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Chris Mueller
Chris Mueller
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Since you actually asked for the moment about the $x$ axis. Calculating the moment of inertia about the $x$ axis is a fair deal more complicated than calculating it about the $z$ axis as in my other answer.

To start with, we will recognize that the symmetry about the $x$ axis lets us only work on the top half and then multiply by a factor of 2 in the end. Now, only looking at the top half we can break the piece up into two sections: $I_1$ is on the left and is a triangle and $I_2$ is on the right and is a right triangle with a circular segmenthypotenuse. Since this is clearly a homework problem, I'm going to skip the algebra steps and just show you the core parts of the problem (i.e. I'm going to use Mathematica to do the brute force algebra and integration).

enter image description here

###$I_1$ Calculation

The moment of inertia is given by $$ I_1=\rho\iint y^2\ dydx, $$ where $\rho$ is the mass density per unit area, which looks simple enough. The difficulty is just in getting the correct limits of the double integral. For a a given position along the $x$ axis, the limits of $y$ range from $0$ to $x\tan(\alpha/2)$. And we will integrate $x$ from 0 to $r_0\cos(\alpha/2)$. This gives $$ \begin{align} I_1&=\rho\int_0^{r_0\cos(\alpha/2)}dx\int_0^{x\tan(\alpha/2)}dy\ y^2\\ &=\rho\int_0^{r_0\cos(\alpha/2)}dx\ x^3\tan^3(\alpha/2)\\ &=\rho\frac{r_0^4}{12}\sin^3(\alpha/2)\cos(\alpha/2) \end{align} $$ This simplifies to $\frac{\rho}{12} bh^3$ which is the well known value for the moment of inertia of a triangle.

$I_2$ Calculation

This one goes the same way as the last one, but the limits and the integration are more difficult. This time $x$ will vary from $r_0\cos(\alpha/2)$ to $r_0$. Over that range $y$ will vary from 0 to $\sqrt{r_0^2-x^2}$. Putting this into the integral gives $$ \begin{align} I_2&=\rho\int_{r_0\cos(\alpha/2)}^{r_0}dx\int_0^{\sqrt{r_0^2-x^2}}dy\ y^2\\ &=\frac{\rho}{3}\int_{r_0\cos(\alpha/2)}^{r_0}dx\ (r_0^2-x^2)^{3/2}\\ &=\frac{\rho}{96}r_0^4\left(6\alpha-8\sin(\alpha)+\sin(2\alpha)\right) \end{align} $$ That last integral was quite tricky, and I ended up just plugging it into Mathematica. I'm sure it is possible to find it in the standard integral tables though.

All together now

Finally, putting everything together and working through some trig identities simplifies the whole thing to $$ 2(I_1+I_2)=\frac{\rho}{8}r_0^4(\alpha-\sin\alpha) $$ Which is again what we expect from the standard table for a circular segment.

Since you actually asked for the moment about the $x$ axis. Calculating the moment of inertia about the $x$ axis is a fair deal more complicated than calculating it about the $z$ axis as in my other answer.

To start with, we will recognize that the symmetry about the $x$ axis lets us only work on the top half and then multiply by a factor of 2 in the end. Now, only looking at the top half we can break the piece up into two sections: $I_1$ is on the left and is a triangle and $I_2$ is on the right and is a circular segment. Since this is clearly a homework problem, I'm going to skip the algebra steps and just show you the core parts of the problem (i.e. I'm going to use Mathematica to do the brute force algebra and integration).

