A classical mechanics problem by Rocco Dalto

Let the axis A : A: the line x = x 0 + a t , y = y 0 + b t , z = z 0 + c t x = x_{0} + at, y = y_{0} +bt, z = z_{0} + ct .

The moment of inertia I I of a lamina S S of unit density about A A is:

I = S ( c ( y y 0 ) b ( z z 0 ) ) 2 + ( a ( z z 0 ) c ( x x 0 ) ) 2 + ( b ( x x 0 ) a ( y y 0 ) ) 2 a 2 + b 2 + c 2 d A I = \int_{S} \int \dfrac{(c(y - y_{0}) - b(z - z_{0}))^2 + (a(z - z_{0}) - c(x - x_{0}))^2 + (b(x - x_{0}) - a(y - y_{0}))^2}{a^2 + b^2 + c^2} \: dA

Refer to previous problem. . .

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1 solution

Rocco Dalto
Dec 17, 2016

Let B : ( x , y , z ) B:(x,y,z) be any point not on the given line and choose the points 0 : ( x 0 , y 0 , z 0 ) 0:(x_{0},y_{0},z_{0}) and A : ( x 0 + a , y 0 + b , z 0 + c ) A:(x_0 + a,y_{0} + b,z_{0} + c) on the given line.

The distance from point B B to the given line is d = O B X O A O A d = \dfrac{|\vec{OB} \: X \: \vec{OA}|}{|\vec{OA}|}

0 A = a i + b j + c k \vec{0A} = a \vec{i} + b \vec{j} + c \vec{k}

0 B = ( x x 0 ) i + ( y y 0 ) j + ( z z 0 ) k \vec{0B} = (x - x_{0}) \vec{i} + (y - y_{0}) \vec{j} + (z - z_{0}) \vec{k}

O B X O A = ( c ( y y 0 ) b ( z z 0 ) ) i + ( a ( z z 0 ) c ( x x 0 ) ) j + ( b ( x x 0 ) a ( y y 0 ) ) k \vec{OB} \: X \: \vec{OA} = (c(y - y_{0}) - b(z - z_{0})) \vec{i} + (a(z - z_{0}) - c(x - x_{0})) \vec{j} + (b(x - x_{0}) - a(y - y_{0})) \vec{k}

d 2 = ( c ( y y 0 ) b ( z z 0 ) ) 2 + ( a ( z z 0 ) c ( x x 0 ) ) 2 + ( b ( x x 0 ) a ( y y 0 ) ) 2 a 2 + b 2 + c 2 \implies d^2 = \dfrac{(c(y - y_{0}) - b(z - z_{0}))^2 + (a(z - z_{0}) - c(x - x_{0}))^2 + (b(x - x_{0}) - a(y - y_{0}))^2}{a^2 + b^2 + c^2}

I = S d 2 d A = \therefore I = \int_{S} \int d^2 \: dA =

S ( c ( y y 0 ) b ( z z 0 ) ) 2 + ( a ( z z 0 ) c ( x x 0 ) ) 2 + ( b ( x x 0 ) a ( y y 0 ) ) 2 a 2 + b 2 + c 2 d A \int_{S} \int \dfrac{(c(y - y_{0}) - b(z - z_{0}))^2 + (a(z - z_{0}) - c(x - x_{0}))^2 + (b(x - x_{0}) - a(y - y_{0}))^2}{a^2 + b^2 + c^2} \: dA

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