Statics and Dynamics Quiz: Parallel Axis Theorem

What this quiz covers

This quiz focuses on Parallel Axis Theorem, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $d_{\text{horiz leg}} = |22.4 - 5| = 17.4$ mm and $d_{\text{vert leg}} = |22.4 - 50| = 27.6$ mm, because the parallel-axis theorem for the vertical centroidal axis uses the vertical (y-direction) offsets between each part's centroid and the composite centroid.
2. $d_{\text{horiz leg}} = |50 - 17.4| = 32.6$ mm and $d_{\text{vert leg}} = |5 - 17.4| = 12.4$ mm, since the parallel-axis distance for a vertical axis is the horizontal offset between each part's centroid and the composite centroidal vertical axis. (correct answer)
3. $d_{\text{horiz leg}} = \sqrt{(50-17.4)^2 + (5-22.4)^2}$ mm and $d_{\text{vert leg}} = \sqrt{(5-17.4)^2 + (45-22.4)^2}$ mm, because the parallel-axis theorem requires the full Euclidean distance between each part's centroid and the composite centroid.
4. $d_{\text{horiz leg}} = 50$ mm and $d_{\text{vert leg}} = 5$ mm, because the parallel-axis distances must be measured from the origin (outer corner datum) rather than from the composite centroidal axis, since the composite centroid location already accounts for the section geometry.

Question 2

1. 1.50 m
2. 0.87 m
3. 1.22 m (correct answer)
4. 6.00 m

Question 3

1. $3mR^2/2$
2. $5mR^2/4$ (correct answer)
3. $mR^2$
4. $mR^2/4$

Question 4

1. $mL^2/9$ (correct answer)
2. $4mL^2/9$
3. $13mL^2/36$
4. $7mL^2/36$

Question 5

1. 7 kg·m²
2. 15 kg·m² (correct answer)
3. 9 kg·m²
4. 12 kg·m²

Question 6

1. 3.0 kg·m²
2. 4.0 kg·m²
3. 1.0 kg·m²
4. 2.0 kg·m² (correct answer)

Question 7

1. $I_{\text{total}} = 3\left(\frac{1}{2}mr^2\right) = \frac{3}{2}mr^2$, because the three parallel-axis terms $mR^2$ cancel by symmetry since the displacement vectors of the three disk centers sum to zero for a symmetric $120°$ arrangement.
2. $I_{\text{total}} = 3\left(\frac{1}{4}mr^2 + mR^2\right) = \frac{3}{4}mr^2 + 3mR^2$, because the relevant centroidal moment for a disk about a diameter axis is $\frac{1}{4}mr^2$, which applies when the spin axis is perpendicular to the disk's own axis.
3. $I_{\text{total}} = 3\left(\frac{1}{2}mr^2 + mR^2\right) = \frac{3}{2}mr^2 + 3mR^2$, because each disk contributes its own centroidal polar moment $\frac{1}{2}mr^2$ plus the parallel-axis shift $mR^2$, and the three shifts add rather than cancel due to the symmetric but non-collinear arrangement. (correct answer)
4. $I_{\text{total}} = \frac{1}{2}(3m)(3r)^2 = \frac{27}{2}mr^2$, because the assembly can be treated as a single equivalent disk of total mass $3m$ and effective radius $3r$ about the central axis, leveraging the rotational symmetry.

Question 8

1. $I_{\bar{x}} = \frac{\pi R^4}{8} - \frac{\pi R^2}{2}\left(\frac{4R}{3\pi}\right)^2 = \frac{\pi R^4}{8} - \frac{4R^4}{9\pi}$, because the $Ad^2$ term evaluates to $\frac{\pi R^2}{2} \cdot \frac{16R^2}{9\pi^2} = \frac{4R^4}{9\pi}$ after canceling one factor of $\pi$.
2. $I_{\bar{x}} = \frac{\pi R^4}{8} + \frac{\pi R^2}{2}\left(\frac{4R}{3\pi}\right)^2 = \frac{\pi R^4}{8} + \frac{8R^4}{9\pi}$, because shifting from a non-centroidal axis to the centroidal axis always adds the $Ad^2$ term rather than subtracting it.
3. $I_{\bar{x}} = \frac{\pi R^4}{8} - \frac{\pi R^2}{2}\left(\frac{4R}{3\pi}\right) = \frac{\pi R^4}{8} - \frac{2R^3}{3}$, because the parallel-axis correction for a reverse shift uses $Ad$ (first power) rather than $Ad^2$.
4. $I_{\bar{x}} = \frac{\pi R^4}{8} - \frac{\pi R^2}{2}\left(\frac{4R}{3\pi}\right)^2 = \frac{\pi R^4}{8} - \frac{8R^4}{9\pi}$, which simplifies to $I_{\bar{x}} = R^4\left(\frac{\pi}{8} - \frac{8}{9\pi}\right)$. (correct answer)