###$I_1$ Calculation

The moment of inertia is given by $$ I_1=\rho\iint y^2\ dydx, $$ where $\rho$ is the mass density per unit area, which looks simple enough. The difficulty is just in getting the correct limits of the double integral. For a a given position along the $x$ axis, the limits of $y$ range from $0$ to $x\tan(\alpha/2)$. And we will integrate $x$ from 0 to $r_0\cos(\alpha/2)$. This gives $$ \begin{align} I_1&=\rho\int_0^{r_0\cos(\alpha/2)}dx\int_0^{x\tan(\alpha/2)}dy\ y^2\\ &=\rho\int_0^{r_0\cos(\alpha/2)}dx\ x^3\tan^3(\alpha/2)\\ &=\rho\frac{r_0^4}{12}\sin^3(\alpha/2)\cos(\alpha/2) \end{align} $$ This simplifies to $\frac{\rho}{12} bh^3$ which is the well known value for the moment of inertia of a triangle.

$I_2$ Calculation

This one goes the same way as the last one, but the limits and the integration are more difficult. This time $x$ will vary from $r_0\cos(\alpha/2)$ to $r_0$. Over that range $y$ will vary from 0 to $\sqrt{r_0^2-x^2}$. Putting this into the integral gives $$ \begin{align} I_2&=\rho\int_{r_0\cos(\alpha/2)}^{r_0}dx\int_0^{\sqrt{r_0^2-x^2}}dy\ y^2\\ &=\frac{\rho}{3}\int_{r_0\cos(\alpha/2)}^{r_0}dx\ (r_0^2-x^2)^{3/2}\\ &=\frac{\rho}{96}r_0^4\left(6\alpha-8\sin(\alpha)+\sin(2\alpha)\right) \end{align} $$ That last integral was quite tricky, and I ended up just plugging it into Mathematica. I'm sure it is possible to find it in the standard integral tables though.

All together now

Finally, putting everything together and working through some trig identities simplifies the whole thing to $$ 2(I_1+I_2)=\frac{\rho}{8}r_0^4(\alpha-\sin\alpha) $$ Which is again what we expect from the standard table.

Since you actually asked for the moment about the $x$ axis. Calculating the moment of inertia about the $x$ axis is a fair deal more complicated than calculating it about the $z$ axis as in my other answer.

To start with, we will recognize that the symmetry about the $x$ axis lets us only work on the top half and then multiply by a factor of 2 in the end. Now, only looking at the top half we can break the piece up into two sections: $I_1$ is on the left and is a triangle and $I_2$ is on the right and is a right triangle with a circular hypotenuse. Since this is clearly a homework problem, I'm going to skip the algebra steps and just show you the core parts of the problem (i.e. I'm going to use Mathematica to do the brute force algebra and integration).

enter image description here

###$I_1$ Calculation

The moment of inertia is given by $$ I_1=\rho\iint y^2\ dydx, $$ where $\rho$ is the mass density per unit area, which looks simple enough. The difficulty is just in getting the correct limits of the double integral. For a a given position along the $x$ axis, the limits of $y$ range from $0$ to $x\tan(\alpha/2)$. And we will integrate $x$ from 0 to $r_0\cos(\alpha/2)$. This gives $$ \begin{align} I_1&=\rho\int_0^{r_0\cos(\alpha/2)}dx\int_0^{x\tan(\alpha/2)}dy\ y^2\\ &=\rho\int_0^{r_0\cos(\alpha/2)}dx\ x^3\tan^3(\alpha/2)\\ &=\rho\frac{r_0^4}{12}\sin^3(\alpha/2)\cos(\alpha/2) \end{align} $$ This simplifies to $\frac{\rho}{12} bh^3$ which is the well known value for the moment of inertia of a triangle.

$I_2$ Calculation

This one goes the same way as the last one, but the limits and the integration are more difficult. This time $x$ will vary from $r_0\cos(\alpha/2)$ to $r_0$. Over that range $y$ will vary from 0 to $\sqrt{r_0^2-x^2}$. Putting this into the integral gives $$ \begin{align} I_2&=\rho\int_{r_0\cos(\alpha/2)}^{r_0}dx\int_0^{\sqrt{r_0^2-x^2}}dy\ y^2\\ &=\frac{\rho}{3}\int_{r_0\cos(\alpha/2)}^{r_0}dx\ (r_0^2-x^2)^{3/2}\\ &=\frac{\rho}{96}r_0^4\left(6\alpha-8\sin(\alpha)+\sin(2\alpha)\right) \end{align} $$ That last integral was quite tricky, and I ended up just plugging it into Mathematica. I'm sure it is possible to find it in the standard integral tables though.

