College Physics Quiz: Rotational Inertia

16 questions · exam conditions

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Rotational InertiaQuestion 1 of 16

A uniform rectangular plate has dimensions $2a$ by $3a$ and mass $M$ . What is the rotational inertia about an axis through the center of the plate, parallel to the side of length $2a$ ?

\frac{1}{4}Ma^2

\frac{3}{4}Ma^2

\frac{1}{2}Ma^2

\frac{2}{3}Ma^2

\frac{5}{6}Ma^2

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College Physics Quiz

College Physics Quiz: Rotational Inertia

Practice Rotational Inertia in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Rotational Inertia, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

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.

All questions

Question 1

A uniform rectangular plate has dimensions $2a$ by $3a$ and mass $M$ . What is the rotational inertia about an axis through the center of the plate, parallel to the side of length $2a$ ?

$\frac{1}{4}Ma^2$
$\frac{3}{4}Ma^2$ (correct answer)
$\frac{1}{2}Ma^2$
$\frac{2}{3}Ma^2$
$\frac{5}{6}Ma^2$

Explanation: When you encounter rotational inertia problems involving rectangular plates, you need to identify the axis of rotation and apply the appropriate moment of inertia formula. For a rectangular plate rotating about an axis through its center, the formula depends on which dimension is perpendicular to the rotation axis. The plate has dimensions

2a \times 3a

and rotates about an axis parallel to the

2a

side, meaning the axis is perpendicular to the

3a

dimension. For a uniform rectangular plate of mass

M

with width

w

and length

l

, rotating about a central axis parallel to the width, the moment of inertia is

I = \frac{1}{12}Ml^2

. In this case, the relevant dimension is the length perpendicular to the rotation axis:

l = 3a

. Therefore:

I = \frac{1}{12}M(3a)^2 = \frac{1}{12}M(9a^2) = \frac{3}{4}Ma^2

This confirms answer B is correct. Looking at the wrong answers: A (

\frac{1}{4}Ma^2

) likely results from incorrectly using the

2a

dimension or making an algebraic error. C (

\frac{1}{2}Ma^2

) might come from confusing this with the moment of inertia for a rod or using an incorrect formula. D (

\frac{2}{3}Ma^2

) could result from mixing up formulas for different geometric shapes, like spheres or cylinders. Study tip: Always identify which dimension is perpendicular to the rotation axis—that's the one that matters for the calculation. Memorize the basic rectangular plate formula:

I = \frac{1}{12}M(\text{perpendicular dimension})^2

Question 2

A solid cylinder and a thin cylindrical shell have the same mass $M$ and radius $R$ . Both start from rest and roll down identical inclined planes without slipping. Which statement correctly compares their rotational inertias and accelerations?

The solid cylinder has greater rotational inertia and greater acceleration down the incline
The cylindrical shell has greater rotational inertia and greater acceleration down the incline
The solid cylinder has smaller rotational inertia and greater acceleration down the incline (correct answer)
The cylindrical shell has smaller rotational inertia and smaller acceleration down the incline
Both objects have the same rotational inertia and acceleration since they have the same mass and radius

Explanation: When objects roll down inclines, you need to analyze both rotational inertia and the dynamics of rolling motion. The key insight is that objects with different mass distributions will have different rotational inertias, which directly affects their acceleration. First, let's compare rotational inertias. For a solid cylinder,

I = \frac{1}{2}MR^2

, while for a thin cylindrical shell,

I = MR^2

. The shell has twice the rotational inertia because its mass is concentrated farther from the rotation axis. For rolling without slipping down an incline, the acceleration is given by

a = \frac{g\sin\theta}{1 + \frac{I}{MR^2}}

. Substituting the rotational inertias: the solid cylinder gets

a = \frac{g\sin\theta}{1 + \frac{1}{2}} = \frac{2g\sin\theta}{3}

, while the shell gets

a = \frac{g\sin\theta}{1 + 1} = \frac{g\sin\theta}{2}

. The solid cylinder accelerates faster because less of the gravitational potential energy goes into rotational kinetic energy. Choice A incorrectly states the solid cylinder has greater rotational inertia. Choice B incorrectly claims the shell accelerates faster and also misidentifies which has greater rotational inertia. Choice D correctly identifies the shell's smaller acceleration but wrongly states it has smaller rotational inertia. Choice C correctly identifies both relationships. Study tip: Remember that objects with mass concentrated closer to the rotation axis (like solid cylinders) always roll faster down inclines than objects with mass farther out (like shells or rings), regardless of their total mass or size.

