Statics Quiz: Polar Moment Of Inertia
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Polar Moment Of InertiaQuestion 1 of 14

Two identical rectangular strips, each with dimensions 60 mm × 10 mm, are arranged to form an L-shaped cross-section. If the polar moment of inertia about the centroid of the individual strip is JstripJ_{strip}, which expression gives the polar moment of inertia about the centroid of the L-shaped section?

Jtotal=2JstripJ_{total} = 2J_{strip} since the strips are identical and simply added together
Jtotal=2Jstrip+2Ad2J_{total} = 2J_{strip} + 2Ad^2 where dd is the distance from each strip's centroid to the L-section's centroid
Jtotal=Jstrip+Ad2J_{total} = J_{strip} + Ad^2 since only one strip needs the parallel axis correction for the L-section geometry
Jtotal=2Jstrip+Ad2J_{total} = 2J_{strip} + Ad^2 where AA is the area of one strip and dd is the average distance correction
Jtotal=4JstripJ_{total} = 4J_{strip} because the L-configuration doubles the effective contribution of each strip's polar moment
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Statics Quiz

Statics Quiz: Polar Moment Of Inertia

Practice Polar Moment Of Inertia in Statics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Polar Moment Of Inertia, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics.

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Question 1

Two identical rectangular strips, each with dimensions 60 mm × 10 mm, are arranged to form an L-shaped cross-section. If the polar moment of inertia about the centroid of the individual strip is JstripJ_{strip}, which expression gives the polar moment of inertia about the centroid of the L-shaped section?

  1. Jtotal=2JstripJ_{total} = 2J_{strip} since the strips are identical and simply added together
  2. Jtotal=2Jstrip+2Ad2J_{total} = 2J_{strip} + 2Ad^2 where dd is the distance from each strip's centroid to the L-section's centroid (correct answer)
  3. Jtotal=Jstrip+Ad2J_{total} = J_{strip} + Ad^2 since only one strip needs the parallel axis correction for the L-section geometry
  4. Jtotal=2Jstrip+Ad2J_{total} = 2J_{strip} + Ad^2 where AA is the area of one strip and dd is the average distance correction
  5. Jtotal=4JstripJ_{total} = 4J_{strip} because the L-configuration doubles the effective contribution of each strip's polar moment
Explanation: When analyzing composite cross-sections like this L-shape, you need to apply the parallel axis theorem to transfer moments of inertia from individual centroids to the overall centroid of the combined section. The polar moment of inertia for the L-shaped section requires calculating where the new centroid lies, then accounting for how far each strip's centroid is from this new location. Since the L-shape has its centroid at a different location than either individual strip, both strips need parallel axis corrections. For each strip, you start with its polar moment of inertia JstripJ_{strip} about its own centroid, then add Ad2Ad^2 where AA is the strip's area and dd is the distance from the strip's centroid to the L-section's centroid. Since you have two strips, each requiring this correction, the total becomes Jtotal=2Jstrip+2Ad2J_{total} = 2J_{strip} + 2Ad^2, making answer B correct. Answer A incorrectly assumes you can simply add the moments without any geometric corrections - this would only work if both strips were already calculated about the same reference point. Answer C makes the error of thinking only one strip needs correction, but both strips are displaced from the L-section's centroid. Answer D attempts to use some kind of "average" correction factor, but the parallel axis theorem requires individual corrections for each component. Study tip: For composite sections, always remember that every component needs its own parallel axis correction unless it's already calculated about the final centroid. Never just add moments of inertia without checking reference points first.

Question 2

A solid circular shaft has a diameter of 50 mm. If the polar moment of inertia is calculated using the standard formula, what happens to the polar moment of inertia if the diameter is increased by a factor of 1.5?

