Q. 49 Moment of Inertia The moment of ... [FREE SOLUTION]

Chapter 7: Q. 49 (page 501)

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
$\begin{aligned} I_{u} = I_{x} \cos^{2} 鈦� 胃 + I_{y} \sin^{2} 鈦� 胃鈭� 2 I_{x y} \sin 鈦� 胃 \cos 鈦� 胃 \\ I_{v} = I_{x} \sin^{2} 鈦� 胃 + I_{y} \cos^{2} 鈦� 胃 + 2 I_{x y} \sin 鈦� 胃 \cos 鈦� 胃 \end{aligned}$
Use Product-to-Sum Formulas to show that
$I_{u} = \frac{I_{x} + I_{y}}{2} + \frac{I_{x} 鈭� I_{y}}{2} \cos 鈦� (2 胃) 鈭� I_{x y} \sin 鈦� (2 胃)$
and
$I_{v} = \frac{I_{x} + I_{y}}{2} - \frac{I_{x} 鈭� I_{y}}{2} \cos 鈦� (2 胃) + I_{x y} \sin 鈦� (2 胃)$

Short Answer

Expert verified

$I_{u} = \frac{I_{x} + I_{y}}{2} + \frac{I_{x} 鈭� I_{y}}{2} \cos 鈦� (2 胃) 鈭� I_{x y} \sin 鈦� (2 胃)$ is true.

$I_{v} = \frac{I_{x} + I_{y}}{2} 鈭� \frac{I_{x} 鈭� I_{y}}{2} \cos 鈦� (2 胃) + I_{x y} \sin 鈦� (2 胃)$ 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" $\sin a \cos b = \frac{1}{2} [\sin (a + b) + \sin (a - b)]$

role="math" localid="1646559354355" $\begin{aligned} I_{u} = I_{x} \cos^{2} 鈦� 胃 + I_{y} \sin^{2} 鈦� 胃鈭� 2 I_{x y} \sin 鈦� 胃 \cos 鈦� 胃 \\ = I_{x} \cos^{2} 鈦� 胃 + I_{y} \sin^{2} 鈦� 胃鈭� 2 I_{x y} [\frac{1}{2} [\sin 鈦� (胃 + 胃) + \sin 鈦� (胃鈭� 胃)]] \\ = I_{x} \cos^{2} 鈦� 胃 + I_{y} \sin^{2} 鈦� 胃鈭� 2 I_{x y} [\frac{1}{2} [\sin 鈦� (2 胃) + \sin 鈦� (0)]] \\ = I_{x} \cos^{2} 鈦� 胃 + I_{y} \sin^{2} 鈦� 胃鈭� I_{x y} [\sin 鈦� (2 胃)] \\ = I_{x} \cos^{2} 鈦� 胃 + I_{y} \sin^{2} 鈦� 胃鈭� I_{x y} \sin 鈦� (2 胃) \end{aligned}$

Step 2. Rewrite the resulting Iu equation

Rewrite the resulting $I_{u}$ equation by applying the trigonometric identity.

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

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

$\begin{aligned} I_{u} = I_{x} \cos^{2} 鈦� 胃 + I_{y} \sin^{2} 鈦� 胃鈭� I_{x y} \sin 鈦� (2 胃) \\ = I_{x} [\frac{1 + \cos 鈦� (2 胃)}{2}] + I_{y} [\frac{1 鈭� \cos 鈦� (2 胃)}{2}] 鈭� I_{x y} \sin 鈦� (2 胃) \\ = \frac{I_{x} + I_{x} \cos 鈦� (2 胃)}{2} + \frac{I_{y} 鈭� I_{y} \cos 鈦� (2 胃)}{2} 鈭� I_{x y} \sin 鈦� (2 胃) \\ = \frac{I_{x}}{2} + \frac{I_{x} \cos 鈦� (2 胃)}{2} + \frac{I_{y}}{2} 鈭� \frac{I_{y} \cos 鈦� (2 胃)}{2} 鈭� I_{x y} \sin 鈦� (2 胃) \\ = \frac{I_{x} + I_{y}}{2} + \frac{I_{x} \cos 鈦� (2 胃)}{2} 鈭� \frac{I_{y} \cos 鈦� (2 胃)}{2} 鈭� I_{x y} \sin 鈦� (2 胃) \\ = \frac{I_{x} + I_{y}}{2} + \cos 鈦� (2 胃) [\frac{I_{x}}{2} 鈭� \frac{I_{y}}{2}] 鈭� I_{x y} \sin 鈦� (2 胃) \\ = \frac{I_{x} + I_{y}}{2} + \frac{I_{x} 鈭� I_{y}}{2} \cos 鈦� (2 胃) 鈭� I_{x y} \sin 鈦� (2 胃) \end{aligned}$

