Question

Please show ALL YOUR WORK and organize it in a logical and neat manner.

Determine by direct integration the moment of inertia of the shaded area with respect to the x-axis (I_x) and the y-axis (I_y).

HINT: Start by calculating the value of k.

NOTE: Make sure to select differential areas parallel to the axis you are calculating the moment about.

Answer 1

We choose a differential strip 'dA" parallel to \(\mathrm{x}-\) axis. \(d A=I \times a^{5}\)

\(I=x\)

We have, \(y=K x^{\frac{5}{2}}\)

When, \(x=a, y=b\)

Hence, \(b=K \times a^{\frac{s}{2}}\)

or, \(K=\frac{b}{a^{\frac{5}{2}}}\)

Now, \(y=\frac{b}{a^{\frac{5}{2}}} \times x^{\frac{5}{2}}\)

\(d y=\frac{5}{2} \frac{b}{a^{\frac{5}{2}}} \times x^{\frac{3}{2}} d x\)

Now, \(d I_{x}=y^{2} d A\)

Integrating from \(y=0\) to \(y=b\)

\(\begin{aligned} I_{z} &=\int_{0}^{b} y^{2} d A \\ &=\int_{0}^{a}\left(\frac{b}{a^{\frac{3}{2}}} \times x^{\frac{3}{2}}\right)^{2} \times x \times \frac{5}{2} \times \frac{b}{a^{\frac{5}{2}}} \times x^{\frac{3}{2}} d x \\ &=\frac{5}{2} \times \frac{b^{2}}{a^{3}} \times \frac{b}{a^{\frac{5}{2}}} \int_{0}^{a} x^{7 s} d x \\ &=\frac{5}{2} \times \frac{b^{3}}{a^{75}}\left[\frac{x^{45}}{8.5}\right]_{0}^{2} \\ &=\frac{5}{2} \times \frac{b^{3}}{a^{3^{3}}} \times \frac{a^{a s}}{8.5} \\ I_{z} &=\frac{5}{17} a b^{3} \end{aligned}\)

Consider an elementary strip of thickness " \(d x^{\prime \prime}\) and length ( \(b-y\) ) which is at a distance of " \(x^{*}\) from y - axis Area of the elemental strip \(d A=(b-y) d x\)

Given that \(y=k x^{32}\) At point \(E(a, b)\) So the point \(E(a, b)\), will satisfy The equation \(y=k x^{52}\)

$$ \begin{array}{l} b=k a^{52} \\ k=\frac{b}{a^{3 / 2}} \end{array} $$

Moment of Inertia of the shaded area bout \(y-\) aris \(I_{y}=\int_{0}^{g} x^{2} d A\)

\(I_{y}=\int_{0}^{a} x^{2}(b-y) d x\)

\(I_{y}=\int_{0}^{a} x^{2}\left[b-k x^{32}\right] d x\)

\(\left[\because y=k x^{32}\right]\)

\(I_{3}=\int_{0}^{a} x^{2}\left[b-\frac{b}{a^{3 / 2}} x^{3 / 7}\right] d x\)

\(I_{j}=\int_{j}^{9}\left(x^{2} b-\frac{b}{a^{3 / 2}} x^{22}\right) d x\)

\(\left.I_{y}=\frac{x^{3} b}{3}-\frac{b}{a^{5}} x^{22}\right]_{5}^{a}\)

\(I_{y}=\frac{a^{3} b}{3}-\frac{b}{a^{32}} \frac{a^{2^{2}}}{\frac{9}{2}+1}\)

\(I_{y}=\frac{11 b a^{3}-6 b a^{3}}{33}\)

\(I_{y}=\frac{5}{33} b a^{3}\)