  1. It increases by a factor of 1.5
  2. It increases by a factor of 2.25
  3. It increases by a factor of 5.06 (correct answer)
  4. It increases by a factor of 3.38
  5. It increases by a factor of 7.59
Explanation: When you encounter problems involving moments of inertia and geometric scaling, the key insight is understanding how area-based properties scale with dimensional changes. The polar moment of inertia depends on the fourth power of the radius. For a solid circular shaft, the polar moment of inertia is J=πd432J = \frac{\pi d^4}{32}, where d is the diameter. This formula shows that J is proportional to the fourth power of the diameter. When the diameter increases by a factor of 1.5, the new polar moment of inertia becomes: Jnew=π(1.5d)432=πd4⋅(1.5)432J_{new} = \frac{\pi (1.5d)^4}{32} = \frac{\pi d^4 \cdot (1.5)^4}{32} Since (1.5)4=5.0625≈5.06(1.5)^4 = 5.0625 ≈ 5.06, the polar moment of inertia increases by a factor of 5.06. Option A (1.5) incorrectly assumes a linear relationship, as if the polar moment scales directly with diameter. Option B (2.25) represents (1.5)2(1.5)^2, which would be correct for second moment of area calculations but not polar moment of inertia. Option D (3.38) represents (1.5)3(1.5)^3, which might come from confusing this with a volume-related property. Remember that geometric properties scale predictably: linear dimensions scale to the first power, areas to the second power, and moments of inertia to the fourth power. Always check the exponent in the formula to determine the scaling factor—raise the dimensional multiplier to that same power.

Question 3

Two circular cross-sections have the same total cross-sectional area. One is solid with radius rr, and the other is hollow with outer radius RR and inner radius 0.8R0.8R. Which statement correctly compares their polar moments of inertia?

  1. The solid shaft has a larger polar moment of inertia because it has no hollow center to weaken it
  2. The hollow shaft has a larger polar moment of inertia because material is distributed farther from the center (correct answer)
  3. Both shafts have identical polar moments of inertia since they have the same cross-sectional area
  4. The comparison depends on the specific material properties and cannot be determined geometrically
  5. The solid shaft has exactly twice the polar moment of inertia due to its continuous cross-section
Explanation: When comparing structural properties like polar moment of inertia, the key insight is that it's not just about how much material you have, but where that material is located relative to the axis of rotation. Polar moment of inertia measures resistance to torsion and depends heavily on the distance of material from the center. Let's verify this with the given information. Both cross-sections have equal areas, so:
  • Solid: A=πr2A = \pi r^2
  • Hollow: A=πR2−π(0.8R)2=πR2(1−0.64)=0.36πR2A = \pi R^2 - \pi(0.8R)^2 = \pi R^2(1 - 0.64) = 0.36\pi R^2
Setting these equal: πr2=0.36πR2\pi r^2 = 0.36\pi R^2, so r=0.6Rr = 0.6R. For polar moments of inertia:
  • Solid: J=πr42=π(0.6R)42=0.065πR4J = \frac{\pi r^4}{2} = \frac{\pi (0.6R)^4}{2} = 0.065\pi R^4
  • Hollow: J=π2[R4−(0.8R)4]=πR42(1−0.41)=0.295πR4J = \frac{\pi}{2}[R^4 - (0.8R)^4] = \frac{\pi R^4}{2}(1 - 0.41) = 0.295\pi R^4
The hollow shaft's polar moment is about 4.5 times larger because its material sits farther from the center. Choice A incorrectly assumes having a solid center automatically means higher stiffness. Choice C falls into the trap of thinking equal areas mean equal moments of inertia – this ignores the critical r4r^4 relationship in the polar moment formula. Choice D suggests material properties matter, but polar moment of inertia is purely geometric. Study tip: Remember that for torsional resistance, material placement trumps material quantity. Hollow sections are often superior because they maximize the distance from the neutral axis.

Question 4

The relationship between the polar moment of inertia JJ and the rectangular area moments of inertia IxI_x and IyI_y for any cross-section is J=Ix+IyJ = I_x + I_y. For a circular cross-section, why are IxI_x and IyI_y equal?