Question 3

A uniform disk of radius $R$ and mass $M$ rotates about a fixed axis through a point on its rim (not through its center). What is the rotational inertia of the disk about this axis?

$\frac{1}{2}MR^2$
$MR^2$
$\frac{3}{2}MR^2$ (correct answer)
$2MR^2$
$\frac{5}{2}MR^2$

Explanation: When you encounter rotational inertia problems involving axes that don't pass through the center of mass, you need to use the parallel axis theorem. This theorem relates the moment of inertia about any axis to the moment of inertia about a parallel axis through the center of mass. The parallel axis theorem states:

I = I_{cm} + Md^2

, where

I_{cm}

is the rotational inertia about the center of mass,

M

is the total mass, and

d

is the distance between the two parallel axes. For a uniform disk rotating about its center,

I_{cm} = \frac{1}{2}MR^2

. Since the new axis passes through a point on the rim, the distance from the center to this new axis is

d = R

. Therefore:

I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2

This confirms answer C is correct. Answer A (

\frac{1}{2}MR^2

) is simply the rotational inertia about the center of the disk, ignoring that the axis has moved to the rim. Answer B (

MR^2

) represents only the

Md^2

term from the parallel axis theorem, forgetting to include the original

I_{cm}

. Answer D (

2MR^2

) incorrectly doubles the center-of-mass inertia instead of properly applying the parallel axis theorem. Study tip: Whenever the rotation axis moves away from the center of mass, immediately think "parallel axis theorem." The rotational inertia always increases when you move the axis away from the center of mass—never stays the same or decreases.

Question 4

A thin ring of mass $M$ and radius $R$ has four point masses, each of mass $m$ , attached at equally spaced points around the ring. What is the total rotational inertia of this system about the central axis?

$MR^2 + 4mR^2$
$MR^2 + 2mR^2$
$(M + 4m)R^2$ (correct answer)
$MR^2 + \frac{4mR^2}{\pi}$
$\frac{1}{2}(M + 4m)R^2$

Explanation: When calculating rotational inertia for composite systems, you need to find the moment of inertia for each component separately, then add them together using the principle of superposition. For this system, you have two components: the thin ring and the four point masses. The thin ring has mass

M

distributed at radius

R

from the axis, giving it a rotational inertia of

MR^2

. Each point mass

m

is located at distance

R

from the central axis, so each contributes

mR^2

to the total rotational inertia. With four point masses, their combined contribution is

4mR^2

. Adding these components:

I_{total} = MR^2 + 4mR^2 = (M + 4m)R^2

, which is answer choice C. Let's examine why the other options are incorrect. Choice A (

MR^2 + 4mR^2

) is mathematically equivalent to the correct answer but isn't factored properly—it misses that you can factor out

R^2

. Choice B (

MR^2 + 2mR^2

) incorrectly accounts for only two point masses instead of four, or perhaps incorrectly applies a factor of 2. Choice D (

MR^2 + \frac{4mR^2}{\pi}

) incorrectly introduces a factor of

\pi

in the denominator, which has no physical basis for point masses at fixed distances. Study tip: For rotational inertia problems, always identify each mass element and its distance from the axis of rotation. The key insight is that rotational inertia is additive—just sum up

mr^2

for each component.

Question 5

A uniform solid cone of mass $M$ , base radius $R$ , and height $h$ rotates about its central axis (the line from apex to center of base). Which expression correctly represents its rotational inertia?

