameter, and the hole in the center has a 1.5 inch diameter. (Sorry for the odd units, this was the way it was given).
Find the moment of inertia of the 45 RPM record[/B]
Homework Equations
[itex]I=1/2MR^2[/itex]
ITotal=IDISC - IHOLE
Area of a circle = [itex]πR^2[/itex]
MHOLE=(MDISC*AREAHOLE) / (AREADISC-AREAHOLE)[/B]
The Attempt at a Solution
Given the fact that the moment of inertia of a disc is ½MR^2, using the information given, I calculated IDISC to be 275.63 g [cm]^2.
IDISC=0.5(45g)([3.5in])^2 = 275.63 g in^2
To find the moment of inertia of the hole, I had to find what the mass would have been if it were included.
Using the MHOLE equation, I got an answer of 2.17g.
MHOLE= (45g*0.5625π[in]^2) / (12.25π in^2 - 0.5625π in^2) = 2.17g
The mass was then used to find the moment of inertia of the hole
IHOLE=0.5(2.17g)([0.75in])^2 = 0.610g in^2
Subtracting these two values
275.63 g in^2 - 0.610 g in^2= 275.02 g in^2
Is this correct? My only concern was with finding the mass of the hole, and whether it should have been added into the mass of the disc when finding IDISC or not. Any feedback is appreciated.
Thank you
Last edited: