Chapter 13: Q. 64 (page 1056)

In Exercises 61–64, let R be the rectangular solid defined by $R = \{(x, y, z) | 0 \leq a_{1} \leq x \leq a_{2}, 0 \leq b_{1} \leq y \leq b_{2}, 0 \leq c_{2} \leq z \leq c_{2}\} .$
Assuming that the density at each point in R is proportional to the distance of the point from the yz-plane, find the first moment of inertia about the x-axis and the radius of gyration about the x-axis.

Short Answer

Expert verified

The first moment of inertia about the x-axis is $I_{x} = k (\frac{(a_{2} - a_{1}) (b_{2} - b_{1}) (c_{2} - c_{1})}{3}) (b_{2}^{2} + b_{1}^{2} + b_{1} b_{2} + c_{2}^{2} + c_{1}^{2} + c_{1} c_{2})$ and the radius of gyration about the x-axis islocalid="1650373580986" $R_{x} = \sqrt{\frac{(b_{2}^{2} + b_{1}^{2} + b_{1} b_{2} + c_{2}^{2} + c_{1}^{2} + c_{1} c_{2})}{3}} .$

Step by step solution

Step 1. Given Information.

The given rectangular solid is defined by $R = \{(x, y, z) | 0 \leq a_{1} \leq x \leq a_{2}, 0 \leq b_{1} \leq y \leq b_{2}, 0 \leq c_{2} \leq z \leq c_{2}\} .$

Step 2. Find the first moment of inertia about the x-axis.

It is given that the density at each point in R is proportional to the distance of the point from the yz-plane, so $ρ (x, y, z) = k x .$

Now, the moment of the inertia about the x-axis is,

$I_{x} = ∭_{T} (y^{2} + z^{2}) ρ (x, y, z) d V I_{x} = \int_{x = a_{1}}^{a_{2}} \int_{y = b_{1}}^{b_{2}} \int_{z = c_{1}}^{c_{2}} (y^{2} + z^{2}) (k x) d x d y d z I_{x} = {k (\frac{x^{2}}{2})}_{x = a_{1}}^{a_{2}} \times {\int_{z = c_{1}}^{c_{2}} (\frac{y^{3}}{3})}_{b_{1}}^{b_{2}} + \int_{x = a_{1}}^{a_{2}} {z^{2} (y)}_{b_{1}}^{b_{2}} d z$

Integrate with respect to 'y'

$I_{x} = {\int_{x = a_{1}}^{a_{2}} (c_{2} - c_{1}) (\frac{y^{3}}{3})}_{y = b_{1}}^{b_{2}} d x + {\int_{x = a_{1}}^{a_{2}} (\frac{1}{3}) (c_{2}^{3} - c_{1}^{3}) (y)}_{y = b_{1}}^{b_{2}} d x I_{x} = {\frac{(c_{2} - c_{1}) (b_{2}^{3} - b_{1}^{3})}{3} (x) |}_{x = a_{1}}^{a_{2}} + {\frac{(c_{2}^{3} - c_{1}^{3}) (b_{2} - b_{1})}{3} (x) |}_{x = a_{1}}^{a_{2}} I_{x} = \frac{(c_{2} - c_{1}) (b_{2}^{3} - b_{1}^{3}) (a_{2} - a_{1})}{3} + \frac{(c_{2}^{3} - c_{1}^{3}) (b_{2} - b_{1}) (a_{2} - a_{1})}{3} I_{x} = k (\frac{(a_{2} - a_{1}) (b_{2} - b_{1}) (c_{2} - c_{1})}{3}) (b_{2}^{2} + b_{1}^{2} + b_{1} b_{2} + c_{2}^{2} + c_{1}^{2} + c_{1} c_{2})$

Step 3. Find the radius of the gyration.

To find the radius of gyration about the x-axis, we have to find the mass of the rectangular solid:

$M = ∭_{T} ρ (x, y, z) d x d y d z M = \int_{x = a_{1}}^{a_{2}} \int_{y = b_{1}}^{b_{2}} \int_{z = c_{1}}^{c_{2}} (k) d x d y d z M = = k (a_{2} - a_{1}) (b_{2} - b_{1}) (c_{2} - c_{1})$

So, the radius of gyration is:

$R_{x} = \sqrt{\frac{I_{x}}{m}} R_{x} = \sqrt{\frac{k (\frac{(a_{2} - a_{1}) (b_{2} - b_{1}) (c_{2} - c_{1})}{3}) (b_{2}^{2} + b_{1}^{2} + b_{1} b_{2} + c_{2}^{2} + c_{1}^{2} + c_{1} c_{2})}{k (a_{2} - a_{1}) (b_{2} - b_{1}) (c_{2} - c_{1})}} R_{x} = \sqrt{\frac{(b_{2}^{2} + b_{1}^{2} + b_{1} b_{2} + c_{2}^{2} + c_{1}^{2} + c_{1} c_{2})}{3}}$