Computing the polar moment of inertia (calculus)

In summary, the common approach to finding the rotational inertia at point O involves using polar coordinates and the formula ##dI_O=r^2\sigma\, dA##, with the resulting value for the triangle being ##\sqrt 3 /2##. However, when using rectangular coordinates, the resulting answer is different. After showing the steps, it is revealed that the mistake lies in using the incorrect limits for integration, with the correct answer being ##\frac{\sqrt 3}2##.
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Leo Liu
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Homework Statement
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Relevant Equations
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Question:
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Diagram:
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So the common approach to this problem is using polar coordinates.
The definition of infinitesimal rotational inertia at O is ##dI_O=r^2\sigma\, dA##. Therefore the r. inertia of the triangle is
$$I_O=\int_{0}^{\pi/3}\int_{0}^{\sec\theta}r^2r\,drd\theta$$
whose value is ##=\sqrt 3 /2##.

But when I used rectangular coordinates to solve this problem, I got a different answer. The steps are shown below.
$$\xcancel{\begin{aligned}
I_O&=\int_{0}^{1}\int_{0}^{2x}x^2+y^2\,dydx\\
&=\int_0^1x^2y+\frac{y^3} 3\Bigg|_0^{2x}dx\\
&=\int_0^12x^3+\frac 8 3x^3\,dx\\
&=\frac{x^4} 2+\frac 2 3x^4\Bigg|_0^1=\frac 7 6
\end{aligned}}$$
Can someone please tell me where my mistakes are? Thanks!

Edit:
$$\begin{aligned}
I_O&=\int_{0}^{1}\int_{0}^{\sqrt 3 x}x^2+y^2\,dydx\\
&=\int_0^1x^2y+\frac{y^3} 3\Bigg|_0^{\sqrt 3 x}dx\\
&=\int_0^1 \sqrt 3 x^3+\sqrt 3 x^3\,dx\\
&=\frac{2\sqrt 3}{4}=\frac{\sqrt 3}2
\end{aligned}$$
 
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Perhaps the slope of the line is not 2 but a little less ... :wink:
 
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Related to Computing the polar moment of inertia (calculus)

1. What is the polar moment of inertia?

The polar moment of inertia is a measure of an object's resistance to rotational motion around its center of mass. It takes into account the distribution of mass around the axis of rotation and is often used in engineering and physics calculations.

2. How is the polar moment of inertia calculated?

The polar moment of inertia is calculated using calculus, specifically by integrating the square of the distance from the axis of rotation to each infinitesimal element of mass in the object. The resulting integral is then multiplied by the density of the object and the limits of integration are determined by the shape and orientation of the object.

3. What is the difference between polar moment of inertia and moment of inertia?

The polar moment of inertia is a measure of an object's resistance to rotational motion around its center of mass, while the moment of inertia is a measure of an object's resistance to rotational motion around a specific axis. The polar moment of inertia is a combination of the moment of inertia around two perpendicular axes.

4. Why is the polar moment of inertia important in engineering?

The polar moment of inertia is important in engineering because it allows for the calculation of an object's torsional stiffness, which is crucial in designing structures and machines that can withstand torsional forces. It is also used in calculating the stress and strain on rotating components, such as shafts and gears.

5. Can the polar moment of inertia be negative?

No, the polar moment of inertia cannot be negative. It is always a positive value because it is calculated by squaring the distance from the axis of rotation, which results in a positive number. A negative value would indicate that the object has a negative mass, which is not physically possible.

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