  1. Because the cross-sectional area is uniformly distributed in all radial directions from the center
  2. Because the circular cross-section has rotational symmetry about any axis through its centroid (correct answer)
  3. Because the moment of inertia formula is the same for both x and y directions in circular geometries
  4. Because the radius is constant at all points on the circumference of the circular boundary
  5. Because circular cross-sections have equal width and height when measured along perpendicular axes
Explanation: When analyzing moments of inertia for cross-sections, you need to understand how geometric symmetry affects these calculations. The polar moment of inertia JJ relates to rectangular moments through J=Ix+IyJ = I_x + I_y, but the key insight here is recognizing why Ix=IyI_x = I_y for circles. For a circular cross-section, IxI_x and IyI_y are equal because of rotational symmetry about any axis through the centroid. This means you can rotate the coordinate system by any angle around the center, and the cross-section looks identical. Since the x and y axes are just perpendicular references through the centroid, rotating them 90° gives you the same geometric distribution of area. Therefore, IxI_x must equal IyI_y. Answer A is incorrect because "uniformly distributed in radial directions" describes area distribution, but moments of inertia depend on the distance squared from the axis, not just uniform distribution. Answer C misses the point entirely – the moment of inertia formula I=∫y2dAI = \int y^2 dA is the same for any axis, but that doesn't explain why the values are equal for different axes. Answer D confuses boundary geometry with the area moment calculation. While the radius is constant on the circumference, moments of inertia integrate over the entire area, not just the boundary. Remember this pattern: whenever you see equal moments of inertia about perpendicular axes, look for rotational symmetry. This concept applies to any shape with rotational symmetry (circles, squares rotated 45°, regular polygons), making it a powerful tool for recognizing when Ix=IyI_x = I_y.

Question 5

When calculating the polar moment of inertia for a composite shape using the parallel axis theorem, which approach is most likely to lead to an error?

  1. Forgetting to calculate the centroid location of the overall composite section before applying parallel axis corrections
  2. Using the wrong sign convention when measuring distances from individual centroids to the composite centroid
  3. Adding the individual polar moments of inertia directly without applying any parallel axis theorem corrections (correct answer)
  4. Calculating the area moments IxI_x and IyI_y first, then applying parallel axis theorem to each, then adding them to get JJ
  5. Using the parallel axis theorem formula J=Jc+Ad2J = J_c + Ad^2 where dd is measured along a coordinate direction rather than as the radial distance
Explanation: When analyzing composite shapes for polar moment of inertia, you must account for how each component's moment changes when referenced to the composite centroid rather than its own centroid. The polar moment of inertia represents a cross-section's resistance to torsion. For composite shapes, you cannot simply add the individual polar moments because each component's tabulated value (J0J_0) is calculated about its own centroid, not the composite centroid where you need the final result. Option C describes exactly this fundamental error—adding individual polar moments directly without parallel axis corrections. This ignores the fact that when you move the reference point from each component's centroid to the composite centroid, you must add Aidi2A_i d_i^2 for each component, where did_i is the distance between centroids. Skipping this step severely underestimates the actual polar moment of inertia. Option A is actually good practice, not an error—you need the composite centroid location to apply corrections properly. Option B represents a calculation mistake with sign conventions, but the parallel axis theorem uses d2d^2, so signs cancel out anyway. Option D describes a perfectly valid alternative approach: calculate IxI_x and IyI_y with their respective parallel axis corrections, then use J=Ix+IyJ = I_x + I_y. The key insight is that moments of inertia are always referenced to a specific axis or point. When you change that reference point, you must apply the parallel axis theorem to account for the additional rotational inertia from the displaced mass.

Question 6

The polar moment of inertia of an equilateral triangle with side length ss about its centroid is J=s424J = \frac{s^4}{24}. For an equilateral triangle with side length 36 mm, what is the polar moment of inertia?