Question 9

1. The expression is correct. It yields approximately $I_x \approx 2.41 \times 10^7 \text{ mm}^4$, obtained by adding each part's own centroidal inertia to the parallel-axis term $A d^2$, with distances measured from each part's centroid to the composite centroid. (correct answer)
2. The expression is incorrect because the parallel-axis distances should be measured from the bottom of the section (datum), not from the composite centroid. The correct distances are $d_F = 110$ mm and $d_W = 50$ mm, yielding $I_x \approx 5.62 \times 10^7 \text{ mm}^4$.
3. The expression is incorrect because the parallel-axis theorem requires squaring the sum of the individual offsets: $I_x = I_{\bar{x},F} + A_F(75.4 + 15.4)^2 + I_{\bar{x},W}$, which gives $I_x \approx 3.54 \times 10^7 \text{ mm}^4$.
4. The expression is correct in form but numerically incorrect because the composite centroid is at 34.6 mm, so $d_F$ should be $110 - 34.6 = 75.4$ mm and $d_W$ should be $34.6 - 50 = -15.4$ mm; since $d_W$ is negative, it should not be squared, yielding $I_x \approx 2.28 \times 10^7 \text{ mm}^4$.

Question 10

1. The second step is valid only if $s \geq d$, because the parallel-axis theorem requires the destination axis to be farther from the centroid than the starting axis, ensuring the moment of inertia increases monotonically with distance.
2. The second step is always valid because the parallel-axis theorem is a general shift theorem between any two parallel axes, provided the shift distance $s$ is measured correctly between the two parallel axes in question.
3. The second step is valid only if $d = 0$, meaning the intermediate axis coincidentally passes through the centroid, in which case the two-step procedure reduces to the standard single application of the theorem.
4. The second step is never valid in this form, because the parallel-axis theorem requires the starting axis to be centroidal. To find $I'$, one must return to the centroid: $I' = \frac{1}{12}mL^2 + m(d+s)^2$, using the total distance from the centroid to the final axis. (correct answer)

Question 11

1. $I_{\text{total}} = 2\left(\frac{1}{2}mR^2\right) + 2m(R)^2 = mR^2 + 2mR^2 = 3mR^2$, because each disk's center is $R$ from the midpoint axis, so each parallel-axis term contributes $m(R)^2$. (correct answer)
2. $I_{\text{total}} = 2\left(\frac{1}{2}mR^2\right) + 2m(2R)^2 = mR^2 + 8mR^2 = 9mR^2$, because the parallel-axis distance for each disk is the full center-to-center separation $2R$, not the half-distance.
3. $I_{\text{total}} = 2\left(\frac{1}{2}mR^2\right) = mR^2$, because the midpoint axis is equidistant from both disk centers and the parallel-axis corrections for the two disks cancel each other out by symmetry.
4. $I_{\text{total}} = \frac{1}{2}(2m)(2R)^2 = 4mR^2$, because the combined system can be treated as a single equivalent disk of total mass $2m$ and effective radius $2R$ about the midpoint axis.

Question 12

1. Neither is correct. The parallel-axis distance should be $\frac{a+b}{4}$, which represents the average of the outer and inner half-dimensions and correctly locates the centroid of the hollow cross-section.
2. The colleague is correct. The parallel-axis distance must reflect the centroid of the material (the hollow walls), which is located at the mid-thickness of the wall, a distance $\frac{a-b}{2}$ from the outer surface.
3. The engineer is correct. The parallel-axis distance is measured from the composite section's centroid to the target axis. The centroid of the hollow tube lies at its geometric center, which is $a/2$ from the outer bottom edge, regardless of the wall thickness or inner dimension $b$. (correct answer)
4. The engineer is correct in distance but uses the wrong area. The area of the hollow section is $a^2 - b^2$, but the parallel-axis theorem for a hollow section requires using the outer area $a^2$ and subtracting the inner contribution $b^2(b/2)^2$ separately.

