Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination — Statics and Dynamics (04-BS-3), December 2014. 3 hours, closed book (one 8.5"×11" self-prepared note sheet permitted). Candidates were required to complete 2 of the 3 questions in Part A (Statics) and 2 of the 3 questions in Part B (Dynamics); every question in both parts is solved below.
Check — figure readings (Questions 1 and 3): (1) In Figure 1's truss only three members meet at joint D (C–D, D–B, D–E) — there is no separate C–E member, and the truss coordinates are A(0,0), B(4,0), C(1,2), D(3,2.6), E(7,4) m. (2) Figure 3A's axial dimension stack (2+4+2+4+2 = 14 in) does not sum to the labelled overall height of 16 in unless the hub's "2 in." label is read as a half-thickness measured from the shaft centreline (giving a full hub thickness of 4 in, which closes the stack exactly: 2+4+4+4+2 = 16 in). Both readings are disclosed here and used consistently in the solutions below.
Part A — Mass moment of inertia and radius of gyration of the flywheel
Given. Axisymmetric steel flywheel, specific weight $\gamma=490\ \text{lb}_f/\text{ft}^3$. Five hollow-cylinder segments stacked along the axis of rotation (see the callout above for the 16 in reconciliation):
Flywheel segments (about the axis of rotation)
Segment
Outer R (in)
Inner r (in)
Axial thickness (in)
Rim (top)
4
2
2
Web plate (top, "1 in thick solid plate")
4
2
1
Hub
3
2
4
Web plate (bottom)
4
2
1
Rim (bottom)
4
2
2
Find. $I_m$ and $k=\sqrt{I_m/M}$ about the axis of rotation.
Figure 3A — flywheel cross-section (axis of rotation horizontal, radii shown to scale).
Approach. Treat every segment as a hollow cylinder about the same central axis; compute each segment's weight from $\gamma\times$ volume, convert to mass, apply $I_m=\tfrac12 m(R^2+r^2)$, and sum. Work in feet so $\gamma$ ($\text{lb}_f/\text{ft}^3$) and $g=32.2\ \text{ft/s}^2$ are dimensionally consistent, giving $I_m$ in $\text{slug}\cdot\text{ft}^2$.
Mass of each segment. $V=\pi(R^2-r^2)t$, $W=\gamma V$, $m=W/g$. Converting each $R,r,t$ to feet ($\div 12$):
Segment
$V$ (ft³)
$m$ (slug)
Rim ×2
0.04363
0.6640 each
Web ×2
0.02182
0.3320 each
Hub
0.03636
0.5533
Total mass $M = 2(0.6640)+2(0.3320)+0.5533 = 2.545\ \text{slug}$ (weight $\approx 81.96\ \text{lb}_f$).
Moment of inertia of each segment about the axis ($I_m=\tfrac12 m(R^2+r^2)$, R,r in ft):
$$I_{\text{rim}} = \tfrac12(0.6640)\!\left[\left(\tfrac{4}{12}\right)^2+\left(\tfrac{2}{12}\right)^2\right] = 0.04611\ \text{slug}\cdot\text{ft}^2 \ \text{(each; ×2 rims)}$$
$$I_{\text{web}} = \tfrac12(0.3320)\!\left[\left(\tfrac{4}{12}\right)^2+\left(\tfrac{2}{12}\right)^2\right] = 0.02305\ \text{slug}\cdot\text{ft}^2 \ \text{(each; ×2 webs)}$$
$$I_{\text{hub}} = \tfrac12(0.5533)\!\left[\left(\tfrac{3}{12}\right)^2+\left(\tfrac{2}{12}\right)^2\right] = 0.02498\ \text{slug}\cdot\text{ft}^2$$
Check: the hub's full axial thickness (4 in, treated as symmetric about the shaft centreline) and the web's radial span (bore to rim inner radius, R=4 in, r=2 in) are read from the printed figure per the reconciliation callout above; the arithmetic method and formula application are exact given these dimensions.
Part B — Centroid of the area under $y=b(1-kx^3)$ by direct integration
Given. Plane area in the first quadrant bounded by the $x$-axis, the $y$-axis, and the curve $y=b(1-kx^3)$, with $y$-intercept $b$ (at $x=0$) and $x$-intercept $a$ (at $x=a$).
Find. The centroid $(\bar x,\bar y)$ in terms of $a$ and $b$, by direct integration.
Figure 3B — area under $y=b(1-x^3/a^3)$ with its centroid.
Approach. Since the $x$-intercept is $a$, the constant $k=1/a^3$, so $y=b(1-x^3/a^3)$. Use single-integration (vertical strip) formulas $A=\int y\,dx$, $\bar x A=\int xy\,dx$, $\bar y A = \int \tfrac{y^2}{2}\,dx$.
First moment about the $y$-axis, for $\bar x$.
$$\int_0^a x\,y\,dx = b\int_0^a\!\left(x-\frac{x^4}{a^3}\right)dx = b\left(\frac{a^2}{2}-\frac{a^2}{5}\right) = \frac{3a^2b}{10}$$
$$\bar x = \frac{3a^2b/10}{3ab/4} = \frac{4a}{10} = \boxed{\frac{2a}{5}}$$
First moment about the $x$-axis, for $\bar y$.
$$\int_0^a \frac{y^2}{2}\,dx = \frac{b^2}{2}\int_0^a\!\left(1-\frac{2x^3}{a^3}+\frac{x^6}{a^6}\right)dx = \frac{b^2}{2}\left(a-\frac{a}{2}+\frac{a}{7}\right) = \frac{9ab^2}{28}$$
$$\bar y = \frac{9ab^2/28}{3ab/4} = \frac{9b}{28}\cdot\frac{4}{3} = \boxed{\frac{3b}{7}}$$