All together now

Finally, putting everything together and working through some trig identities simplifies the whole thing to $$ 2(I_1+I_2)=\frac{\rho}{8}r_0^4(\alpha-\sin\alpha) $$ Which is again what we expect from the standard table for a circular segment.

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Chris Mueller
Chris Mueller
  • 6.5k
  • 2
  • 28
  • 56

Since you actually asked for the moment about the $x$ axis. Calculating the moment of inertia about the $x$ axis is a fair deal more complicated than calculating it about the $z$ axis as in my other answer.

To start with, we will recognize that the symmetry about the $x$ axis lets us only work on the top half and then multiply by a factor of 2 in the end. Now, only looking at the top half we can break the piece up into two sections: $I_1$ is on the left and is a triangle and $I_2$ is on the right and is a portion of a circlecircular segment. Since this is clearly a homework problem, I'm going to skip the algebra steps and just show you the core parts of the problem (i.e. I'm going to use Mathematica to do the brute force algebra and integration).

###$I_1$ Calculation

The moment of inertia is given by $$ I_1=\rho\iint y^2\ dydx, $$ where $\rho$ is the mass density per unit area, which looks simple enough. The difficulty is just in getting the correct limits of the double integral. For a a given position along the $x$ axis, the limits of $y$ range from $0$ to $x\tan(\alpha/2)$. And we will integrate $x$ from 0 to $r_0\cos(\alpha/2)$. This gives $$ \begin{align} I_1&=\rho\int_0^{r_0\cos(\alpha/2)}dx\int_0^{x\tan(\alpha/2)}dy\ y^2\\ &=\rho\int_0^{r_0\cos(\alpha/2)}dx\ x^3\tan^3(\alpha/2)\\ &=\rho\frac{r_0^4}{12}\sin^3(\alpha/2)\cos(\alpha/2) \end{align} $$ This simplifies to $\rho r_0^4/12 bh^3$$\frac{\rho}{12} bh^3$ which is the well known value for the moment of inertia of a triangle.

$I_2$ Calculation

This one goes the same way as the last one, but the limits and the integration are more difficult. This time $x$ will vary from $r_0\cos(\alpha/2)$ to $r_0$. Over that range $y$ will vary from 0 to $\sqrt{r_0^2-x^2}$. Putting this into the integral gives $$ \begin{align} I_2&=\rho\int_{r_0\cos(\alpha/2)}^{r_0}dx\int_0^{\sqrt{r_0^2-x^2}}dy\ y^2\\ &=\frac{\rho}{3}\int_{r_0\cos(\alpha/2)}^{r_0}dx\ (r_0^2-x^2)^{3/2}\\ &=\frac{\rho}{96}r_0^4\left(6\alpha-8\sin(\alpha)+\sin(2\alpha)\right) \end{align} $$ That last integral was quite tricky, and I ended up just plugging it into Mathematica. I'm sure it is possible to find it in the standard integral tables though.

All together now

Finally, putting everything together and working through some trig identities simplifies the whole thing to $$ 2(I_1+I_2)=\frac{\rho}{8}r_0^4(\alpha-\sin\alpha) $$ Which is again what we expect from the standard table.

Since you actually asked for the moment about the $x$ axis. Calculating the moment of inertia about the $x$ axis is a fair deal more complicated than calculating it about the $z$ axis as in my other answer.

To start with, we will recognize that the symmetry about the $x$ axis lets us only work on the top half and then multiply by a factor of 2 in the end. Now, only looking at the top half we can break the piece up into two sections: $I_1$ is on the left and is a triangle and $I_2$ is on the right and is a portion of a circle. Since this is clearly a homework problem, I'm going to skip the algebra steps and just show you the core parts of the problem (i.e. I'm going to use Mathematica to do the brute force algebra and integration).

