AP Physics C Mechanics Quiz: Rotational Inertia

20 questions · exam conditions

0:00

Rotational InertiaQuestion 1 of 20

Two point masses, $m_1 = 2.0$ kg and $m_2 = 3.0$ kg, are located at $x_1 = -1.0$ m and $x_2 = 2.0$ m, respectively, on the x-axis. What is the rotational inertia of this system about the y-axis?

5.0 \, \text{kg} \cdot \text{m}^2

8.0 \, \text{kg} \cdot \text{m}^2

14 \, \text{kg} \cdot \text{m}^2

17 \, \text{kg} \cdot \text{m}^2

← Back to quizzes

AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Rotational Inertia

Practice Rotational Inertia in AP Physics C Mechanics 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 AP Physics C Mechanics.

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

Two point masses, $m_1 = 2.0$ kg and $m_2 = 3.0$ kg, are located at $x_1 = -1.0$ m and $x_2 = 2.0$ m, respectively, on the x-axis. What is the rotational inertia of this system about the y-axis?

$5.0 \, \text{kg} \cdot \text{m}^2$
$8.0 \, \text{kg} \cdot \text{m}^2$
$14 \, \text{kg} \cdot \text{m}^2$ (correct answer)
$17 \, \text{kg} \cdot \text{m}^2$

Explanation: The rotational inertia of a system of point masses is given by the sum

I = \sum m_i r_i^2

, where

r_i

is the perpendicular distance of each mass from the axis of rotation. Here, the axis is the y-axis, so the distances are the absolute values of the x-coordinates.

I = m_1 r_1^2 + m_2 r_2^2 = (2.0 \, \text{kg})(-1.0 \, \text{m})^2 + (3.0 \, \text{kg})(2.0 \, \text{m})^2 = 2.0 \, \text{kg} \cdot \text{m}^2 + 12.0 \, \text{kg} \cdot \text{m}^2 = 14 \, \text{kg} \cdot \text{m}^2

Question 2

A thin hoop of mass $M$ and radius $R$ has two small beads, each of mass $m$ , attached to its rim at opposite ends of a diameter. What is the total rotational inertia of the hoop-beads system about an axis perpendicular to the plane of the hoop and passing through its center?

$MR^2$
$MR^2 + 2mR^2$ (correct answer)
$(M+2m)R$
$MR^2 + mR^2$

Explanation: Rotational inertias are additive. The total rotational inertia is the sum of the inertia of the hoop and the inertias of the two beads. The hoop's inertia is

I_{hoop} = MR^2

. Each bead is a point mass at a distance R from the axis, so each has an inertia of

I_{bead} = mR^2

. The total inertia is

I_{total} = I_{hoop} + I_{bead1} + I_{bead2} = MR^2 + mR^2 + mR^2 = MR^2 + 2mR^2

Question 3

Three identical point masses $m$ are fixed at the vertices of an equilateral triangle with side length $s$ .

What is the rotational inertia of the system about an axis that passes through one of the masses and is perpendicular to the plane of the triangle?

$ms^2$
$2ms^2$ (correct answer)
$3ms^2$
$\frac{3}{2}ms^2$

Explanation: Let the axis pass through mass 1. Its distance from the axis is

r_1 = 0

. The other two masses, mass 2 and mass 3, are at a distance

s

from mass 1. The total rotational inertia is the sum of the individual inertias:

I = \sum m_i r_i^2 = m(0)^2 + m(s)^2 + m(s)^2 = 0 + ms^2 + ms^2 = 2ms^2

Question 4

The rotational inertia of a rigid body of mass $M$ about a given axis is often expressed as $I = Mk^2$ , where $k$ is the radius of gyration. What is the physical significance of the radius of gyration $k$ ?

It is the average distance of the particles of the body from the axis of rotation.
It is the distance from the axis where a single point mass $M$ would have the same rotational inertia. (correct answer)
It is the distance from the center of mass to a parallel axis that doubles the rotational inertia.
It is the radius of a thin hoop with mass $M$ that has the same rotational inertia as the body.

