Question

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

If the density at each point in $T$is proportional to 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 $T$about the $x$- and $y$-axes.

Short answer

Radius of gyration is $R_y=\sqrt{\frac{3}{5}}\text{and}{R}_x=\sqrt{\frac{1}{5}}$

Mass of lamina is $m=\frac{2}{3}k$

Moments are $I_x=\frac{2}{15}k$about$y$axis and about$x$axis is$I_y=\frac{2}{5}k$

Step-by-step solution

Given Information

Density at every point is proportional to point's distance from$y$axis.

Moment of Inertia about  axis

$x$It is given by

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

Putting limits

$I_y={\int }_0^1{\int }_{-x}^x{x}^2\rho (x,y)dydx$

$I_y={\int }_0^1{\int }_{-x}^x{x}^{2}kxdydx\left[\rho \right(x,y)=kx]$

$I_y=k{\int }_0^1{\int }_{-x}^x{x}^{3}dydx$

Solving inner integral first

$I_y=k{\int }_0^1[y{]}_{-x}^x{x}^{3}dx$

$I_y=k{\int }_0^1[x-(-x\left)\right]x^{3}dx$

$I_y=k{\int }_0^{1}2x^{4}dx$

Integrating wrt $x$

$I_y=2k{\left[\frac{x^5}{5}\right]}_0^1=\frac{2}{5}k$

Moment of Inertia about y axis

Similarly

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

$I_x={\int }_0^1{\int }_{-x}^x{y}^2\rho (x,y)dydx$

$I_x={\int }_0^1{\int }_{-x}^x{y}^{2}kxdydx\left[\rho \right(x,y)=kx]$

$I_x=k{\int }_0^{1}x{\left[\frac{y^3}{3}\right]}_{-x}^{x}dx$

$I_x=k{\int }_0^{1}x\left[\frac{x^3-(-x{)}^3}{3}\right]dx$

role="math" localid="1653986937609" $I_x=k{\int }_0^1\left[\frac{2x^4}{3}\right]dx$

$I_x={\left[\frac{2x^5}{15}\right]}_0^{1}k=\frac{2}{15}k$

Calculating Mass of Lamina

Mass of lamina is given by $m={\iint }_{\Omega }\rho (x,y)dA$

$m={\int }_0^1{\int }_{-x}^{x}kxdydx$

$m=k{\int }_0^1[y{]}_{-x}^{x}xdx$

$m=k{\int }_0^{1}2x^{2}dx$

$m=\frac{2k}{3}{\left[x^3\right]}_0^1$

$m=\frac{2}{3}k$

Find radius of gyration

About both axis, it is given by

$R_y=\sqrt{\frac{I_y}{m}}\text{and}{R}_x=\sqrt{\frac{I_x}{m}}$

$R_y=\sqrt{\frac{\frac{2}{5}k}{\frac{2}{3}k}}\text{and}{R}_x=\sqrt{\frac{\frac{2}{15}k}{\frac{2}{3}k}}$

$R_y=\sqrt{\frac{3}{5}}\text{and}{R}_x=\sqrt{\frac{1}{5}}$

$R_y=\sqrt{\frac{3}{5}}\text{and}{R}_x=\sqrt{\frac{1}{5}}$