###$I_1$ Calculation

The moment of inertia is given by $$ I_1=\rho\iint y^2\ dydx, $$ where $\rho$ is the mass density per unit area, which looks simple enough. The difficulty is just in getting the correct limits of the double integral. For a a given position along the $x$ axis, the limits of $y$ range from $0$ to $x\tan(\alpha/2)$. And we will integrate $x$ from 0 to $r_0\cos(\alpha/2)$. This gives $$ \begin{align} I_1&=\rho\int_0^{r_0\cos(\alpha/2)}dx\int_0^{x\tan(\alpha/2)}dy\ y^2\\ &=\rho\int_0^{r_0\cos(\alpha/2)}dx\ x^3\tan^3(\alpha/2)\\ &=\rho\frac{r_0^4}{12}\sin^3(\alpha/2)\cos(\alpha/2) \end{align} $$ This simplifies to $\rho r_0^4/12 bh^3$ which is the well known value for the moment of inertia of a triangle.

$I_2$ Calculation

This one goes the same way as the last one, but the limits and the integration are more difficult. This time $x$ will vary from $r_0\cos(\alpha/2)$ to $r_0$. Over that range $y$ will vary from 0 to $\sqrt{r_0^2-x^2}$. Putting this into the integral gives $$ \begin{align} I_2&=\rho\int_{r_0\cos(\alpha/2)}^{r_0}dx\int_0^{\sqrt{r_0^2-x^2}}dy\ y^2\\ &=\frac{\rho}{3}\int_{r_0\cos(\alpha/2)}^{r_0}dx\ (r_0^2-x^2)^{3/2}\\ &=\frac{\rho}{96}r_0^4\left(6\alpha-8\sin(\alpha)+\sin(2\alpha)\right) \end{align} $$ That last integral was quite tricky, and I ended up just plugging it into Mathematica. I'm sure it is possible to find it in the standard integral tables though.

All together now

Finally, putting everything together and working through some trig identities simplifies the whole thing to $$ 2(I_1+I_2)=\frac{\rho}{8}r_0^4(\alpha-\sin\alpha) $$ Which is again what we expect from the standard table.

Since you actually asked for the moment about the $x$ axis. Calculating the moment of inertia about the $x$ axis is a fair deal more complicated than calculating it about the $z$ axis as in my other answer.

To start with, we will recognize that the symmetry about the $x$ axis lets us only work on the top half and then multiply by a factor of 2 in the end. Now, only looking at the top half we can break the piece up into two sections: $I_1$ is on the left and is a triangle and $I_2$ is on the right and is a circular segment. Since this is clearly a homework problem, I'm going to skip the algebra steps and just show you the core parts of the problem (i.e. I'm going to use Mathematica to do the brute force algebra and integration).

###$I_1$ Calculation

The moment of inertia is given by $$ I_1=\rho\iint y^2\ dydx, $$ where $\rho$ is the mass density per unit area, which looks simple enough. The difficulty is just in getting the correct limits of the double integral. For a a given position along the $x$ axis, the limits of $y$ range from $0$ to $x\tan(\alpha/2)$. And we will integrate $x$ from 0 to $r_0\cos(\alpha/2)$. This gives $$ \begin{align} I_1&=\rho\int_0^{r_0\cos(\alpha/2)}dx\int_0^{x\tan(\alpha/2)}dy\ y^2\\ &=\rho\int_0^{r_0\cos(\alpha/2)}dx\ x^3\tan^3(\alpha/2)\\ &=\rho\frac{r_0^4}{12}\sin^3(\alpha/2)\cos(\alpha/2) \end{align} $$ This simplifies to $\frac{\rho}{12} bh^3$ which is the well known value for the moment of inertia of a triangle.