  1. 5.83×104 mm45.83 \times 10^4 \text{ mm}^4
  2. 1.17×105 mm41.17 \times 10^5 \text{ mm}^4
  3. 8.75×104 mm48.75 \times 10^4 \text{ mm}^4
  4. 7.00×104 mm47.00 \times 10^4 \text{ mm}^4 (correct answer)
  5. 9.33×104 mm49.33 \times 10^4 \text{ mm}^4
Explanation: When you encounter polar moment of inertia problems, you're dealing with a measure of how mass is distributed around a point - in this case, the centroid of the triangle. The key is recognizing that you have a direct formula and simply need to substitute the given values carefully. Given the formula J=s424J = \frac{s^4}{24} and side length s=36s = 36 mm, you substitute directly: J=(36)424=1,679,61624=69,984 mm4J = \frac{(36)^4}{24} = \frac{1,679,616}{24} = 69,984 \text{ mm}^4 This rounds to 7.00×104 mm47.00 \times 10^4 \text{ mm}^4, which is answer D. Let's examine why the other options are incorrect: Option A (5.83×1045.83 \times 10^4) results from using s3s^3 instead of s4s^4 in the calculation - a common error when students misremember the formula's power relationship. Option B (1.17×1051.17 \times 10^5) comes from using the wrong denominator, likely 12 instead of 24. This might happen if you confuse this formula with similar moment of inertia expressions for other geometric properties. Option C (8.75×1048.75 \times 10^4) appears when students use an incorrect constant in the denominator, perhaps 18 or 20, showing incomplete memorization of the specific formula. Remember: polar moment of inertia formulas are geometry-specific and must be memorized exactly. Always double-check that you're using the correct power (here, s4s^4) and the right constant (here, 24). These problems test both formula recall and careful arithmetic - write out each step to avoid calculation errors.

Question 7

Two circular shafts are welded end-to-end to form a stepped shaft. The first section has diameter 20 mm and length 100 mm; the second has diameter 30 mm and length 150 mm. When analyzing torsional properties, which statement about the polar moments of inertia is correct?

  1. The overall polar moment of inertia is the sum of the individual sections' polar moments since they are in series
  2. The effective polar moment of inertia is dominated by the smaller section since it creates the bottleneck for torsional resistance
  3. Each section retains its individual polar moment of inertia, and torsional analysis must consider each section separately (correct answer)
  4. The polar moment of inertia should be calculated using the average diameter of the two sections for the combined length
  5. The larger section's polar moment of inertia becomes negligible compared to the smaller section due to the step change
Explanation: When analyzing stepped shafts in torsion, you're dealing with a composite structure where different sections have different geometric properties. The key insight is that torsional deformation and stress vary along the shaft's length based on each section's individual properties. Answer C is correct because each section of a stepped shaft maintains its own polar moment of inertia based on its individual diameter: J=πd432J = \frac{\pi d^4}{32}. The 20mm section has its polar moment, and the 30mm section has its own different value. When solving torsional problems, you must analyze each section separately using its specific properties, then combine the results (like angles of twist) appropriately. Answer A incorrectly suggests adding polar moments like resistors in series. This fundamentally misunderstands that polar moment of inertia is a geometric property of each cross-section, not a quantity that combines algebraically along the shaft length. Answer B confuses the concept of limiting factors in structural analysis. While the smaller section will indeed experience higher shear stress for the same applied torque, this doesn't mean you use only its polar moment for the entire shaft analysis. Answer D represents a common but incorrect averaging approach. You cannot average diameters and treat the stepped shaft as uniform - this ignores the physics of how torsional stress and deformation actually occur in each distinct section. Remember: stepped shafts require section-by-section analysis. Each cross-section's properties govern the behavior at that location, so calculate polar moments individually and analyze each section using its own geometry.

Question 8

In the parallel axis theorem for polar moment of inertia, J=Jc+Ad2J = J_c + Ad^2, the term Ad2Ad^2 represents what physical concept?

