CE Board Exam Randomizer

Question 1

Moments of Inertia

Area and mass moments of inertia, polar moment, parallel-axis theorems, and radius of gyration.

Answer

The area is

$$A=\int_0^3(9-x^2)\,dx=18.$$

For the $x$-coordinate:

$$\bar{x}=\frac{1}{18}\int_0^3x(9-x^2)\,dx=\frac{1}{18}\left[\frac{9x^2}{2}-\frac{x^4}{4}\right]_0^3=\frac{9}{8}$$

For the $y$-coordinate:

$$\bar{y}=\frac{1}{18}\cdot\frac{1}{2}\int_0^3(9-x^2)^2\,dx=\frac{18}{5}$$

$$\boxed{\left(\frac{9}{8},\frac{18}{5}\right)}$$

Answer

At $y=2$, $x=4$. Use vertical strips from $x=0$ to $x=4$. The strip height is $2-\frac{x^2}{8}$.

$$I_y=\int_0^4x^2\left(2-\frac{x^2}{8}\right)dx$$

$$=\left[\frac{2x^3}{3}-\frac{x^5}{40}\right]_0^4=\boxed{17.07}$$

Answer

Use horizontal strips. Since $x=\frac{y^2}{4}$, the strip length is $\frac{y^2}{4}$.

$$I_x=\int_0^2y^2\left(\frac{y^2}{4}\right)dy=\frac{1}{4}\int_0^2y^4\,dy$$

$$I_x=\frac{1}{4}\left[\frac{y^5}{5}\right]_0^2=\boxed{1.6}$$

Answer

The area is

$$A=\int_0^2\frac{y^2}{4}\,dy=\frac{2}{3}.$$

From the previous result, $I_x=1.6$.

$$k_x=\sqrt{\frac{I_x}{A}}=\sqrt{\frac{1.6}{2/3}}=\boxed{1.55}$$

Answer

For $y^2=9x$, use horizontal strips from $y=0$ to $6$, with $x$ from $y^2/9$ to 4. Moment of inertia about the x-axis:
$I_x=\int_0^6 y^2\left(4-\frac{y^2}{9}\right)dy$
$I_x=4\frac{6^3}{3}-\frac{1}{9}\frac{6^5}{5}$
$\boxed{576/5}$

Answer

For the area between $y=2x$ and $y=x^2$, intersections are $x=0$ and $x=2$. Moment of inertia about the y-axis:
$I_y=\int_0^2 x^2(2x-x^2)dx$
$I_y=\left[\frac{x^4}{2}-\frac{x^5}{5}\right]_0^2$
$\boxed{1.6}$

Answer

For the same area, $A=\int_0^2(2x-x^2)dx=4/3$ and $\bar{x}=1$. Also $I_y=1.6=8/5$. The centroidal value is:
$I_{\bar{y}}=I_y-A\bar{x}^2$
$I_{\bar{y}}=\frac{8}{5}-\frac{4}{3}$
$\boxed{4/15}$

Answer

For $x^2=4y$, the first-quadrant region has $0\le y\le1$ and $0\le x\le2\sqrt{y}$. For moment of inertia about the x-axis,
$$I_x=\int y^2\,dA$$
Using horizontal strips:
$$I_x=\int_0^1 y^2(2\sqrt{y})dy$$
$$I_x=2\int_0^1y^{5/2}dy=2\left(\frac{2}{7}\right)=\frac{4}{7}$$
Therefore, $\boxed{I_x=4/7}$.

Answer

For $x^2=4y$, the first-quadrant region bounded by $y=1$ and the y-axis has $0\le y\le1$ and $0\le x\le2\sqrt{y}$. For the moment of inertia about the y-axis,
$$I_y=\int x^2\,dA$$
$$I_y=\int_0^1\int_0^{2\sqrt{y}}x^2\,dx\,dy$$
$$I_y=\int_0^1\frac{(2\sqrt{y})^3}{3}dy=\frac{8}{3}\int_0^1y^{3/2}dy$$
$$I_y=\frac{8}{3}\left(\frac{2}{5}\right)=\frac{16}{15}$$
Therefore, $\boxed{I_y=16/15}$.

