Q 34. - Calculus

Question

Let $T_2$ be triangular region with vertices $(1,0),(2,1)\text{, and }(2,-1)$

If the density at each point in T2 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 $T_2$ about the $x$- and $y$-axes.

Step-by-Step Solution

Answer

The moment of is $I_y=\frac{49}{10}k$ and $I_x=\frac{49}{30}k$

Mass of lamina 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 $(1,0),(2,1)\text{, and }(2,-1)$

$\rho (x,y)=kx$

Step 2: Finding y -

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

Putting limits

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

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

$I_y=k{\int }_1^2{\int }_{-x+1}^{x-1}{x}^{3}dydx$

Solving inner integral first

$I_y=k{\int }_1^2[y{]}_{-x+1}^{x-1}{x}^{3}dx$

$I_y=k{\int }_1^2[x-1-(-x+1\left)\right]x^{3}dx$

$I_y=k{\int }_1^{2}2\left(x^4-x^3\right)dx$

$I_y=2k{\left[\frac{x^5}{5}-\frac{x^4}{4}\right]}_1^2$

$I_y=2k\left[\left(\frac{2^5}{5}-\frac{2^4}{4}\right)-\left(\frac{1}{5}-\frac{1}{4}\right)\right]$

$I_y=\frac{49}{10}k$

Step 3: Finding x -

Similarly

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

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

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

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

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

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

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

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

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

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

$I_x=\frac{49}{30}k$

Step 4: Find Mass of Lamina

Formula is $m={\iint }_{\Omega }\rho (x,y)dA$

Putting limits

$m={\int }_1^2{\int }_{-x+1}^{x-1}kxdydx$

Solving inner integral first

$m=k{\int }_1^2[y{]}_{-x+1}^{x-1}xdx$

$m=k{\int }_1^{2}2\left(x^2-x\right)dx$

$m=2k{\left[\frac{x^3}{3}-\frac{x^2}{2}\right]}_1^2$

$m=2k\left[\left(\frac{2^3}{3}-\frac{2^2}{2}\right)-\left(\frac{1}{3}-\frac{1}{2}\right)\right]$

$m=\frac{5}{3}k$

Step 5: Radius of Gyration

Radius of gyration is given by

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

$R_y=\sqrt{\frac{\frac{49}{10}k}{\frac{5}{3}k}}\text{ and }{R}_x=\sqrt{\frac{\frac{49}{30}k}{\frac{5}{3}k}}$

$R_y=\sqrt{\frac{147}{50}}\text{ and }{R}_x=\sqrt{\frac{147}{150}}$

Hence,

$R_y=\sqrt{\frac{147}{50}}\text{ and }{R}_x=\sqrt{\frac{147}{150}}$