$I_2$ Calculation

This one goes the same way as the last one, but the limits and the integration are more difficult. This time $x$ will vary from $r_0\cos(\alpha/2)$ to $r_0$. Over that range $y$ will vary from 0 to $\sqrt{r_0^2-x^2}$. Putting this into the integral gives $$ \begin{align} I_2&=\rho\int_{r_0\cos(\alpha/2)}^{r_0}dx\int_0^{\sqrt{r_0^2-x^2}}dy\ y^2\\ &=\frac{\rho}{3}\int_{r_0\cos(\alpha/2)}^{r_0}dx\ (r_0^2-x^2)^{3/2}\\ &=\frac{\rho}{96}r_0^4\left(6\alpha-8\sin(\alpha)+\sin(2\alpha)\right) \end{align} $$ That last integral was quite tricky, and I ended up just plugging it into Mathematica. I'm sure it is possible to find it in the standard integral tables though.

All together now

Finally, putting everything together and working through some trig identities simplifies the whole thing to $$ 2(I_1+I_2)=\frac{\rho}{8}r_0^4(\alpha-\sin\alpha) $$ Which is again what we expect from the standard table.

Source Link
Chris Mueller
Chris Mueller
  • 6.5k
  • 2
  • 28
  • 56

Since you actually asked for the moment about the $x$ axis. Calculating the moment of inertia about the $x$ axis is a fair deal more complicated than calculating it about the $z$ axis as in my other answer.

To start with, we will recognize that the symmetry about the $x$ axis lets us only work on the top half and then multiply by a factor of 2 in the end. Now, only looking at the top half we can break the piece up into two sections: $I_1$ is on the left and is a triangle and $I_2$ is on the right and is a portion of a circle. Since this is clearly a homework problem, I'm going to skip the algebra steps and just show you the core parts of the problem (i.e. I'm going to use Mathematica to do the brute force algebra and integration).

###$I_1$ Calculation

The moment of inertia is given by $$ I_1=\rho\iint y^2\ dydx, $$ where $\rho$ is the mass density per unit area, which looks simple enough. The difficulty is just in getting the correct limits of the double integral. For a a given position along the $x$ axis, the limits of $y$ range from $0$ to $x\tan(\alpha/2)$. And we will integrate $x$ from 0 to $r_0\cos(\alpha/2)$. This gives $$ \begin{align} I_1&=\rho\int_0^{r_0\cos(\alpha/2)}dx\int_0^{x\tan(\alpha/2)}dy\ y^2\\ &=\rho\int_0^{r_0\cos(\alpha/2)}dx\ x^3\tan^3(\alpha/2)\\ &=\rho\frac{r_0^4}{12}\sin^3(\alpha/2)\cos(\alpha/2) \end{align} $$ This simplifies to $\rho r_0^4/12 bh^3$ which is the well known value for the moment of inertia of a triangle.

$I_2$ Calculation

This one goes the same way as the last one, but the limits and the integration are more difficult. This time $x$ will vary from $r_0\cos(\alpha/2)$ to $r_0$. Over that range $y$ will vary from 0 to $\sqrt{r_0^2-x^2}$. Putting this into the integral gives $$ \begin{align} I_2&=\rho\int_{r_0\cos(\alpha/2)}^{r_0}dx\int_0^{\sqrt{r_0^2-x^2}}dy\ y^2\\ &=\frac{\rho}{3}\int_{r_0\cos(\alpha/2)}^{r_0}dx\ (r_0^2-x^2)^{3/2}\\ &=\frac{\rho}{96}r_0^4\left(6\alpha-8\sin(\alpha)+\sin(2\alpha)\right) \end{align} $$ That last integral was quite tricky, and I ended up just plugging it into Mathematica. I'm sure it is possible to find it in the standard integral tables though.

All together now

Finally, putting everything together and working through some trig identities simplifies the whole thing to $$ 2(I_1+I_2)=\frac{\rho}{8}r_0^4(\alpha-\sin\alpha) $$ Which is again what we expect from the standard table.