Answer

The region is under $y=x^2/8$ from $x=0$ to $x=4$. For moment of inertia about the x-axis,
$$I_x=\int y^2\,dA$$
Using vertical strips:
$$I_x=\int_0^4\int_0^{x^2/8}y^2\,dy\,dx$$
$$I_x=\int_0^4\frac{1}{3}\left(\frac{x^2}{8}\right)^3dx=\int_0^4\frac{x^6}{1536}dx$$
$$I_x=\frac{4^7}{7(1536)}=1.52$$
Therefore, $\boxed{I_x=1.52}$.

Answer

The region is in the first quadrant under $y=x^2/8$ from $x=0$ to $x=4$. For the moment of inertia about the y-axis,
$$I_y=\int x^2\,dA$$
Using vertical strips:
$$I_y=\int_0^4 x^2\left(\frac{x^2}{8}\right)dx$$
$$I_y=\frac{1}{8}\int_0^4x^4dx=\frac{1}{8}\left[\frac{x^5}{5}\right]_0^4$$
$$I_y=25.6$$
Therefore, $\boxed{I_y=25.6}$.

Answer

In the first quadrant, the region is $0\le x\le1$ and $0\le y\le2\sqrt{x}$. For moment of inertia about the x-axis,
$$I_x=\int y^2\,dA$$
Using vertical strips:
$$I_x=\int_0^1\int_0^{2\sqrt{x}}y^2\,dy\,dx$$
$$I_x=\int_0^1\frac{(2\sqrt{x})^3}{3}dx=\frac{8}{3}\int_0^1x^{3/2}dx$$
$$I_x=\frac{8}{3}\left(\frac{2}{5}\right)=\frac{16}{15}=1.067$$
Therefore, $\boxed{I_x=1.067}$.

Answer

In the first quadrant, $0\le x\le1$ and $0\le y\le2\sqrt{x}$. For the moment of inertia about the y-axis,
$$I_y=\int x^2\,dA$$
Using vertical strips:
$$I_y=\int_0^1 x^2(2\sqrt{x})dx=2\int_0^1x^{5/2}dx$$
$$I_y=2\left[\frac{2}{7}x^{7/2}\right]_0^1=\frac{4}{7}=0.571$$
Therefore, $\boxed{I_y=0.571}$.

Answer

In the first quadrant, $0\le y\le2$ and $0\le x\le y^2/4$. For the moment of inertia about the x-axis,
$$I_x=\int y^2\,dA$$
Using horizontal strips:
$$I_x=\int_0^2 y^2\left(\frac{y^2}{4}\right)dy$$
$$I_x=\frac{1}{4}\int_0^2y^4dy=\frac{1}{4}\left[\frac{y^5}{5}\right]_0^2$$
$$I_x=1.6$$
Therefore, $\boxed{I_x=1.6}$.

Answer

In the first quadrant, $0\le y\le2$ and $0\le x\le y^2/4$. For the moment of inertia about the y-axis,
$$I_y=\int x^2\,dA$$
Using horizontal strips:
$$I_y=\int_0^2\int_0^{y^2/4}x^2\,dx\,dy$$
$$I_y=\int_0^2\frac{1}{3}\left(\frac{y^2}{4}\right)^3dy=\int_0^2\frac{y^6}{192}dy$$
$$I_y=\frac{128}{1344}=0.095$$
Therefore, $\boxed{I_y=0.095}$.