$\frac{1}{2}MR^2$
$\frac{3}{10}MR^2$ (correct answer)
$\frac{2}{5}MR^2$
$\frac{1}{4}MR^2$
$\frac{3}{5}MR^2$

Explanation: When you encounter rotational inertia problems involving geometric shapes, you need to integrate over the mass distribution since different parts of the object are at different distances from the rotation axis. For a solid cone rotating about its central axis, we set up the problem by considering thin circular disks perpendicular to the axis. At distance

z

from the apex, each disk has radius

r = \frac{Rz}{h}

(from similar triangles) and thickness

dz

. The mass of each disk is

dm = \rho \pi r^2 dz = \frac{3M}{\pi R^2 h} \cdot \pi \left(\frac{Rz}{h}\right)^2 dz = \frac{3MR^2z^2}{h^3}dz

. Each disk's rotational inertia about the axis is

dI = \frac{1}{2}r^2 dm = \frac{1}{2}\left(\frac{Rz}{h}\right)^2 \cdot \frac{3MR^2z^2}{h^3}dz

. Integrating from

z = 0

z = h

I = \int_0^h \frac{3MR^4z^4}{2h^5}dz = \frac{3MR^4}{2h^5} \cdot \frac{h^5}{5} = \frac{3}{10}MR^2

. Choice A (

\frac{1}{2}MR^2

) is the moment of inertia for a thin ring or hollow cylinder, where all mass is at radius

R

. Choice C (

\frac{2}{5}MR^2

) corresponds to a solid sphere about its center. Choice D (

\frac{1}{4}MR^2

) represents a solid disk or cylinder about its central axis. Study tip: Memorize the standard rotational inertia formulas for common shapes, but understand that cones and other tapered objects always require integration. The factor is typically between the hollow and solid versions of simpler shapes.

Question 6

A student has two objects with the same mass $M$ and outer radius $R$ : a solid disk and a ring. Both objects are placed on an inclined plane and released simultaneously. If both roll without slipping, what is the ratio of their rotational inertias, $I_{ring}/I_{disk}$ ?

$\frac{1}{2}$
$1$
$2$ (correct answer)
$\frac{3}{2}$
$4$

Explanation: When you encounter rolling motion problems, the key concept is that different shapes with the same mass and radius have different rotational inertias due to how their mass is distributed relative to the rotation axis. For a solid disk, the mass is distributed throughout the entire area, so its rotational inertia is

I_{disk} = \frac{1}{2}MR^2

. For a ring, all the mass is concentrated at the outer edge at distance

R

from the center, giving

I_{ring} = MR^2

. The ratio is therefore:

\frac{I_{ring}}{I_{disk}} = \frac{MR^2}{\frac{1}{2}MR^2} = \frac{MR^2 \cdot 2}{MR^2} = 2

Answer choice (A)

\frac{1}{2}

incorrectly inverts the ratio—this would be

I_{disk}/I_{ring}

. Choice (B)

1

suggests both objects have equal rotational inertia, which ignores how mass distribution affects rotational properties. Choice (D)

\frac{3}{2}

might result from confusing the ring formula with that of a solid sphere, whose rotational inertia is

\frac{2}{5}MR^2

. The correct answer is (C)

2

. Remember this pattern: objects with mass concentrated farther from the rotation axis have larger rotational inertias. A ring has all its mass at maximum distance

R

, while a disk averages closer to the center. This is why the ring will actually roll down the incline more slowly—it has more rotational inertia to overcome. Always visualize where the mass sits relative to the rotation axis.

Question 7

A flywheel consists of a solid cylinder (mass $M_1$ , radius $R_1$ ) surrounded by a thin ring (mass $M_2$ , radius $R_2 = 2R_1$ ) that is rigidly connected to the cylinder. When this flywheel rotates about its central axis, what is its total rotational inertia?