Explanation: The radius of gyration

k

is a conceptual distance. It is defined as the radial distance from the axis of rotation at which the entire mass (

M

) of the body could be concentrated to produce the same rotational inertia (

I

) as the actual body. This concept simplifies comparisons of rotational inertia for different objects.

Question 5

An engineer is designing a flywheel to store a large amount of rotational energy for a given mass and angular velocity. To maximize the flywheel's rotational inertia, how should the mass be distributed?

Concentrated as much as possible near the axis of rotation.
Concentrated as much as possible at the outer rim, far from the axis. (correct answer)
Distributed uniformly as a solid disk.
Shaped into a solid sphere to minimize air resistance.

Explanation: Rotational inertia is given by the sum or integral of

mr^2

r^2 dm

. Since the inertia depends on the square of the distance

r

from the axis of rotation, placing the mass as far as possible from the axis will maximize the rotational inertia for a given total mass. This is why flywheels are often designed like spoked wheels with a heavy rim.

Question 6

A solid sphere, a thin-walled hollow sphere, a solid disk, and a thin hoop all have the same mass $M$ and the same outer radius $R$ . They are all to be rotated about an axis passing through their centers of mass. Which object has the greatest rotational inertia?

The solid sphere
The solid disk
The thin-walled hollow sphere
The thin hoop (correct answer)

Explanation: Rotational inertia depends on how mass is distributed relative to the axis of rotation. The farther the mass is from the axis, the greater the rotational inertia. The formulas are: Hoop (

MR^2

), Hollow Sphere (

\frac{2}{3}MR^2

), Solid Disk (

\frac{1}{2}MR^2

), and Solid Sphere (

\frac{2}{5}MR^2

). Since

1 > 2/3 > 1/2 > 2/5

, the thin hoop has the greatest rotational inertia because all of its mass is concentrated at the maximum possible radius

R

Question 7

A uniform solid sphere of mass $M$ and radius $R$ has a rotational inertia $I = \frac{2}{5}MR^2$ about an axis through its center. What is its rotational inertia about an axis that is tangent to its surface?

$\frac{2}{5}MR^2$
$\frac{3}{5}MR^2$
$MR^2$
$\frac{7}{5}MR^2$ (correct answer)

Explanation: Using the parallel axis theorem,

I' = I_{cm} + Md^2

. The axis tangent to the surface is parallel to the axis through the center, and the distance

d

between these axes is the radius

R

. Therefore,

I' = \frac{2}{5}MR^2 + M(R)^2 = (\frac{2}{5} + 1)MR^2 = \frac{7}{5}MR^2

Question 8

A uniform thin rod of mass $M$ and length $L$ has a rotational inertia about its center of mass given by $I_{cm} = \frac{1}{12}ML^2$ . What is its rotational inertia about an axis perpendicular to the rod and passing through one of its ends?

$\frac{1}{12}ML^2$
$\frac{1}{6}ML^2$
$\frac{1}{4}ML^2$
$\frac{1}{3}ML^2$ (correct answer)

Explanation: The parallel axis theorem states that

I = I_{cm} + Md^2

, where

d

is the distance between the center of mass axis and the parallel axis of rotation. For a rod rotating about its end, the distance

d

from the center to the end is

L/2

. Thus,

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

Question 9

A thin rod of length $L$ lies along the x-axis from $x=0$ to $x=L$ . Its linear mass density varies with position according to the function $\lambda(x) = \beta x^2$ , where $\beta$ is a positive constant. Which of the following integrals represents the rotational inertia of the rod about an axis passing through the origin ( $x=0$ ) and perpendicular to the rod?