  1. The additional rotational resistance created by moving the reference axis away from the material's geometric center
  2. The correction factor needed to account for the increased lever arm of each area element about the new axis
  3. The mathematical compensation required because polar moment of inertia is not an additive property when axes are displaced
  4. The additional second moment contribution from treating the entire cross-sectional area as concentrated at its centroid (correct answer)
  5. The geometric scaling factor that relates polar moments of inertia calculated about different parallel axes through similar shapes
Explanation: The parallel axis theorem is fundamentally about how moments of inertia change when you shift your reference axis. Understanding what each term represents helps you grasp the underlying physics. The correct answer is D. The term Ad2Ad^2 comes from a clever mathematical insight: instead of calculating how each tiny area element's distance changes when you move the axis, you can get the same result by imagining all the cross-sectional area AA is concentrated at a single point (the centroid) located distance dd from the new axis. This concentrated area contributes Ad2Ad^2 to the polar moment of inertia. It's essentially treating the entire shape as a point mass for this additional contribution. Answer A incorrectly describes this as "rotational resistance." While Ad2Ad^2 does increase the total moment of inertia, it's not about resistance—it's a geometric property based on area distribution. Answer B mentions "lever arm correction," but Ad2Ad^2 isn't correcting individual lever arms. Each area element already has its proper distance accounted for in the full calculation. Answer C suggests polar moment of inertia isn't additive, which is wrong. Moments of inertia are additive properties—that's exactly why the parallel axis theorem works by adding Jc+Ad2J_c + Ad^2. Remember this pattern: in parallel axis problems, the added term always represents the "point mass equivalent" contribution. This same concept appears in linear moment of inertia (I=Ic+Ad2I = I_c + Ad^2) and helps you understand why moving away from centroids always increases moments of inertia.

Question 9

A cross-section has polar moment of inertia J=850×103 mm4J = 850 \times 10^3 \text{ mm}^4 about its centroid. If this same cross-section's polar moment of inertia about a parallel axis 25 mm away is 1.45×106 mm41.45 \times 10^6 \text{ mm}^4, what is the cross-sectional area?

  1. 840 mm2840 \text{ mm}^2
  2. 960 mm2960 \text{ mm}^2 (correct answer)
  3. 1080 mm21080 \text{ mm}^2
  4. 720 mm2720 \text{ mm}^2
  5. 1200 mm21200 \text{ mm}^2
Explanation: When you encounter problems involving moments of inertia about different axes, you're dealing with the parallel axis theorem. This fundamental relationship connects the moment of inertia about a centroidal axis to that about any parallel axis. The parallel axis theorem for polar moments of inertia states: Jp=Jc+Ad2J_p = J_c + Ad^2, where JpJ_p is the polar moment about the parallel axis, JcJ_c is the polar moment about the centroidal axis, AA is the cross-sectional area, and dd is the distance between axes. Substituting the given values: 1.45×106=850×103+A(25)21.45 \times 10^6 = 850 \times 10^3 + A(25)^2 Solving for area: 1.45×106−850×103=625A1.45 \times 10^6 - 850 \times 10^3 = 625A 600×103=625A600 \times 10^3 = 625A A=960 mm2A = 960 \text{ mm}^2 This confirms answer (B) is correct. Looking at the wrong answers: (A) 840 mm² results from incorrectly using d=20d = 20 mm instead of 25 mm. (C) 1080 mm² comes from using d=30d = 30 mm, suggesting a misreading of the given distance. (D) 720 mm² occurs when students mistakenly subtract the parallel axis term instead of adding it, reversing the theorem's logic. Study tip: Always write out the parallel axis theorem explicitly before substituting values. The key insight is that moments of inertia are always larger about non-centroidal axes, so Jp>JcJ_p > J_c. If your calculation suggests otherwise, check your setup immediately.

Question 10

Which of the following statements about polar moment of inertia is fundamentally incorrect?

