Rotational Inertia - AP Physics C Mechanics

1. An ice skater is spinning with her arms outstretched. When she pulls her arms in close to her body, her angular velocity increases. Which statement best explains the change in her rotational inertia?
   Answer: Her rotational inertia decreases because she brings mass closer to the axis of rotation. Rotational inertia depends on the distribution of mass relative to the axis of rotation. By pulling her arms in, the skater moves mass from a larger radius to a smaller radius. Since rotational inertia is proportional to the square of the distance from the axis ($I = \sum mr^2$), bringing mass closer to the axis decreases her total rotational inertia.
2. 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?
   Answer: $2ma^2$ 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.
3. 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?
   Answer: $\frac{7}{5}MR^2$ 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, $I' = \frac{2}{5}MR^2 + M(R)^2 = (\frac{2}{5} + 1)MR^2 = \frac{7}{5}MR^2$.
4. 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?
   Answer: $\int_0^L \beta x^4 \,dx$ 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 $I = \int_0^L x^2 (\beta x^2 \,dx) = \int_0^L \beta x^4 \,dx$.
5. A bicycle wheel can be modeled as a thin rim of mass $M$ and radius $R$ with $N$ identical spokes, each of mass $m$ and length $R$. If the spokes are treated as thin rods extending radially from the hub to the rim, what is the moment of inertia of the wheel about its central axis?
   Answer: $MR^2 + \frac{NmR^2}{3}$ When you encounter moment of inertia problems involving composite objects, break the system into simpler components and use the principle of superposition—the total moment of inertia equals the sum of each component's moment of inertia about the same axis. This bicycle wheel has two distinct components: a thin rim and N spokes. For the thin rim of mass M and radius R, all mass is concentrated at distance R from the central axis, so its moment of inertia is $I_{rim} = MR^2$. Each spoke is a thin rod of mass m and length R, rotating about an axis through one end (the hub). Using the standard formula for a rod rotating about its end, each spoke contributes $I_{spoke} = \frac{mR^2}{3}$. Since there are N identical spokes, their total contribution is $N \cdot \frac{mR^2}{3} = \frac{NmR^2}{3}$. The total moment of inertia is: $I_{total} = MR^2 + \frac{NmR^2}{3}$, confirming answer B. Choice A incorrectly uses $\frac{mR^2}{12}$, which is the moment of inertia for a rod rotating about its center, not its end. Choice C treats each spoke as if it's a point mass at the rim ($mR^2$), ignoring that mass is distributed along the spoke's length. Choice D uses $\frac{mR^2}{6}$, which has no basis in standard moment of inertia formulas. Remember: always identify the axis of rotation first, then match each component to the correct standard formula. The moment of inertia depends critically on both the mass distribution and the specific rotation axis.
6. 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?
   Answer: $MR^2 + \frac{mr^2}{2}$ 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.
7. Which of the following represents the correct SI units for rotational inertia?
   Answer: $\text{kg} \cdot \text{m}^2$ The basic formula for the rotational inertia of a point mass is $I = mr^2$. The SI unit for mass ($m$) is the kilogram (kg) and the SI unit for distance ($r$) is the meter (m). Therefore, the SI units for rotational inertia are $kg \cdot m^2$.
8. 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 $I$?
   Answer: $16I$ 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 $I' = \frac{1}{2}M'(2R)^2 = \frac{1}{2}(4M)(4R^2) = 16(\frac{1}{2}MR^2) = 16I$.
9. 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?
   Answer: $\frac{1}{2}MR^2 - \frac{1}{2}mr^2$ 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 $I_{annulus} = I_{large} - I_{removed} = \frac{1}{2}MR^2 - \frac{1}{2}mr^2$.
10. 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?
    Answer: The axis passing through the body's center of mass. 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.
11. 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 $dm$?
    Answer: $dm = \frac{M}{L} \,dx$ 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, $dm = \frac{M}{L} \,dx$.
12. 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 $k$?
    Answer: It is the distance from the axis where a single point mass $M$ would have the same rotational inertia. 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.
13. 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?
    Answer: $2ms^2$ 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: $I = \sum m_i r_i^2 = m(0)^2 + m(s)^2 + m(s)^2 = 0 + ms^2 + ms^2 = 2ms^2$.
14. 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?
    Answer: $\frac{15}{16}I_0$ 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 $I_{final} = \frac{15/32}{1/2}I_0 = \frac{15}{16}I_0$.
15. 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?
    Answer: $14 \, \text{kg} \cdot \text{m}^2$ 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. $I = m_1 r_1^2 + m_2 r_2^2 = (2.0 \, \text{kg})(-1.0 \, \text{m})^2 + (3.0 \, \text{kg})(2.0 \, \text{m})^2 = 2.0 \, \text{kg} \cdot \text{m}^2 + 12.0 \, \text{kg} \cdot \text{m}^2 = 14 \, \text{kg} \cdot \text{m}^2$.
16. 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?
    Answer: $MR^2 + 2mR^2$ 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 $I_{total} = I_{hoop} + I_{bead1} + I_{bead2} = MR^2 + mR^2 + mR^2 = MR^2 + 2mR^2$.
17. 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?
    Answer: They have the same inertia, but Object A has a greater rotational inertia. 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, $I_A > I_B$.
18. 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?
    Answer: Concentrated as much as possible at the outer rim, far from the axis. Rotational inertia is given by the sum or integral of $mr^2$ or $r^2 dm$. 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.
19. Rotational inertia is the physical quantity that quantifies an object's...
    Answer: resistance to a change in its rotational motion. Rotational inertia ($I$) is the rotational analog of mass (inertia). Just as mass measures an object's resistance to a change in its linear velocity (i.e., its resistance to linear acceleration), rotational inertia measures an object's resistance to a change in its angular velocity (i.e., its resistance to angular acceleration).
20. 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 $dr$?
    Answer: $dm = \sigma (2\pi r \,dr)$ 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, $dm = \sigma \,dA = \sigma(2\pi r \,dr)$.

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