Question 13

1. $I_x = \frac{1}{12}(120)(80)^3 - \left[\frac{\pi}{4}(20)^4 + \pi(20)^2(20)^2\right] \text{ mm}^4$, because the circular hole's centroid is 20 mm above the rectangle's centroid, so the parallel-axis shift uses $d = 20$ mm. (correct answer)
2. $I_x = \frac{1}{12}(120)(80)^3 - \left[\frac{\pi}{4}(20)^4 + \pi(20)^2(30)^2\right] \text{ mm}^4$, because the horizontal-axis parallel-axis theorem requires using the full radial offset of 30 mm rather than only the vertical component.
3. $I_x = \frac{1}{12}(120)(80)^3 - \left[\frac{\pi}{4}(20)^4 + \pi(20)^2(1300)\right] \text{ mm}^4$, because the distance used in the parallel-axis theorem is the total distance $d = \sqrt{30^2 + 20^2} = \sqrt{1300}$ mm from the rectangle's centroid to the circle's centroid.
4. $I_x = \frac{1}{12}(120)(80)^3 - \frac{\pi}{4}(20)^4 \text{ mm}^4$, because the circle's own centroidal axis is already parallel to the x-axis and the parallel-axis theorem only applies when there is a vertical offset between centroids.

Question 14

1. Step (3) is invalid because the reverse parallel-axis theorem subtracts $A_{\text{hemi}}\,d$, not $A_{\text{hemi}}\,d^2$, when converting from a boundary (flat-face) axis to the interior centroidal axis. The correct result is $I_{\bar{x},\text{hemi}} = \frac{mR^2}{5} - \frac{m}{2}\cdot\frac{3R}{8}$.
2. Step (3) is valid. $I_{\bar{x},\text{hemi}} = \frac{mR^2}{5} - \frac{9mR^2}{128} = mR^2\left(\frac{128 - 45}{640}\right) = \frac{83mR^2}{640}$, obtained by correctly subtracting $A_{\text{hemi}}\,d^2$ from the flat-face inertia. (correct answer)
3. Step (3) is invalid because the centroid offset $\bar{y} = 3R/8$ is measured from the flat face, but the parallel-axis theorem requires the distance from the sphere's center, which is $R - 3R/8 = 5R/8$. The correct expression is $I_{\bar{x},\text{hemi}} = \frac{mR^2}{5} - \frac{m}{2}\left(\frac{5R}{8}\right)^2$.
4. Step (3) is valid in form but uses the wrong mass. The hemisphere's mass is $m/2$, so the base inertia from step (2) should be expressed as $\frac{2}{5}\cdot\frac{m}{2}\cdot R^2 = \frac{mR^2}{5}$ before applying the correction, giving $I_{\bar{x},\text{hemi}} = \frac{mR^2}{5} - \frac{m}{2}\left(\frac{3R}{8}\right)^2 = \frac{83mR^2}{640}$.

Question 15

1. The second application is invalid because the parallel-axis theorem must start from a centroidal axis. Since $I_{\text{edge}}$ is not a centroidal value, using it as the base for another shift double-counts part of the centroid-to-edge distance. The correct result is found in one step: $I_{\text{target}} = \frac{bh^3}{12} + bh(h)^2 = \frac{13bh^3}{12}$, using the total distance $d = h$ from the centroid to the target axis. (correct answer)
2. The second application is valid because the parallel-axis theorem can be applied repeatedly between any two parallel axes, regardless of whether the intermediate axis is centroidal, provided each incremental shift distance is measured correctly between the two axes in question.
3. The second application is invalid because the target axis lies outside the cross-section, and the parallel-axis theorem only applies to axes that pass through or within the body's material boundaries.
4. The second application is valid because once the moment of inertia about the edge is known, applying the theorem again with $d = h/2$ correctly accounts for the additional shift, and the result $\frac{bh^3}{3} + bh(h/2)^2 = \frac{5bh^3}{12}$ is confirmed by the additive nature of area moments.