$\frac{1}{2}M_1 R_1^2 + M_2 R_2^2$ (correct answer)
$\frac{1}{2}M_1 R_1^2 + \frac{1}{2}M_2 R_2^2$
$\frac{1}{2}(M_1 + M_2)R_1^2$
$\frac{1}{2}M_1 R_1^2 + 2M_2 R_1^2$
$M_1 R_1^2 + 4M_2 R_1^2$

Explanation: When you encounter rotational motion problems involving composite objects, remember that rotational inertia is additive - you can find each component's moment of inertia separately, then sum them together. For this flywheel, you have two distinct rotating components: a solid cylinder and a thin ring. Each has its own standard formula for rotational inertia about its central axis. The solid cylinder contributes

\frac{1}{2}M_1 R_1^2

(the standard formula for a solid cylinder). The thin ring contributes

M_2 R_2^2

(the standard formula for a thin ring, where all mass is concentrated at radius

R_2

). Since

R_2 = 2R_1

, you can substitute: the ring's contribution becomes

M_2(2R_1)^2 = 4M_2 R_1^2

. The total rotational inertia is

\frac{1}{2}M_1 R_1^2 + M_2 R_2^2

, which is choice A. Choice B incorrectly applies the

\frac{1}{2}

factor to the ring - this factor only appears in the solid cylinder formula, not the thin ring formula. Choice C treats the entire system as a single solid cylinder with combined mass, ignoring the different mass distributions. Choice D correctly identifies the cylinder's contribution but makes an error with the ring's contribution by writing

2M_2 R_1^2

instead of

4M_2 R_1^2

. Always memorize the standard rotational inertia formulas for common shapes, and remember that for composite objects, you simply add the individual contributions - don't try to treat them as a single shape with modified properties.

Question 8

A wheel consists of eight identical uniform rods, each of mass $m$ and length $L$ , arranged like spokes radiating from a central hub. What is the rotational inertia of this wheel about the central axis (perpendicular to the plane of the wheel)?

$\frac{8}{3}mL^2$ (correct answer)
$\frac{2}{3}mL^2$
$\frac{2}{3}mL^2$
$4mL^2$
$\frac{1}{3}mL^2$

Explanation: When analyzing rotational inertia problems involving composite objects, you need to consider each component separately and apply the principle of superposition - the total rotational inertia equals the sum of individual contributions. For a single uniform rod rotating about an axis through one end (perpendicular to its length), the rotational inertia is

\frac{1}{3}mL^2

. This is a standard formula you should memorize. Since the wheel has eight identical rods arranged as spokes from the central hub, each rod rotates about an axis through its end. The total rotational inertia is simply:

I_{total} = 8 \times \frac{1}{3}mL^2 = \frac{8}{3}mL^2

Looking at the wrong answers: Answer B gives

\frac{2}{3}mL^2

, which would be correct for just two rods, not eight - this represents a counting error. Answer C is identical to B (likely a typo in the original question). Answer D gives

4mL^2

, which might result from incorrectly using

\frac{1}{2}mL^2

per rod - but that's the formula for rotation about the center of a rod, not about its end. The key insight is recognizing that each spoke rotates about the central axis, which passes through one end of each rod. Many students mistakenly try to use the parallel axis theorem or formulas for rotation about the center of mass, but the geometry here directly gives you rotation about the end. Study tip: Master the standard rotational inertia formulas for basic shapes (rod about end vs. center, disk, etc.) and remember that composite objects simply require adding up individual contributions.

Question 9

A compound object consists of a solid cylinder (mass $M_c$ , radius $R$ ) with a thin disk (mass $M_d$ , radius $2R$ ) attached to one end, sharing the same axis. What is the rotational inertia of this compound object about their common axis?

$\frac{1}{2}M_c R^2 + 2M_d R^2$ (correct answer)
$\frac{1}{2}M_c R^2 + \frac{1}{2}M_d (2R)^2$
$\frac{1}{2}(M_c + M_d)R^2$
$\frac{1}{2}M_c R^2 + M_d R^2$
$\frac{1}{2}M_c R^2 + \frac{1}{4}M_d R^2$

Explanation: When you encounter compound objects in rotational motion problems, the key principle is that rotational inertias are additive. Each component of the object contributes independently to the total rotational inertia about the common axis. For this compound object, you need to find the rotational inertia of each component separately, then add them together. The solid cylinder has rotational inertia

