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Moment of inertia of a half disk about an axis

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1. Homework Statement
Consider a half disk (of uniform density) with the flat end lying on the x-axis, symmetric about the y-axis (i.e. being cut into two quarters by the y-axis). Calculate the moments of inertia about each of the axes.

2. Homework Equations
$$I_{rr}=\sum_{i}m_ir_i^2$$

3. The Attempt at a Solution
I just need some making sure that I'm setting up the problem correctly.

The distance to the x-axis from any point on the disk is ##y##. Or, alternatively, ##r\sin{\phi}.## So we find that, $$I_{xx}=\int_{0}^{R}\int_{0}^{\pi}(r^2\sin^2{\phi})\rho\,rdrd\phi,$$ where ##\rho## is the mass density per unit area. However, my instructor has in his notes that, $$I_{xx}=\int_{0}^{R}\int_{0}^{\pi}(r^2\cos^2{\phi})\rho\,rdrd\phi.$$ I'm not sure why that would be, as the distance from any point to the axis of rotation (x in the case of ##I_{xx}## is ##y=r\sin{\rho}.##
 

tnich

Homework Helper
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1. Homework Statement
Consider a half disk (of uniform density) with the flat end lying on the x-axis, symmetric about the y-axis (i.e. being cut into two quarters by the y-axis). Calculate the moments of inertia about each of the axes.

2. Homework Equations
$$I_{rr}=\sum_{i}m_ir_i^2$$

3. The Attempt at a Solution
I just need some making sure that I'm setting up the problem correctly.

The distance to the x-axis from any point on the disk is ##y##. Or, alternatively, ##r\sin{\phi}.## So we find that, $$I_{xx}=\int_{0}^{R}\int_{0}^{\pi}(r^2\sin^2{\phi})\rho\,rdrd\phi,$$ where ##\rho## is the mass density per unit area. However, my instructor has in his notes that, $$I_{xx}=\int_{0}^{R}\int_{0}^{\pi}(r^2\cos^2{\phi})\rho\,rdrd\phi.$$ I'm not sure why that would be, as the distance from any point to the axis of rotation (x in the case of ##I_{xx}## is ##y=r\sin{\rho}.##
I think you have it right. Maybe the instructor wrote it down wrong.
 

kuruman

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Have you and your instructor defined angle ##\phi## the same way or is one ##90^o - \phi## of the other?
... in the case of ##I_{xx}## is ##y=r \sin \rho##.
You mean ##y=r \sin \phi##.
 
73
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Have you and your instructor defined angle ##\phi## the same way or is one ##90^o - \phi## of the other?

You mean ##y=r \sin \phi##.
Yes -- and not that I know of.
 

kuruman

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OK, then you need to ask yourself, "does it make any difference?" Is the area under the ##\cos^2(\phi)## curve from ##0## to ##\pi## different from the area under the ##\sin^2(\phi)## curve? Draw the curves and see for yourself.
 
73
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OK, then you need to ask yourself, "does it make any difference?" Is the area under the ##\cos^2(\phi)## curve from ##0## to ##\pi## different from the area under the ##\sin^2(\phi)## curve? Draw the curves and see for yourself.
Yeah, it wouldn't make a difference. The integral evaluates to the same result either way.
 

kuruman

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Yeah, it wouldn't make a difference. The integral evaluates to the same result either way.
Yes, and if you want to satisfy your curiosity, you may have to ask your instructor about the definition of ##\phi##, otherwise leave it alone.
 

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