Q 30. - Calculus

Question

Let $T$ be rectangular 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 $x$-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.

Step-by-Step Solution

Answer

Moment of Inertia is $I_y=\frac{k}{5}$ and $I_x=\frac{k}{10}$

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

Radius of gyration is $R_x=\sqrt{\frac{3}{10}},R_y=\sqrt{\frac{3}{5}}$

Step 1: Given Information

Vertices of triangular region are $(0,0),(1,1),\text{ and }(1,-1)$

Density  $\rho (x,y)=ky$

Step 2: MOI about y axis

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

Putting limits $I_y={\int }_0^1{\int }_{-x}^x{x}^{2}kydydx \left[\rho \right(x,y)=ky]$

As per figure

$I_y=2{\int }_0^1{\int }_0^x{x}^{2}kydydx$

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

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

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

$I_y=\frac{k}{5}$

Step 3: MOI about x axis

It is expressed as:

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

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

As region is symmetric about $x$ axis

$I_x=2{\int }_0^1{\int }_0^x{y}^{2}kydydx$

$I_x=2k{\int }_0^1{\int }_0^x{y}^{3}dydx$

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

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

$I_x=\frac{k}{10}$

Step 4: Mass of triangular region

It is given by $m=2{\int }_0^1{\int }_0^{x}kydydx \left[m={\iint }_{\Omega }\rho (x,y)dA\right]$

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

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

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

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

Step 3: Radius of gyration

It is given by

$R_x=\sqrt{\frac{I_x}{m}}\text{ and }{R}_y=\sqrt{\frac{I_y}{m}}$

$R_x=\sqrt{\frac{k/10}{k/3}}\text{ and }{R}_y=\sqrt{\frac{k/5}{k/3}}$

$R_x=\sqrt{\frac{3}{10}},R_y=\sqrt{\frac{3}{5}}$