I_c = \frac{1}{2}M_c R^2

about its central axis. The thin disk has rotational inertia

I_d = \frac{1}{2}M_d r^2

where

r

is its radius. Since the disk's radius is

2R

, its rotational inertia becomes

I_d = \frac{1}{2}M_d (2R)^2 = \frac{1}{2}M_d \cdot 4R^2 = 2M_d R^2

. The total rotational inertia is therefore

\frac{1}{2}M_c R^2 + 2M_d R^2

, which is answer A. Answer B incorrectly leaves the disk's rotational inertia as

\frac{1}{2}M_d (2R)^2

without simplifying the

(2R)^2 = 4R^2

term. Answer C treats the compound object as if it were a single cylinder with combined mass

(M_c + M_d)

and radius

R

, ignoring the different geometry of the disk component. Answer D correctly calculates the cylinder's contribution but incorrectly uses

M_d R^2

for the disk, missing both the

\frac{1}{2}

factor and the fact that the disk's radius is

2R

, not

R

. Remember: for compound objects, calculate each component's rotational inertia using its own mass and geometry, then sum them up. Don't try to treat the entire object as a single standard shape.

Question 10

Two identical uniform rods, each of mass $M$ and length $L$ , are connected end-to-end to form a single rod of length $2L$ . What is the rotational inertia of this combined rod about an axis through the junction point, perpendicular to the rod?

$\frac{1}{6}ML^2$
$\frac{1}{3}ML^2$
$\frac{2}{3}ML^2$ (correct answer)
$\frac{4}{3}ML^2$
$\frac{5}{6}ML^2$

Explanation: When you encounter rotational inertia problems involving composite objects, the key is recognizing that you can treat each component separately and then combine their contributions using the parallel axis theorem when necessary. For this problem, you have two identical rods joined end-to-end, with the rotation axis at their junction point. Each rod has mass

M

and length

L

, and the axis passes through one end of each rod (the junction). The rotational inertia of a uniform rod about an axis through one end is

\frac{1}{3}ML^2

. Since you have two identical rods, each contributing equally to the total rotational inertia, the combined system has:

I_{total} = I_{rod1} + I_{rod2} = \frac{1}{3}ML^2 + \frac{1}{3}ML^2 = \frac{2}{3}ML^2

Looking at the wrong answers: Option A (

\frac{1}{6}ML^2

) would result from incorrectly using the rotational inertia formula for a rod about its center (

\frac{1}{12}ML^2

) and somehow doubling it. Option B (

\frac{1}{3}ML^2

) represents the rotational inertia of just one rod, forgetting to account for the second rod. Option D (

\frac{4}{3}ML^2

) might come from incorrectly applying the parallel axis theorem or using the wrong reference formula. Remember: for composite objects rotating about a common axis, simply add the individual rotational inertias when each component rotates about the same axis. Always identify what standard formula applies to each component before combining them.

Question 11

Two identical uniform rods, each of mass $M$ and length $L$ , are welded together to form an L-shape (perpendicular to each other, meeting at their ends). What is the rotational inertia of this L-shaped object about an axis through the junction point, perpendicular to the plane of the L?

$\frac{1}{6}ML^2$
$\frac{2}{3}ML^2$ (correct answer)
$\frac{1}{3}ML^2$
$\frac{4}{3}ML^2$
$\frac{5}{6}ML^2$

Explanation: When you encounter rotational inertia problems involving composite objects, you need to find the moment of inertia for each component about the specified axis, then add them together. For a uniform rod of mass

M

and length

L

, the moment of inertia about one end is

\frac{1}{3}ML^2

. Since both rods in this L-shape are identical and each rotates about an axis through one of its ends (the junction point), each rod contributes

\frac{1}{3}ML^2

to the total rotational inertia. The total rotational inertia is simply the sum:

I_{total} = \frac{1}{3}ML^2 + \frac{1}{3}ML^2 = \frac{2}{3}ML^2

, which is answer B. Let's examine why the other options are incorrect: Answer A (

\frac{1}{6}ML^2

) would be the moment of inertia of just one rod about its center, not about its end, and it only accounts for one rod instead of two. Answer C (

\frac{1}{3}ML^2

) represents the moment of inertia of only one rod about its end, forgetting that there are two rods in the L-shape. Answer D (

\frac{4}{3}ML^2

) might result from incorrectly using

\frac{2}{3}ML^2

for each rod (which would be the moment about the center, not the end) and then doubling it. Study tip: Always remember that for composite rigid bodies, rotational inertias simply add together. Make sure you're using the correct formula for each component's geometry and axis of rotation.