  1. Polar moment of inertia has units of length to the fourth power in consistent unit systems
  2. For any cross-section, the polar moment of inertia equals the sum of two perpendicular area moments of inertia
  3. Polar moment of inertia is always larger than either individual rectangular area moment of inertia for the same section
  4. The polar moment of inertia represents the resistance of a cross-section to torsional deformation under applied torque (correct answer)
  5. Material located farther from the centroid contributes more significantly to the polar moment of inertia than material closer to the center
Explanation: When you encounter polar moment of inertia questions, you're dealing with a geometric property that's crucial for understanding both bending and torsional behavior of structural members. The key insight here is recognizing what polar moment of inertia actually represents versus what it's commonly used for in engineering calculations. While polar moment of inertia J=∫r2 dAJ = \int r^2 \, dA appears in torsional stress formulas, it fundamentally represents a geometric property of the cross-section, not the material's resistance to deformation. Answer D is incorrect because it confuses the geometric property with mechanical behavior. Polar moment of inertia is purely a function of geometry - specifically, how area is distributed relative to a point. The actual resistance to torsional deformation depends on both this geometric property AND the material's shear modulus. It's like saying that the cross-sectional area of a rod represents its resistance to axial force - the geometry contributes, but you need the material properties too. Answer A is correct - JJ has units of length4\text{length}^4 since we integrate r2r^2 over area. Answer B correctly states the perpendicular axis theorem: J=Ix+IyJ = I_x + I_y for any point. Answer C is mathematically sound since J=Ix+IyJ = I_x + I_y, making JJ necessarily larger than either individual moment of inertia (assuming both are positive). Remember this distinction: geometric properties describe shape and size, while resistance to deformation requires both geometry and material properties. Don't let familiar applications blur the fundamental definitions.

Question 11

For a cross-section where Ix=450×103 mm4I_x = 450 \times 10^3 \text{ mm}^4 and Iy=280×103 mm4I_y = 280 \times 10^3 \text{ mm}^4, what is the polar moment of inertia about the centroid?

  1. 170×103 mm4170 \times 10^3 \text{ mm}^4
  2. 365×103 mm4365 \times 10^3 \text{ mm}^4
  3. 730×103 mm4730 \times 10^3 \text{ mm}^4 (correct answer)
  4. 530×103 mm4530 \times 10^3 \text{ mm}^4
  5. 630×103 mm4630 \times 10^3 \text{ mm}^4
Explanation: When you encounter polar moment of inertia problems, you're dealing with a cross-section's resistance to torsional (twisting) deformation. The key relationship to remember is that the polar moment of inertia JJ equals the sum of the two rectangular moments of inertia about perpendicular axes through the centroid. The fundamental formula is: J=Ix+IyJ = I_x + I_y This relationship exists because the polar moment of inertia represents how area is distributed relative to a point (the centroid), while IxI_x and IyI_y represent area distribution relative to the x and y axes respectively. When you add them together, you get the total effect about the centroidal point. Substituting the given values: J=450×103+280×103=730×103 mm4J = 450 \times 10^3 + 280 \times 10^3 = 730 \times 10^3 \text{ mm}^4 This confirms answer C is correct. Looking at the wrong answers: A (170×103170 \times 10^3) represents the difference Ix−IyI_x - I_y, which has no physical meaning for polar moment calculations. B (365×103365 \times 10^3) is exactly half the correct answer, suggesting someone might have incorrectly averaged the two values instead of adding them. D (530×103530 \times 10^3) appears to be a calculation error, possibly from misreading one of the given values. Study tip: Always remember J=Ix+IyJ = I_x + I_y for polar moment of inertia. This is one of the most straightforward formulas in statics—resist the urge to overthink it or apply more complex relationships when this simple addition is all that's needed.

Question 12

A thin-walled circular tube has a mean radius of 25 mm and wall thickness of 2 mm. Using the thin-wall approximation, what is the polar moment of inertia?

