Q 44. - Calculus

Question

Let $R$ be rectangular region having vertices $(0,0),(b,0),(0,h),\text{ and }(b,h)$

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

Step-by-Step Solution

Answer

Moment is $I_y=\frac{1}{5}kh{b}^5,I_x=\frac{1}{9}k{h}^3{b}^3$

Mass is $m=\frac{1}{3}kh{b}^3$

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

Step 1: Given Information

Let the vertices of rectangular region is $(0,0),(b,0),(0,h)\text{ and }(b,h)$

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

Step 2: Calculating I y

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

Put limits $I_y={\int }_0^b{\int }_0^h{x}^2\rho (x,y)dydx$

$I_y={\int }_0^b{\int }_0^h{x}^{2}k{x}^{2}dydx \left[\rho (x,y)=k{x}^2\right]$

$I_y=k{\int }_0^b{\int }_0^h{x}^{4}dydx$

Solving inner integral first

$I_y=k{\int }_0^b[y{]}_0^h{x}^{4}dx=k{\int }_0^b[h]x^{4}dx=kh{\int }_0^b{x}^{4}dx$

Solving further

$I_y=kh{\left[\frac{x^5}{5}\right]}_0^b$

$I_y=kh\left[\frac{b^5}{5}\right]$

$I_y=\frac{1}{5}kh{b}^5$

Step 3: Calculating I x

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

Solving as same in above step

$I_x={\int }_0^b{\int }_0^h{y}^2\rho (x,y)dydx$

$I_x={\int }_0^b{\int }_0^h{y}^{2}k{x}^{2}dydx \left[\rho (x,y)=k{x}^2\right]$

$I_x=k{\int }_0^b{x}^2{\left[\frac{y^3}{3}\right]}_0^{h}dx$

$I_x=k{\int }_0^b{x}^2\left[\frac{h^3}{3}\right]dx=\frac{1}{3}k{h}^3{\int }_0^b{x}^{2}dx$

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

Hence, $I_x=\frac{1}{9}k{h}^3{b}^3$

Step 4: Calculating Mass of Lamina

The mass is given by $m={\iint }_{\Omega }\rho (x,y)dA$

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

$m={\int }_0^b{\int }_0^{h}k{x}^{2}dydx$

Solving inner integral

$m={\int }_0^{b}k{x}^2[y{]}_0^{h}dx$

$m={\int }_0^{b}k{x}^{2}hdx=kh{\int }_0^b{x}^{2}dx$

Solving further

$m=kh{\left[\frac{x^3}{3}\right]}_0^b$

$\Rightarrow m=\frac{1}{3}kh{b}^3$

Step 5: Radius of Gyration

It is given by

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

$R_y=\sqrt{\frac{\frac{1}{5}kh{b}^5}{\frac{1}{3}kh{b}^3}}\text{ and }{R}_x=\sqrt{\frac{\frac{1}{9}k{h}^3{b}^3}{\frac{1}{3}kh{b}^3}}$

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

Putting values

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