$\int_0^L \beta x^2 \,dx$
$\int_0^L \beta x^3 \,dx$
$\int_0^L \beta x^4 \,dx$ (correct answer)
$\int_0^L \beta x^5 \,dx$

Explanation: The formula for rotational inertia of a continuous body is

I = \int r^2 \,dm

. For this rod rotating about the origin, the distance from the axis is

r=x

. The differential mass element is

dm = \lambda(x) \,dx = (\beta x^2) \,dx

. Substituting these into the integral gives

I = \int_0^L x^2 (\beta x^2 \,dx) = \int_0^L \beta x^4 \,dx

Question 10

A uniform solid disk has mass $M$ , radius $R$ , and a rotational inertia $I = \frac{1}{2}MR^2$ about its central axis. A second disk is constructed from the same material and has the same thickness, but its radius is $2R$ . What is the rotational inertia of the second disk in terms of $I$ ?

$4I$
$8I$
$16I$ (correct answer)
$32I$

Explanation: Let the density be

\rho

and thickness be

t

. The mass of the original disk is

M = \rho(\pi R^2 t)

. The mass of the second disk is

M' = \rho(\pi (2R)^2 t) = 4\rho(\pi R^2 t) = 4M

. The rotational inertia of the second disk is

I' = \frac{1}{2}M'(2R)^2 = \frac{1}{2}(4M)(4R^2) = 16(\frac{1}{2}MR^2) = 16I

Question 11

The rotational inertia of a rigid body is measured about several different axes of rotation, all of which are parallel to one another. About which axis will the body's rotational inertia have the minimum possible value?

The axis passing through the body's center of mass. (correct answer)
The axis passing through the body's geometric center.
The axis passing through the point on the body farthest from the center of mass.
The axis about which the body has the greatest angular momentum.

Explanation: According to the parallel axis theorem,

I = I_{cm} + Md^2

, where

I_{cm}

is the rotational inertia about the center of mass, and

I

is the inertia about a parallel axis a distance

d

away. Since

M

and

d^2

are always non-negative, the term

Md^2

is always greater than or equal to zero. The minimum value of

I

occurs when

d=0

, which corresponds to the axis passing through the center of mass.

Question 12

To calculate the rotational inertia of a uniform solid disk of radius $R$ and surface mass density $\sigma$ about its central axis, one can use the integral $I = \int r^2 \,dm$ . What is the correct expression for the differential mass element $dm$ corresponding to a thin ring of radius $r$ and width $dr$ ?

$dm = \sigma (\pi r^2)$
$dm = \sigma (2\pi r \,dr)$ (correct answer)
$dm = \sigma (r \,dr)$
$dm = \frac{M}{\pi R^2} (\pi r^2)$

Explanation: The mass element

dm

is the product of the surface mass density

\sigma

and the differential area element

dA

. For a thin ring of radius

r

and width

dr

, the area can be approximated by its circumference multiplied by its width, so

dA = (2\pi r)dr

. Therefore,

dm = \sigma \,dA = \sigma(2\pi r \,dr)

Question 13

A uniform solid disk of mass $M$ and radius $R$ has an initial rotational inertia $I_0$ about its central axis. A smaller concentric circular piece of radius $R/2$ is removed from the disk. The mass of the material removed is $M/4$ . What is the rotational inertia of the remaining annular ring?

$\frac{3}{4}I_0$
$\frac{7}{8}I_0$
$\frac{15}{16}I_0$ (correct answer)
$\frac{1}{2}I_0$

Explanation: The initial rotational inertia is

I_0 = \frac{1}{2}MR^2

. The removed piece is a disk of mass

M_{rem} = M/4

and radius

R_{rem} = R/2

. Its rotational inertia about the center is

I_{rem} = \frac{1}{2}M_{rem}R_{rem}^2 = \frac{1}{2}(\frac{M}{4})(\frac{R}{2})^2 = \frac{1}{2}\frac{M}{4}\frac{R^2}{4} = \frac{1}{32}MR^2

. The final inertia is the initial inertia minus the removed inertia:

I_{final} = I_0 - I_{rem} = \frac{1}{2}MR^2 - \frac{1}{32}MR^2 = \frac{15}{32}MR^2

. To express this in terms of

I_0

, we have

I_{final} = \frac{15/32}{1/2}I_0 = \frac{15}{16}I_0

Question 14

Object A is a thin hoop of mass $M$ and radius $R$ . Object B is a solid disk, also of mass $M$ and radius $R$ . Which statement correctly compares their inertia (mass) and their rotational inertia about an axis through their centers of mass?

