Q 37. - Calculus

Question

Let  be triangular region with vertices $(1,0),(2,1)\text{, and }(2,-1)$

If the density at each point in $T_2$ is proportional to the square of the point’s distance from the $y$-axis, find the

moments of inertia about the $x$- and $y$-axes. Use these

answers to find the radii of gyration of $T_2$ about the

$x$- and $y$-axes.

Step-by-Step Solution

Answer

The moment of inertia is $I_y=\frac{129}{15}k, I_x=\frac{49}{30}k$

The mass is $m=\frac{5}{3}k$

Radius of gyration is $R_y=\sqrt{\frac{147}{50}}\text{ and }{R}_x=\sqrt{\frac{147}{150}}$

Step 1: Given Information

Vertices of triangle are $(1,0),(2,1)\text{, and }(2,-1)$

$\rho (x,y)=k{x}^2$

Step 2: Calculating I Y

It can be calculated as $I_y={\iint }_{\Omega }{x}^2\rho (x,y)dA$

Putting limits gives

$I_y={\int }_1^2{\int }_{-x+1}^{x-1}{x}^2\rho (x,y)dydx$

$I_y={\int }_1^2{\int }_{-x+1}^{x-1}{x}^{2}k{x}^{2}dydx \left[\rho (x,y)=k{x}^2\right]$

$I_y=k{\int }_1^2{\int }_{-x+1}^{x-1}{x}^{4}dydx$

Solving inner integral

$I_y=k{\int }_1^2[y{]}_{-x+1}^{x-1}{x}^{4}dx=k{\int }_1^2[x-1-(-x+1)]x^{4}dxk{\int }_1^{2}2\left(x^5-x^4\right)dx$

Solving further

$I_y=2k{\left[\frac{x^6}{6}-\frac{x^5}{5}\right]}_1^2=2k\left[\left(\frac{2^6}{6}-\frac{2^5}{5}\right)-\left(\frac{1}{6}-\frac{1}{5}\right)\right]=\frac{129}{15}k$

$\frac{49}{30}k$

Step 3: Calculating I x

Similarly, $I_x={\iint }_{\Omega }{y}^2\rho (x,y)dA$

Putting limits

$I_x={\int }_1^2{\int }_{-x+1}^{x-1}{y}^2\rho (x,y)dydx$

$I_x={\int }_1^2{\int }_{-x+1}^{x-1}{y}^{2}k{x}^{2}dydx \left[\rho (x,y)=k{x}^2\right]$

$I_x=k{\int }_1^2{x}^2{\left[\frac{y^3}{3}\right]}_{-x+1}^{x-1}dx=k{\int }_1^2{x}^2\left[\frac{(x-1{)}^3-(-x+1{)}^3}{3}\right]dx=k{\int }_1^2{x}^2\left[\frac{2(x-1{)}^3}{3}dx\right.$

$I_x=\frac{2}{3}k{\int }_1^2\left[x^5-3x^4+3x^3-x^2\right]dx$

Solving further

$I_x=\frac{2}{3}k{\left[\frac{x^6}{6}-\frac{3x^5}{5}+\frac{3x^4}{4}-\frac{x^3}{3}\right]}_1^2$

$I_x=\frac{2}{3}k\left[\left(\frac{2^5}{5}-\frac{3\times 2^4}{4}+\frac{3\times 2^3}{3}-\frac{2^2}{2}\right)-\left(\frac{1}{5}-\frac{3}{4}+\frac{3}{3}-\frac{1}{2}\right)\right]$

$I_x=\frac{2}{3}k\left[\frac{49}{20}\right]=\frac{49}{30}k$