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bphys348
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1. Question:
What is the moment of inertia of a thin, flat plate in the shape of a semicircle rotating about the straight side (Fig. 12.28)? The mass of the plate is M and the radius is R
The diagram is attached for reference.
Moment of Inertia: I = ∫R2dm
Perpendicular Axis Theorem: Iz = Ix + Iy
I made attempts to solve this is a couple of ways...
First attempt: Using I = ∫R2dm
I chose the y-axis to be parallel to the distance from the axis of rotation to mass elements on the disk.
I defined the distance from the rotation axis y = rsinΘ, where r is the distance of the mass element from the center.
In my integral for I, I substituted for R2, r2sin2Θ
dm = (M/A)*dA where A=(πR2)/2 and dA=drdΘ
Plugging into the integral, I get I = 2M/(πR2) ∫r2dr from o to R * ∫sin2ΘdΘ from 0 to π
Solving, I get MR/3
Now, I know there is something fundamentally wrong with process because I realize I should have some multiple of R2... Was hoping someone could help shed some light
Attempt 2
I believe the problem becomes much simpler if I use the perpendicular axis theorem but I still failed to get the correct answer this way.
Iz = Ix + Iy
Ix would mean the half disk coincides with the radial center of the disk. Ix (for a full circle) = (MR2)/2 but we have of a disk, so I believe Ix= (MR2)/4
Iy would mean the half disk rotation about the y-axis intersecting the radial center. Iy (for a full circle) = (MR2)/4 but this is a half disk to Iy = (MR2)/8
Ix + Iy = Iz, for which I get (3MR2)/8
Book says the solution is (MR2)/4 so I'm a little frustrated at this point.
Any help would be great, thanks!
What is the moment of inertia of a thin, flat plate in the shape of a semicircle rotating about the straight side (Fig. 12.28)? The mass of the plate is M and the radius is R
The diagram is attached for reference.
Homework Equations
Moment of Inertia: I = ∫R2dm
Perpendicular Axis Theorem: Iz = Ix + Iy
The Attempt at a Solution
I made attempts to solve this is a couple of ways...
First attempt: Using I = ∫R2dm
I chose the y-axis to be parallel to the distance from the axis of rotation to mass elements on the disk.
I defined the distance from the rotation axis y = rsinΘ, where r is the distance of the mass element from the center.
In my integral for I, I substituted for R2, r2sin2Θ
dm = (M/A)*dA where A=(πR2)/2 and dA=drdΘ
Plugging into the integral, I get I = 2M/(πR2) ∫r2dr from o to R * ∫sin2ΘdΘ from 0 to π
Solving, I get MR/3
Now, I know there is something fundamentally wrong with process because I realize I should have some multiple of R2... Was hoping someone could help shed some light
Attempt 2
I believe the problem becomes much simpler if I use the perpendicular axis theorem but I still failed to get the correct answer this way.
Iz = Ix + Iy
Ix would mean the half disk coincides with the radial center of the disk. Ix (for a full circle) = (MR2)/2 but we have of a disk, so I believe Ix= (MR2)/4
Iy would mean the half disk rotation about the y-axis intersecting the radial center. Iy (for a full circle) = (MR2)/4 but this is a half disk to Iy = (MR2)/8
Ix + Iy = Iz, for which I get (3MR2)/8
Book says the solution is (MR2)/4 so I'm a little frustrated at this point.
Any help would be great, thanks!