Moment of inertia of a half disk about an axis

[vbrasic]

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.

Homework Equations

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

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 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.

Homework Equations

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

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]

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 $y=r \sin \phi$.

 

[vbrasic]

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]

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.

 

[vbrasic]

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]

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.