Answer

The line $x=1$ intersects $y^2=4x$ at $y=\pm2$. Using horizontal strips, the width is
$$1-\frac{y^2}{4}$$
Thus the moment of inertia about the x-axis is
$$I_x=\int_{-2}^{2}y^2\left(1-\frac{y^2}{4}\right)dy$$
Because the region is symmetric:
$$I_x=2\int_0^2\left(y^2-\frac{y^4}{4}\right)dy$$
$$I_x=2\left[\frac{y^3}{3}-\frac{y^5}{20}\right]_0^2=2.13$$
Therefore, $\boxed{I_x=2.13}$.

Answer

In the first quadrant, $y=2\sqrt{x}$ from $x=0$ to $x=1$. For a vertical strip, the moment of inertia about the x-axis is
$I_x=\int_0^1 \frac{1}{3}y^3\,dx$
$I_x=\int_0^1 \frac{1}{3}(2\sqrt{x})^3\,dx=\frac{8}{3}\int_0^1 x^{3/2}\,dx$
$I_x=\frac{8}{3}\left(\frac{2}{5}\right)=\frac{16}{15}$
$\boxed{I_x\approx1.07}$
The printed prompt mentions both axes, but the keyed answer matches the x-axis value.

Answer

$$I_x=\frac{bh^3}{12}=\frac{300(500)^3}{12}=3.125\times10^9\text{ mm}^4$$

Answer: $I_x=3.125\times10^9$ mm⁴.

Question 2

Area Moments of Inertia (Second Moments of Area)

$$I_x=\int y^2\,dA$$

$$I_y=\int x^2\,dA$$

$$I_{xy}=\int xy\,dA$$

Polar Moment of Inertia

$$J_O=\int r^2\,dA$$

$$J_O=I_x+I_y$$

Parallel-Axis Theorem (Area)

$$I_x=I_{\bar{x}}+A d^2$$

$$I_y=I_{\bar{y}}+A d^2$$

Parallel-Axis Theorem (Polar)

$$J_O=J_C+A d^2$$

Radius of Gyration

$$k_x=\sqrt{\frac{I_x}{A}}$$

$$k_y=\sqrt{\frac{I_y}{A}}$$

$$k_O=\sqrt{\frac{J_O}{A}}$$

Mass Moment of Inertia

$$I=\int r^2\,dm$$

$$I_x=\int (y^2+z^2)\,dm$$

$$I_y=\int (x^2+z^2)\,dm$$

$$I_z=\int (x^2+y^2)\,dm$$

Parallel-Axis Theorem (Mass)

$$I_O=I_C+M d^2$$

Question 3

Centroid of a Parabolic Region

Find the centroid of the region in the first quadrant bounded by $y=9-x^2$, the $x$-axis, and the $y$-axis.

Question 4

Moment of Inertia About the y-Axis for a Parabola

Find $I_y$ of the region bounded by $x^2=8y$, $y=2$, and $x=0$.

Question 5

Moment of Inertia About the x-Axis for a Sideways Parabola

Find $I_x$ of the first-quadrant region bounded by $y^2=4x$, $y=2$, and the $y$-axis.

Question 6

Radius of Gyration From Area Moment

For the same region bounded by $y^2=4x$, $y=2$, and the $y$-axis, find the radius of gyration about the $x$-axis.

Question 7

Question Bank: q451

MSTE - Integral Calculus / Moment of Inertia of a Plane Area / Engr. Janclyde Espinosa (Clidez)

Find the moment of inertia of the first-quadrant area bounded by y²=9x and x=4 with respect to the x-axis.

Answer:

576/5
567/5
584/5
548/5

Question 8

Question Bank: q452

MSTE - Integral Calculus / Moment of Inertia of a Plane Area / Engr. Janclyde Espinosa (Clidez)

Find I_y for the area bounded by y=x² and y=2x.

Answer:

1.6
1.8
2.0
2.2

Question 9

Question Bank: q453

MSTE - Integral Calculus / Moment of Inertia of a Plane Area / Engr. Janclyde Espinosa (Clidez)

Determine the moment of inertia about the centroidal y-axis of the area bounded by y=x² and y=2x.

Answer:

4/15
3/16
7/16
8/15

Question 10