A homogeneous rectangular lamina has constant area density ρ. Find the moment of inertia of the lamina about one corner

Question

asked Jul 16, 2021 72.9k views

1 Answer

Puchu · Answer 1 · 2021-07-22T01:41:21+0000

Answer:

$I_(corner) =(\rho _(ab))/(3)(a^2+b^2)$

Explanation:

By applying the concept of calculus;

the moment of inertia of the lamina about one corner
I_(corner) is:

$I_(corner) = \int\limits \int\limits_R (x^2+y^2) \rho d A \\ \\ I_(corner) = \int\limits^a_0\int\limits^b_0 \rho(x^2+y^2) dy dx$

where :

(a and b are the length and the breath of the rectangle respectively )

$I_(corner) = \rho \int\limits^a_0 {x^2y}+ (y^3)/(3) |^ {^ b}_(_0) \, dx$

$I_(corner) = \rho \int\limits^a_0 (bx^2 + (b^3)/(3))dx$

$I_(corner) = \rho [(bx^3)/(3)+ (b^3x)/(3)]^ {^ a} _(_0)$

$I_(corner) = \rho [(a^3b)/(3)+ (ab^3)/(3)]$

$I_(corner) =(\rho _(ab))/(3)(a^2+b^2)$

Thus; the moment of inertia of the lamina about one corner is
$I_(corner) =(\rho _(ab))/(3)(a^2+b^2)$