Question

Moment of Inertia The moment of inertia I of an object is a measure of how easy it is to rotate the object about some fixed point. In engineering mechanics, it is sometimes necessary to compute moments of inertia with respect to a set of rotated axes. These moments are given by the equations

$\begin{array}{r}{I}_u=I_x{\cos }^2\theta +I_y{\sin }^2\theta -2I_{xy}\sin \theta \cos \theta \\ I_v=I_x{\sin }^2\theta +I_y{\cos }^2\theta +2I_{xy}\sin \theta \cos \theta \end{array}$

Use Product-to-Sum Formulas to show that

$I_u=\frac{I_x+I_y}{2}+\frac{I_x-I_y}{2}\cos \left(2\theta \right)-I_{xy}\sin \left(2\theta \right)$

and

$I_v=\frac{I_x+I_y}{2}-\frac{I_x-I_y}{2}\cos \left(2\theta \right)+I_{xy}\sin \left(2\theta \right)$

Short answer

$I_u=\frac{I_x+I_y}{2}+\frac{I_x-I_y}{2}\cos \left(2\theta \right)-I_{xy}\sin \left(2\theta \right)$is true.

$I_v=\frac{I_x+I_y}{2}-\frac{I_x-I_y}{2}\cos \left(2\theta \right)+I_{xy}\sin \left(2\theta \right)$ is true.

Step-by-step solution

Step 1. Write the given equation for Iu

First rewrite the given equation for $I_u$by applying the trigonometric identity

role="math" localid="1646559925103" $\sin a\cos b=\frac{1}{2}\left[\sin (a+b)+\sin (a-b)\right]$

role="math" localid="1646559354355" $\begin{array}{r}{I}_u=I_x{\cos }^2\theta +I_y{\sin }^2\theta -2I_{xy}\sin \theta \cos \theta \\ =I_x{\cos }^2\theta +I_y{\sin }^2\theta -2I_{xy}\left[\frac{1}{2}[\sin (\theta +\theta )+\sin (\theta -\theta \left)\right]\right]\\ =I_x{\cos }^2\theta +I_y{\sin }^2\theta -2I_{xy}\left[\frac{1}{2}[\sin (2\theta )+\sin (0\left)\right]\right]\\ =I_x{\cos }^2\theta +I_y{\sin }^2\theta -I_{xy}[\sin (2\theta \left)\right]\\ =I_x{\cos }^2\theta +I_y{\sin }^2\theta -I_{xy}\sin \left(2\theta \right)\end{array}$

Step 2. Rewrite the resulting Iu equation

Rewrite the resulting $I_u$equation by applying the trigonometric identity.

localid="1646560062672" ${\cos }^2\left(x\right)=\frac{1+\cos \left(2x\right)}{2}$

localid="1646560076367" ${\sin }^2\left(x\right)=\frac{1-\cos \left(2x\right)}{2}$

$\begin{array}{r}{I}_u=I_x{\cos }^2\theta +I_y{\sin }^2\theta -I_{xy}\sin \left(2\theta \right)\\ =I_x\left[\frac{1+\cos \left(2\theta \right)}{2}\right]+I_y\left[\frac{1-\cos \left(2\theta \right)}{2}\right]-I_{xy}\sin \left(2\theta \right)\\ =\frac{I_x+I_x\cos \left(2\theta \right)}{2}+\frac{I_y-I_y\cos \left(2\theta \right)}{2}-I_{xy}\sin \left(2\theta \right)\\ =\frac{I_x}{2}+\frac{I_x\cos \left(2\theta \right)}{2}+\frac{I_y}{2}-\frac{I_y\cos \left(2\theta \right)}{2}-I_{xy}\sin \left(2\theta \right)\\ =\frac{I_x+I_y}{2}+\frac{I_x\cos \left(2\theta \right)}{2}-\frac{I_y\cos \left(2\theta \right)}{2}-I_{xy}\sin \left(2\theta \right)\\ =\frac{I_x+I_y}{2}+\cos \left(2\theta \right)\left[\frac{I_x}{2}-\frac{I_y}{2}\right]-I_{xy}\sin \left(2\theta \right)\\ =\frac{I_x+I_y}{2}+\frac{I_x-I_y}{2}\cos \left(2\theta \right)-I_{xy}\sin \left(2\theta \right)\end{array}$

This verifies that the equationlocalid="1646560409930" $I_u=\frac{I_x+I_y}{2}+\frac{I_x-I_y}{2}\cos \left(2\theta \right)-I_{xy}\sin \left(2\theta \right)$is true.

Step 3. Rewrite the given equation for Iv

Apply the trigonometric identity $\sin a\cos b=\frac{1}{2}\left[\sin (a+b)+\sin (a-b)\right]$

$\begin{array}{r}{I}_v=I_x{\sin }^2\theta +I_y{\cos }^2\theta +2I_{xy}\sin \theta \cos \theta \\ =I_x{\sin }^2\theta +I_y{\cos }^2\theta +2I_{xy}\left[\frac{1}{2}[\sin (\theta +\theta )+\sin (\theta -\theta \left)\right]\right]\\ =I_x{\sin }^2\theta +I_y{\cos }^2\theta +2I_{xy}\left[\frac{1}{2}[\sin (2\theta )+\sin (0\left)\right]\right]\\ =I_x{\sin }^2\theta +I_y{\cos }^2\theta +I_{xy}[\sin (2\theta \left)\right]\\ =I_x{\sin }^2\theta +I_y{\cos }^2\theta +I_{xy}\sin \left(2\theta \right)\end{array}$

Step 4. Rewrite the resulting Iv equation

Rewrite the resulting $I_v$equation by applying the trigonometric identity.

${\cos }^2\left(x\right)=\frac{1+\cos \left(2x\right)}{2}$

${\sin }^2\left(x\right)=\frac{1-\cos \left(2x\right)}{2}$

$\begin{array}{r}{I}_v=I_x{\sin }^2\theta +I_y{\cos }^2\theta +I_{xy}\sin \left(2\theta \right)\\ =I_x\left[\frac{1-\cos \left(2\theta \right)}{2}\right]+I_y\left[\frac{1+\cos \left(2\theta \right)}{2}\right]+I_{xy}\sin \left(2\theta \right)\\ =\frac{I_x-I_x\cos \left(2\theta \right)}{2}+\frac{I_y+I_y\cos \left(2\theta \right)}{2}+I_{xy}\sin \left(2\theta \right)\\ =\frac{I_x}{2}-\frac{I_x\cos \left(2\theta \right)}{2}+\frac{I_y}{2}+\frac{I_y\cos \left(2\theta \right)}{2}+I_{xy}\sin \left(2\theta \right)\\ =\frac{I_x+I_y}{2}-\frac{I_x\cos \left(2\theta \right)}{2}+\frac{I_y\cos \left(2\theta \right)}{2}+I_{xy}\sin \left(2\theta \right)\\ =\frac{I_x+I_y}{2}+\cos \left(2\theta \right)\left[-\frac{I_x}{2}+\frac{I_y}{2}\right]+I_{xy}\sin \left(2\theta \right)\\ =\frac{I_x+I_y}{2}+\frac{-I_x+I_y}{2}\cos \left(2\theta \right)+I_{xy}\sin \left(2\theta \right)\\ =\frac{I_x+I_y}{2}-\frac{I_x-I_y}{2}\cos \left(2\theta \right)+I_{xy}\sin \left(2\theta \right)\end{array}$

localid="1646560482416" $I_v=\frac{I_x+I_y}{2}-\frac{I_x-I_y}{2}\cos \left(2\theta \right)+I_{xy}\sin \left(2\theta \right)$is TRUE.