Question 12

A uniform solid hemisphere of mass $M$ and radius $R$ sits on a table with its flat face down. What is its rotational inertia about an axis through its center of mass, parallel to the flat face?

$\frac{2}{5}MR^2$
$\frac{1}{5}MR^2$
$\frac{83}{320}MR^2$ (correct answer)
$\frac{2}{3}MR^2$
$\frac{1}{2}MR^2$

Explanation: When calculating rotational inertia for complex shapes like hemispheres, you need to use integration or apply the parallel axis theorem to known results. This problem tests your understanding of how rotational inertia depends on both the mass distribution and the axis location. For a uniform solid hemisphere rotating about an axis through its center of mass parallel to the flat face, we start with the rotational inertia of a complete solid sphere about its center:

I_{sphere} = \frac{2}{5}MR^2

. However, a hemisphere has different mass distribution and a different center of mass location than a sphere. The center of mass of a hemisphere is located at

\frac{3R}{8}

from the flat face along the symmetry axis. To find the rotational inertia about an axis through this center of mass parallel to the flat face, we must account for how the mass is distributed relative to this new axis. Through integration in spherical coordinates (considering only the upper hemisphere), the result is

I = \frac{83}{320}MR^2

, making C correct. Option A (

\frac{2}{5}MR^2

) is the rotational inertia of a complete solid sphere about its center—a common mistake when students forget they're dealing with only half the sphere. Option B (

\frac{1}{5}MR^2

) might seem logical as "half" of option A, but rotational inertia doesn't scale linearly with mass when the axis location changes. Option D (

\frac{2}{3}MR^2

) is the rotational inertia of a hollow sphere, showing confusion about mass distribution. Remember: rotational inertia problems for irregular shapes often require integration or looking up tabulated results—don't assume simple proportional relationships.

Question 13

A uniform square plate of mass $M$ and side length $a$ rotates about an axis through one corner, perpendicular to the plane of the plate. What is its rotational inertia about this axis?

$\frac{1}{6}Ma^2$
$\frac{1}{3}Ma^2$
$\frac{2}{3}Ma^2$ (correct answer)
$\frac{5}{6}Ma^2$
$\frac{4}{3}Ma^2$

Explanation: When you encounter rotational inertia problems involving rotation about different axes, the key is recognizing whether you can use standard formulas or need integration. For a square plate rotating about a corner rather than its center, you'll need to apply the parallel axis theorem. First, find the rotational inertia about the center. For a uniform square plate of mass

M

and side

a

rotating about its center perpendicular to the plane,

I_{center} = \frac{1}{6}Ma^2

. Next, apply the parallel axis theorem:

I = I_{center} + Md^2

, where

d

is the distance from the center of mass to the new axis. The center of mass of a square is at its geometric center, so the distance from center to corner is

d = \frac{a\sqrt{2}}{2}

. Therefore:

I_{corner} = \frac{1}{6}Ma^2 + M\left(\frac{a\sqrt{2}}{2}\right)^2 = \frac{1}{6}Ma^2 + M\frac{2a^2}{4} = \frac{1}{6}Ma^2 + \frac{1}{2}Ma^2 = \frac{2}{3}Ma^2

Choice A (

\frac{1}{6}Ma^2

) is just the rotational inertia about the center—you'd get this if you forgot to apply the parallel axis theorem. Choice B (

\frac{1}{3}Ma^2

) might result from incorrectly calculating the distance or making an arithmetic error. Choice D (

\frac{5}{6}Ma^2

) could come from adding the parallel axis term incorrectly. Remember: whenever the rotation axis changes from the center of mass, use the parallel axis theorem. The new inertia is always larger than the center-of-mass value.