They have the same inertia, but Object A has a greater rotational inertia. (correct answer)
They have the same inertia and the same rotational inertia because their mass and radius are identical.
Object A has a greater inertia and a greater rotational inertia because its mass is on the outside.
Object B has a greater rotational inertia because its mass is distributed throughout its area.

Explanation: Inertia in translational motion is simply mass, so both objects have the same inertia (

M

). Rotational inertia depends on the distribution of that mass. For the hoop (Object A), all mass is at radius

R

, so

I_A = MR^2

. For the disk (Object B), the mass is distributed from the center to the edge, resulting in a smaller rotational inertia,

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

. Thus,

I_A > I_B

Question 15

A thin, uniform rod of mass $M$ and length $L$ is oriented along the x-axis from $x = -L/2$ to $x = +L/2$ . To calculate its rotational inertia about the y-axis using $I = \int x^2 dm$ , which is the correct expression for the mass element $dm$ ?

$dm = M \,dx$
$dm = \frac{M}{L} \,dx$ (correct answer)
$dm = \frac{L}{M} \,dx$
$dm = \frac{M}{L} x \,dx$

Explanation: For a uniform rod, the linear mass density (

\lambda

) is constant and equal to the total mass divided by the total length, so

\lambda = M/L

. A small segment of the rod of length

dx

has a mass

dm

. The relationship is

dm = \lambda \,dx

. Therefore,

dm = \frac{M}{L} \,dx

Question 16

A non-uniform object of mass $M$ has a rotational inertia $I_{cm}$ about an axis through its center of mass. When rotated about a parallel axis a distance $d$ away, its rotational inertia is $I$ . The mass $M$ of the object can be found using which expression?

$M = \frac{I}{d^2}$
$M = \frac{I - I_{cm}}{d}$
$M = \frac{I - I_{cm}}{d^2}$ (correct answer)
$M = \frac{I_{cm}}{d^2}$

Explanation: The parallel axis theorem is given by

I = I_{cm} + Md^2

. To solve for the mass

M

, we can rearrange the equation. First, subtract

I_{cm}

from both sides:

I - I_{cm} = Md^2

. Then, divide by

d^2

M = \frac{I - I_{cm}}{d^2}

Question 17

A large, uniform solid disk has mass $M$ and radius $R$ . A smaller, concentric solid disk of mass $m$ and radius $r$ ( $r < R$ ) is removed from its center. What is the rotational inertia of the remaining annular object about the central axis?

$\frac{1}{2}(M-m)(R-r)^2$
$\frac{1}{2}(M-m)R^2$
$\frac{1}{2}MR^2 - \frac{1}{2}mr^2$ (correct answer)
$\frac{1}{2}M(R^2 - r^2)$

Explanation: The principle of superposition applies to rotational inertia. The rotational inertia of the remaining annulus is the rotational inertia of the original large disk minus the rotational inertia of the smaller disk that was removed. The inertia of the large disk is

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

. The inertia of the removed disk is

I_{removed} = \frac{1}{2}mr^2

. Thus, the inertia of the annulus is

I_{annulus} = I_{large} - I_{removed} = \frac{1}{2}MR^2 - \frac{1}{2}mr^2

Question 18

A thin spherical shell of mass $M$ and radius $R$ has a small hole drilled through it. A thin rod of mass $m$ and length $2R$ is inserted through the hole so that it passes through the center of the sphere, with equal lengths extending on both sides. What is the moment of inertia of this composite object about an axis through the center, perpendicular to the rod?

