Moment of Inertia of a Ring about an axis inclined at $\frac {\pi}{4}$ radians with normal to plane of ring [closed]

I have a thin Ring of mass $M$ and radius $R$, I have to find it's moment of Inertia about an axis passing through it's centre and at an angle of $\frac{\pi}{4}$ radians with the normal to the plane of the ring.

My approach:

I am trying to use perpendicular axes theorem. Suppose I place three mutually perpendicular axes on the centre so that one is the diameter, another one is perpendicular to it (also diameter), but on the plane of the ring, while the third is parallel to the normal, now I rotate the axes such that the one of them remains the diameter while the other two are mutually inclined at $\frac{\pi}{4}$ to the normal.

Now, as I know the moment of Inertia about a diameter ($\frac{MR^2}{2}$) So the required moment of Inertia (say $I$) must be :

$I+I = \frac{MR^2}{2}$ from perpendicular axes theorem, So, $I = \frac{Mr^2}{4}$.

Is my way of thinking correct in this case?

closed as off-topic by ACuriousMind♦Sep 2 '18 at 8:42

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I don't think it works that way unfortunately.

Let me propose a more general approach: let $\hat{x}$, $\hat{y}$, $\hat{z}$ be the normal axes to the ring (two diameters and normal respectively), now let $\hat{x}'$, $\hat{y}'$, $\hat{z}'$ be the rotated axes. We want to compute $I_{z'}$. Now write down the change of co-ordinates matrix, which is simply a rotation about the $\hat{y}$ axis:

$$\Lambda = \left[ \begin{matrix} \cos\theta && 0 && -\sin\theta \\ 0 && 1 && 0 \\ \sin\theta && 0 && \cos\theta \end{matrix} \right]$$

This matrix is such that $\vec{x}' = \Lambda \vec{x}$

Now we can easily build the inertial tensor for in the normal co-ordinates because it is diagonal:

$$I = \left[ \begin{matrix} \frac{1}{2}MR^2 && 0 && 0 \\ 0 && \frac{1}{2}MR^2 && 0 \\ 0 && 0 && MR^2 \end{matrix} \right]$$

Now since I is a tensor, it transforms as a tensor, so in the new co-ordinates (the "prime" ones) it is given by $I'_{ij} = \Lambda_{ik}\Lambda_{jl} I_{kl} \rightarrow I' = \Lambda I \Lambda^T$. A simple calculation shows:

$$I' = \left[ \begin{matrix} MR^2(\frac{\cos^2{\theta}}{2}+\sin^2{\theta}) && 0 && \frac{MR^2}{2}\cos{\theta}\sin{\theta} \\ 0 && \frac{MR^2}{2} && 0 \\ \frac{MR^2}{2}\cos{\theta}\sin{\theta} && 0 && MR^2(\frac{\sin^2{\theta}}{2}+\cos^2{\theta}) \end{matrix} \right]$$

So as you can see, when $\theta=\pi/4$ you have $I_{x'x'} = I_{z'z'} = \frac{3}{4}MR^2$.

Simple rule The trace of the inertial tensor is invariant under change of co-ordinates. In normal co-ordinates it is $2MR^2$ (just the sum of the three diagonal components). In our rotated co-ordinates, since when $\theta = \pi/4$ symmetries suggest $I_{x'x'}=I_{z'z'}$ and $I_{y'y'}=I_{yy}=MR^2/2$ (the $y$ axis is the rotation axis and so it doesn't change), you can impose trace invariance and get the result.

• Thanks, but isn't there a simple way using symmetry or some theorem? – drake01 Oct 11 '17 at 1:01
• I can't think to any simpler way than what I've called "Simple rule", namely computing the trace ($2MR^2$) and then use trace invariance and symmetry to write $I+I+MR^2/2 = 2MR^2$. – Matteo Oct 11 '17 at 14:06
• Thanks, It's my fault that I don't understand it as I am in high school. BTW can you tell me why my approach doesn't work, it seems logical though. – drake01 Oct 11 '17 at 14:54
• Sorry, my approach was a bit too hard for high-school! Anyway, if you check wikipedia, the "perpendicular axes theorem" can be applied if the object lies in the plane defined by two of the three axes. In this case, you can use it in $\hat{x},\hat{y},\hat{z}$ cohordinates, but not in the prime ones. – Matteo Oct 11 '17 at 17:36