Find moment of Inertia of the shaded are as shown in figure about its references x and y axls.

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written 3.5 years ago by

teamques10 ★ 68k

(All measurements in cm)

Sr. No.	Part	Area	X-Co-ord of C.G. (y1)	Y-Co-ord of C.G. (y1)	A x1	A y1
$1)$	Full Square
$L=10, \ B=10$	$L\times B$

$= 10 \times 10$ $=100$ | $5$ | $5$ | $500.0000$ | $500.0000$ | | $(2)$ | Cut Triangle $B=3, \ H=6$ | $-0.5 \ BH$ $= -0.5 \times 3 \times 3$ $=9$ | $\frac {B}{3} = \frac {3}{3} = 1$ | $\frac {H}{3} = \frac {6}{3}=2$ | $-9.0000$ | $-18.0000$ | | $3)$ | Cut Semicircle radius (R) =5 | $-0.5 \pi R^2$ $=-0.5 \times 5^2 \pi$ $= -39.27$ | $10 - \frac {4R}{3\pi} = 10 - \frac {4\times 5}{3\pi}$ $= 7.8779$ | $5$ | $-309.3665$ | $-196.3500$ | | $4)$ | Cut Quarter Circle radius (r)=4 | $-0.25 \pi r^2$ $=-0.25 \times 4^2 \pi$ $= - 12.5664$ | $\frac {4r}{3\pi} = \frac {4 \times 4}{3\pi}$ $=1.6977$ | $10 - \frac {4r}{3\pi} = 10 - \frac {4 \times 4}{3\pi}$ $= 8.3023$ | $- 21.3340$ | $- 104.3306$ | | | Total | 39.1636 | | | 160.2995 | 181.3194 | $\overline {x} = \dfrac {\sum Ax}{\sum A} = \dfrac {160.2995}{39.1636} = 4.0931 \ cm$

$\overline {y} = \dfrac {\sum A_y}{\sum A} = \dfrac {181.194}{39.1636} = 4.6298 \ cm$

$\therefore \ Centroid \ (4.0931, 4.6298)$

Let hx and hy denote the distance of the C.G. of the part from its reference X and Y axes respectively.

	X-axis	Y-axis
Sr. No.	Part	MI about C.G. (M1)	$h_x = y1$	Net MI = M1+Ahx2	MI about C.G (M2)	$h_y = x_1$	Net MI = M2+Ahy2
$1)$	Full Square L=10	$\frac {L^4}{12} = \frac {10^4}{12}$
$=833.3333$	$5$ \| $833.3333+100 \times 5^2$ $= 3333.3333$	$\frac {L^4}{12} = \dfrac {10^4}{12}$
$= 833.3333$	$5$ \| $833.3333+100\times 5^2\ = 3333.3333$ \| \| $2)$ \| Cut Triangle B=3, H=6 \| $\frac {BH^3}{36} = \frac {3\times 6^3}{36}$ $=18$	$2$ \| $-(18+9 \times 2^2)$ $=-54$	$\frac {B^3 H}{36} = \frac {3^3 \times 6}{36}$
$=4.5$	$1$ \| $-(4.5 + 9\times 1^2)$ $=-13.5$
$3)$	Cut Semicircle radius (R)=5	$\frac {\pi R^4}{8} = \frac {5^4 \pi}{8}$
$= 245.4369$	$5$ \| $= (245.4369 + 39.27 \times 5^2)$ $= -1227.1869$	$0.1098 \ R^4$

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