moment of inertia - 6

[logistic_guy]

Find the moment of inertia of a thin uniform hoop of radius $\displaystyle R$, mass $\displaystyle M$, and width $\displaystyle w$ when the rotation is through its central diameter.

 

[khansaheb]

Find the moment of inertia of a thin uniform hoop of radius $\displaystyle R$, mass $\displaystyle M$, and width $\displaystyle w$ when the rotation is through its central diameter.

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[logistic_guy]

This one is tough. It took me more than one hour to understand the geometry of the problem!



Let us assume that we placed this thin hoop on the plane so that its central axis is in the direction of the $\displaystyle z$-axis. Now this problem wants us to rotate it about the central diameter that is about the $\displaystyle x$-axis or the $\displaystyle y$-axis. Let us assume that we will rotate about the $\displaystyle x$-axis, then any perpendicular distance from the mass to the axis of rotation (which is the $\displaystyle x$-axis) in this case will be in two directions. One time through the $\displaystyle y$ direction and one time through $\displaystyle z$ direction (One case the two ends that touches the $\displaystyle x$-axis).

In the case of the $\displaystyle y$ direction, any perpendicular distance to the axis of rotation is $\displaystyle y = R_0\sin\theta$. And we can use it to integrate like this $\displaystyle \int R^2 \ dm = \int (R_0\sin\theta)^2 \ dm$ if we want using $\displaystyle R_0$ as the radius of the hoop. But we already know that when an object is rotating about its central diameter, we can use the Perpendicular-axis Theorem combined with the previous result we have done in the last post. There we got the moment of inertia about the central axis of a thin hoop which was:

$\displaystyle I_z = MR^2$

Then, by the theorem we get:

$\displaystyle I_{z_0} = I_{x_0} + I_{y_0} = I_{x_0} + I_{x_0} = 2I_{x_0} = MR^2$

Which gives:

$\displaystyle I_{x_0} = \frac{1}{2}MR^2$

$\displaystyle \textcolor{red}{\text{Note}}$: This results is for a thin hoop without a width $\displaystyle w$! Therefore, with the present of the width $\displaystyle w$ in the hoop, we have to combine this result with the next case.

In the case of the $\displaystyle z$ direction, there is a width $\displaystyle w$ in the hoop which suggests integrating from $\displaystyle z = -\frac{w}{2}$ to $\displaystyle z = \frac{w}{2}$.

Let us call this moment of inertia $\displaystyle I_{\text{ends}}$ (They are in fact not just two ends but about the whole hoop. I am just giving a visualization about the two ends that touches the $\displaystyle x$-axis directly.) This picture or visualization can be seen when the thin hoop is fixed on the plane.

$\displaystyle I_{\text{ends}} = \int R^2 \ dm = \int z^2 \sigma \ dA = \int_{-\omega/2}^{\omega/2} z^2 \sigma 2\pi R \ dz = 2\int_{0}^{\omega/2} z^2 \sigma 2\pi R \ dz$

Solving this gives:

$\displaystyle I_{\text{ends}} = 4\sigma \pi R \frac{z^3}{3}\bigg|_{0}^{\frac{\omega}{2}} = 4\sigma \pi R \frac{w^3}{(8)3} = \sigma \pi R \frac{w^3}{6}$

We know that $\displaystyle \sigma = \frac{M}{A}$, then

$\displaystyle I_{\text{ends}} = \frac{M}{A} \pi R \frac{w^3}{6} = \frac{M}{2\pi R \omega} \pi R \frac{w^3}{6} = M\frac{w^2}{12}$

Finally, we have the moment of inertial about the central diameter of a thin hoop with width $\displaystyle w$ which is:

$\displaystyle I_x = I_{x_0} + I_{\text{ends}} = \textcolor{blue}{\frac{1}{2}MR^2 + \frac{1}{12}Mw^2}$