Q 41. - Calculus

Question

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

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

Step-by-Step Solution

Answer

The moment of inertia is $I_y=\frac{1}{4}kh{b}^4$ and $I_x=\frac{1}{6}k{b}^2{h}^3$.

The mass is $m=\frac{1}{2}kh{b}^2$

The radius of gyration is $R_y=\frac{b}{\sqrt{2}}\text{ and }{R}_x=\frac{h}{\sqrt{3}}$

Step 1: Given Information

It is given that vertices of rectangular region is $(0,0),(b,0),(0,h)\text{ and }(b,h)$

$\rho (x,y)=kx$

Step 2: Calculation of I y

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

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

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

$I_y=k{\int }_0^b{\int }_0^h{x}^{3}dydx$

Solving inner integral

$I_y=k{\int }_0^b[y{]}_0^h{x}^{3}dx=k{\int }_0^b[h]x^{3}dx=kh{\int }_0^b{x}^{3}dx$

$I_y=kh{\left[\frac{x^4}{4}\right]}_0^b=kh\left[\frac{b^4}{4}\right]$

Hence, $I_y=\frac{1}{4}kh{b}^4$

Step 3: Calculating I x

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

Imposing limits

$I_x={\int }_0^b{\int }_0^h{y}^2\rho (x,y)dydx={\int }_0^b{\int }_0^h{y}^{2}kxdydx=k{\int }_0^{b}x{\left[\frac{y^3}{3}\right]}_0^{h}dx$

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

$I_x=\frac{1}{6}k{b}^2{h}^3$

Step 4: Mass of Lamina

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

As $\rho (x,y)=kx$

$\Rightarrow m={\int }_0^b{\int }_0^{h}kxdydx$

$m={\int }_0^{b}kx[y{]}_0^{h}dx$

$m={\int }_0^{b}kxhdx$

$m=kh{\int }_0^{b}xdx$

Solving further

$m=kh{\left[\frac{x^2}{2}\right]}_0^b$

Mass is $m=\frac{1}{2}kh{b}^2$

Step 5: Radius of Gyration

Radius of gyration is $R_y=\sqrt{\frac{I_y}{m}}\text{ and }{R}_x=\sqrt{\frac{I_x}{m}}$

$R_y=\sqrt{\frac{\frac{1}{4}kh{b}^4}{\frac{1}{2}kh{b}^2}}\text{ and }{R}_x=\sqrt{\frac{\frac{1}{6}k{b}^2{h}^3}{\frac{1}{2}kh{b}^2}}$

$\Rightarrow R_y=\frac{b}{\sqrt{2}}\text{ and }{R}_x=\frac{h}{\sqrt{3}}$