$\frac{2MR^2}{3} + \frac{mR^2}{3}$ (correct answer)
$\frac{2MR^2}{3} + \frac{mR^2}{12}$
$\frac{2MR^2}{3} + \frac{mR^2}{6}$
$\frac{2MR^2}{5} + \frac{mR^2}{3}$

Explanation: The spherical shell has moment of inertia

I_{shell} = \frac{2MR^2}{3}

about any axis through its center. The rod of length

2R

has moment of inertia

I_{rod} = \frac{m(2R)^2}{12} = \frac{4mR^2}{12} = \frac{mR^2}{3}

about an axis through its center perpendicular to its length. Total:

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

. Choice B incorrectly uses the rod's moment about its end. Choice C uses an incorrect factor. Choice D uses the wrong moment of inertia for a solid sphere instead of a shell.

Question 19

Three identical point masses, each of mass $m$ , are located at the vertices of an equilateral triangle with side length $a$ . The system rotates about an axis passing through one vertex and perpendicular to the plane of the triangle. What is the moment of inertia of this system?

$\frac{ma^2}{2}$
$ma^2$
$\frac{3ma^2}{2}$
$2ma^2$ (correct answer)

Explanation: When you encounter rotational motion problems involving moment of inertia, you need to identify the axis of rotation and calculate how mass is distributed relative to that axis. For point masses, the moment of inertia is simply

I = \sum mr^2

, where

r

is each mass's perpendicular distance from the rotation axis. In this problem, the axis passes through one vertex of the equilateral triangle. This means one mass sits directly on the axis (

r = 0

), while the other two masses are each a distance

a

away from the axis (since

a

is the side length of the triangle). The moment of inertia calculation becomes:

Mass at the vertex: $I_1 = m(0)^2 = 0$
Two masses at distance $a$ : $I_2 = 2m(a)^2 = 2ma^2$
Total: $I_{total} = 0 + 2ma^2 = 2ma^2$

Choice A (

\frac{ma^2}{2}

) incorrectly assumes all three masses contribute equally at some average distance. Choice B (

ma^2

) miscounts by including only one of the off-axis masses or using an incorrect distance. Choice C (

\frac{3ma^2}{2}

) attempts to average all three masses but fails to recognize that the on-axis mass contributes zero to the moment of inertia. Remember: when calculating moment of inertia, masses on the rotation axis contribute nothing (

r = 0

). Always identify which masses actually have perpendicular distance from the axis before calculating.

Question 20

A yo-yo consists of two identical solid disks of mass $M$ and radius $R$ connected by a thin cylindrical axle of mass $m$ and radius $r$ . What is the moment of inertia of the yo-yo about its central axis?

$MR^2 + mr^2$
$\frac{MR^2}{2} + \frac{mr^2}{2}$
$MR^2 + \frac{mr^2}{2}$ (correct answer)
$\frac{MR^2}{2} + mr^2$

Explanation: When you encounter rotational inertia problems involving composite objects, you need to identify each component and apply the correct moment of inertia formula for its geometry, then sum all contributions. A yo-yo has three components rotating about the central axis: two solid disks and one solid cylinder (the axle). For any solid disk or cylinder rotating about its central axis, the moment of inertia is

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

, where

m

is the mass and

r

is the radius. For this yo-yo:

Each solid disk: $I_{disk} = \frac{1}{2}MR^2$
Two disks contribute: $2 \times \frac{1}{2}MR^2 = MR^2$
The cylindrical axle: $I_{axle} = \frac{1}{2}mr^2$
Total: $I_{total} = MR^2 + \frac{1}{2}mr^2$

This matches answer choice C. Answer choice A (

MR^2 + mr^2

) incorrectly uses

I = mr^2

for the axle, which would apply if all the axle's mass were concentrated at radius

r

rather than distributed throughout the cylinder. Answer choice B (

\frac{MR^2}{2} + \frac{mr^2}{2}

) correctly applies the solid cylinder formula to both components but fails to account for having two disks instead of one. Answer choice D (

\frac{MR^2}{2} + mr^2

) makes both errors: it forgets about the second disk and uses the wrong formula for the axle. Study tip: Always write down the standard moment of inertia formulas for common shapes and carefully count how many of each component you have in composite objects.