Question 14

A uniform ladder of mass $M$ and length $L$ leans against a frictionless wall. At the moment when the ladder makes a $60°$ angle with the horizontal floor, what is its rotational inertia about the contact point with the floor?

$\frac{1}{12}ML^2$
$\frac{1}{3}ML^2$ (correct answer)
$\frac{1}{4}ML^2$
$\frac{5}{12}ML^2$
$\frac{2}{3}ML^2$

Explanation: When you encounter rotational inertia problems involving objects rotating about different points, you need to apply the parallel axis theorem. This theorem relates the moment of inertia about any axis to the moment of inertia about the center of mass. For a uniform rod (like this ladder), the rotational inertia about its center is

I_{cm} = \frac{1}{12}ML^2

. However, the ladder is rotating about its contact point with the floor, not its center. The parallel axis theorem states:

I = I_{cm} + Md^2

, where

d

is the distance from the center of mass to the new rotation axis. Since the ladder's center of mass is at its midpoint, the distance from the floor contact point to the center is

\frac{L}{2}

. Therefore:

I = \frac{1}{12}ML^2 + M(\frac{L}{2})^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \frac{1}{12}ML^2 + \frac{3}{12}ML^2 = \frac{4}{12}ML^2 = \frac{1}{3}ML^2

The correct answer is B. Choice A (

\frac{1}{12}ML^2

) is the rotational inertia about the center of mass, not the contact point. Choice C (

\frac{1}{4}ML^2

) represents only the

Md^2

term from the parallel axis theorem, missing the center-of-mass contribution. Choice D (

\frac{5}{12}ML^2

) likely results from incorrectly adding the terms or using wrong distances. Remember: whenever the rotation axis isn't through the center of mass, use the parallel axis theorem. The 60° angle mentioned is a red herring—rotational inertia depends only on mass distribution and axis location, not orientation.

Question 15

A thin uniform ring of mass $M$ and radius $R$ has four point masses, each of mass $m$ , attached at equally spaced points around its circumference. If $m = \frac{M}{8}$ , what is the rotational inertia of this system about an axis through the center of the ring, perpendicular to its plane?

$\frac{3MR^2}{2}$ (correct answer)
$\frac{5MR^2}{4}$
$\frac{9MR^2}{8}$
$\frac{10MR^2}{8}$

Explanation: The ring contributes

I_{ring} = MR^2

. Each point mass contributes

I_{point} = mR^2 = \frac{M}{8}R^2

. With four point masses:

I_{points} = 4 \cdot \frac{MR^2}{8} = \frac{MR^2}{2}

. Total:

I = MR^2 + \frac{MR^2}{2} = \frac{2MR^2 + MR^2}{2} = \frac{3MR^2}{2}

. Choice B (

\frac{5MR^2}{4}

) would result from incorrectly calculating the point mass contribution. Choice C (

\frac{9MR^2}{8}

) would result from errors in combining fractions. Choice D (

\frac{10MR^2}{8} = \frac{5MR^2}{4}

) is the same as choice B with different representation.

Question 16

A uniform semicircular disk of mass $M$ and radius $R$ rotates about an axis perpendicular to its plane. For which axis location does the rotational inertia have the value $\frac{MR^2}{4}$ ?

Through the center of the original full circle from which the semicircle was cut (correct answer)
Through the geometric center (centroid) of the semicircular disk
Through the midpoint of the straight edge of the semicircle
Through a point on the curved edge, diametrically opposite to the midpoint of the straight edge

Explanation: The rotational inertia of a full circular disk about its center is

\frac{1}{2}MR^2

. For a semicircular disk about the center of the original full circle, by symmetry,

I = \frac{1}{2} \cdot \frac{1}{2}MR^2 = \frac{MR^2}{4}

. Choice B is wrong because the centroid of a semicircle is at distance

\frac{4R}{3\pi}

from the straight edge, and using parallel axis theorem would give a different value. Choice C is wrong because rotation about the midpoint of the straight edge would give

I = \frac{1}{2}MR^2

(same as full disk about center). Choice D is wrong because this point is at distance

R

from the geometric center, leading to a larger rotational inertia due to the parallel axis theorem.