This verifies that the equationlocalid="1646560409930" $I_{u} = \frac{I_{x} + I_{y}}{2} + \frac{I_{x} 鈭� I_{y}}{2} \cos 鈦� (2 胃) 鈭� I_{x y} \sin 鈦� (2 胃)$ is true.

Step 3. Rewrite the given equation for Iv

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

$\begin{aligned} I_{v} = I_{x} \sin^{2} 鈦� 胃 + I_{y} \cos^{2} 鈦� 胃 + 2 I_{x y} \sin 鈦� 胃 \cos 鈦� 胃 \\ = I_{x} \sin^{2} 鈦� 胃 + I_{y} \cos^{2} 鈦� 胃 + 2 I_{x y} [\frac{1}{2} [\sin 鈦� (胃 + 胃) + \sin 鈦� (胃鈭� 胃)]] \\ = I_{x} \sin^{2} 鈦� 胃 + I_{y} \cos^{2} 鈦� 胃 + 2 I_{x y} [\frac{1}{2} [\sin 鈦� (2 胃) + \sin 鈦� (0)]] \\ = I_{x} \sin^{2} 鈦� 胃 + I_{y} \cos^{2} 鈦� 胃 + I_{x y} [\sin 鈦� (2 胃)] \\ = I_{x} \sin^{2} 鈦� 胃 + I_{y} \cos^{2} 鈦� 胃 + I_{x y} \sin 鈦� (2 胃) \end{aligned}$

Step 4. Rewrite the resulting Iv equation

Rewrite the resulting $I_{v}$ equation by applying the trigonometric identity.

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

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

$\begin{aligned} I_{v} = I_{x} \sin^{2} 鈦� 胃 + I_{y} \cos^{2} 鈦� 胃 + I_{x y} \sin 鈦� (2 胃) \\ = I_{x} [\frac{1 鈭� \cos 鈦� (2 胃)}{2}] + I_{y} [\frac{1 + \cos 鈦� (2 胃)}{2}] + I_{x y} \sin 鈦� (2 胃) \\ = \frac{I_{x} 鈭� I_{x} \cos 鈦� (2 胃)}{2} + \frac{I_{y} + I_{y} \cos 鈦� (2 胃)}{2} + I_{x y} \sin 鈦� (2 胃) \\ = \frac{I_{x}}{2} 鈭� \frac{I_{x} \cos 鈦� (2 胃)}{2} + \frac{I_{y}}{2} + \frac{I_{y} \cos 鈦� (2 胃)}{2} + I_{x y} \sin 鈦� (2 胃) \\ = \frac{I_{x} + I_{y}}{2} 鈭� \frac{I_{x} \cos 鈦� (2 胃)}{2} + \frac{I_{y} \cos 鈦� (2 胃)}{2} + I_{x y} \sin 鈦� (2 胃) \\ = \frac{I_{x} + I_{y}}{2} + \cos 鈦� (2 胃) [鈭� \frac{I_{x}}{2} + \frac{I_{y}}{2}] + I_{x y} \sin 鈦� (2 胃) \\ = \frac{I_{x} + I_{y}}{2} + \frac{鈭� I_{x} + I_{y}}{2} \cos 鈦� (2 胃) + I_{x y} \sin 鈦� (2 胃) \\ = \frac{I_{x} + I_{y}}{2} 鈭� \frac{I_{x} 鈭� I_{y}}{2} \cos 鈦� (2 胃) + I_{x y} \sin 鈦� (2 胃) \end{aligned}$

localid="1646560482416" $I_{v} = \frac{I_{x} + I_{y}}{2} 鈭� \frac{I_{x} 鈭� I_{y}}{2} \cos 鈦� (2 胃) + I_{x y} \sin 鈦� (2 胃)$ is TRUE.

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Short Answer

Step by step solution

Step 1. Write the given equation for Iu

Step 2. Rewrite the resulting Iu equation

Step 3. Rewrite the given equation for Iv

Step 4. Rewrite the resulting Iv equation

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