  1. 3.93×103 mm43.93 \times 10^3 \text{ mm}^4
  2. 9.82×104 mm49.82 \times 10^4 \text{ mm}^4 (correct answer)
  3. 1.96×105 mm41.96 \times 10^5 \text{ mm}^4
  4. 7.85×104 mm47.85 \times 10^4 \text{ mm}^4
  5. 4.91×104 mm44.91 \times 10^4 \text{ mm}^4
Explanation: When you encounter thin-walled circular tubes in statics problems, you're dealing with torsional mechanics where the polar moment of inertia determines the structure's resistance to twisting. The thin-wall approximation simplifies calculations by assuming all material is concentrated at the mean radius. For a thin-walled circular tube, the polar moment of inertia formula is: J=2πrm3tJ = 2\pi r_m^3 t where rmr_m is the mean radius and tt is the wall thickness. Substituting the given values: J=2π(25)3(2)=2π(15,625)(2)=98,175 mm4J = 2\pi (25)^3 (2) = 2\pi (15,625)(2) = 98,175 \text{ mm}^4 This equals 9.82×104 mm49.82 \times 10^4 \text{ mm}^4, confirming answer B. Looking at the incorrect options: A (3.93×103 mm43.93 \times 10^3 \text{ mm}^4) appears to result from using the wrong formula or forgetting the factor of 2π. C (1.96×105 mm41.96 \times 10^5 \text{ mm}^4) is exactly double the correct answer, suggesting someone might have applied the formula twice or used an incorrect coefficient. D (7.85×104 mm47.85 \times 10^4 \text{ mm}^4) is close but seems to stem from a calculation error, possibly dropping the factor of 2 in the formula. Study tip: Always remember that thin-wall polar moments use J=2πrm3tJ = 2\pi r_m^3 t, not the solid shaft formula. The key is recognizing when the thin-wall approximation applies (typically when wall thickness is much smaller than radius) and using the mean radius consistently throughout your calculation.

Question 13

A cross-section consists of two identical circular sectors, each with radius RR and central angle π/2\pi/2, arranged so that their straight edges are perpendicular and meet at the common center. What is the polar moment of inertia about the center point?

  1. πR48\frac{\pi R^4}{8}
  2. πR44\frac{\pi R^4}{4} (correct answer)
  3. πR42\frac{\pi R^4}{2}
  4. 3πR48\frac{3\pi R^4}{8}
Explanation: For a circular sector with radius RR and angle θ\theta about its center: J=R4θ4J = \frac{R^4 \theta}{4}. For each sector with θ=π/2\theta = \pi/2: Jsector=R4π/24=πR48J_{sector} = \frac{R^4 \pi/2}{4} = \frac{\pi R^4}{8}. Since we have two identical sectors: Jtotal=2⋅πR48=πR44J_{total} = 2 \cdot \frac{\pi R^4}{8} = \frac{\pi R^4}{4}. Choice A gives value for single sector, choice C would be for a complete disk, choice D reflects an error in angle calculation or sector formula application.

Question 14

A hollow circular shaft has outer radius RR and inner radius R/3R/3. If this shaft is compared to a solid circular shaft of radius rr that has the same polar moment of inertia, what is the ratio r/Rr/R?

  1. 80814\sqrt[4]{\frac{80}{81}} (correct answer)
  2. 26274\sqrt[4]{\frac{26}{27}}
  3. 8081\sqrt{\frac{80}{81}}
  4. 2627\sqrt{\frac{26}{27}}
Explanation: For the hollow shaft: J1=π2[R4−(R/3)4]=πR42[1−181]=πR42⋅8081J_1 = \frac{\pi}{2}[R^4 - (R/3)^4] = \frac{\pi R^4}{2}[1 - \frac{1}{81}] = \frac{\pi R^4}{2} \cdot \frac{80}{81}. For the solid shaft: J2=πr42J_2 = \frac{\pi r^4}{2}. Setting equal: πr42=πR42⋅8081\frac{\pi r^4}{2} = \frac{\pi R^4}{2} \cdot \frac{80}{81}. Therefore: r4=R4⋅8081r^4 = R^4 \cdot \frac{80}{81}, so r/R=80814r/R = \sqrt[4]{\frac{80}{81}}. Choice B uses wrong fraction (26/27), choices C and D use square root instead of fourth root.