Question

13. -/2 points Larson8 14.4.028. Consider the following. Right triangle Verify the given moments of inertia and find and . Assume that each lamina has a density of P l. This region is a common shape used in engineering

Answer 1

given

lamina has a density $\rho$ = 1

the moment of inertia about x - axis is

I_x = $\fn_phv \int \int_{D}^{}$ ( y²$\rho$( x,y ) dA

= $\fn_phv \int_{0}^{b}\int_{0}^{-hx/b+h}$ y²(1) dy dx

= (1/3) $\fn_phv \int_{0}^{b}$ ( -hx/b + h )³ dx

= ( 1/12 ) ( - b/h ) ( -hx/b + h )⁴|_o^b

I_x = ( 1/12 ) b h³

I_y = $\fn_phv \int \int_{D}^{}$ ( x²$\rho$( x,y ) dA

= $\fn_phv \int_{0}^{b}\int_{0}^{-hx/b+h}$ x²(1) dy dx

= (1/3) $\fn_phv \int_{0}^{b}$ ( -hx/b + h )³ dx

= ( 1/12 ) ( - b/h ) ( -hx/b + h )⁴|_o^b

I_y = ( 1/12 ) b³ h

m = $\fn_phv \int \int_{D}^{}$ $\rho$ ( x,y ) dA

= $\fn_phv \int_{0}^{b}\int_{0}^{-hx/b+h}$ (1) dy dx

= $\fn_phv \int_{0}^{b}$ ( -hx/b + h ) dx

= b h / 2

$\fn_phv \bar{x}$ = ( I_y / m )^1/2

= ( ( 1/12 ) b³ h / bh/2 )^1/2

$\fn_phv \bar{x}$ = b / $\fn_phv \sqrt{6}$

$\fn_phv \bar{y}$ = ( I_x / m )^1/2

= ( ( 1/12 ) b h³ / bh/2 )^1/2

$\fn_phv \bar{y}$ = h / $\fn_phv \sqrt{6}$

so $\fn_phv \bar{x}$ = b / $\fn_phv \sqrt{6}$ and $\fn_phv \bar{y}$ = h / $\fn_phv \sqrt{6}$