Rotational Kinetic Energy - AP Physics C Mechanics

1. A thin rod of length $L$ and mass $M$ rotates about an axis perpendicular to the rod through one end. A point mass $m$ is attached to the rod at distance $d$ from the rotation axis. If the system rotates with angular velocity $\omega$, what is the total rotational kinetic energy?
   Answer: $\frac{1}{2}\left(\frac{1}{3}ML^2 + md^2\right)\omega^2$ The total moment of inertia is the sum of individual moments of inertia about the same axis. For a thin rod rotating about one end: $I_{rod} = \frac{1}{3}ML^2$. For a point mass at distance $d$: $I_{mass} = md^2$. Total: $I_{total} = \frac{1}{3}ML^2 + md^2$. Rotational KE = $\frac{1}{2}I_{total}\omega^2$. Choice B uses the wrong moment of inertia for a rod about its center ($\frac{1}{12}ML^2$). Choice C incorrectly separates the kinetic energy terms. Choice D combines multiple errors.
2. A solid disk and a thin hoop of the same mass and radius are released from rest at the same height on an inclined plane and roll without slipping. How do their total kinetic energies and translational kinetic energies compare upon reaching the bottom of the incline?
   Answer: Both have the same total kinetic energy, but the disk has greater translational kinetic energy. By conservation of energy, both objects convert the same amount of gravitational potential energy ($mgh$) into total kinetic energy. Thus, their total kinetic energies are equal. The hoop has a larger rotational inertia ($MR^2$) than the disk ($\frac{1}{2}MR^2$), so for a given total kinetic energy, a larger fraction of the hoop's energy is rotational. Consequently, the disk will have a smaller fraction of its energy as rotational and a larger fraction as translational, meaning it will have a greater translational speed and translational kinetic energy.
3. A solid cylinder starts from rest at height $H$ and rolls without slipping down ramp A. It then rolls without slipping up ramp B. Assume there are no energy losses due to friction. What is the maximum vertical height $H'$ the cylinder reaches on ramp B?
   Answer: $H'$ is equal to $H$. Since the cylinder rolls without slipping on both ramps and there are no dissipative forces, total mechanical energy is conserved throughout the entire process. The initial potential energy $MgH$ is converted to total kinetic energy (translational and rotational) at the bottom, and this total kinetic energy is fully converted back to potential energy $MgH'$ on the second ramp. Therefore, $MgH = MgH'$, which means $H' = H$.
4. A solid ball starts from rest at height $H$ on an incline and rolls without slipping to the bottom. It then encounters a second, frictionless incline. What is the maximum vertical height $H'$ the ball reaches on the second incline?
   Answer: $H'$ is less than $H$. On the first incline, the initial potential energy $MgH$ is converted into both translational and rotational kinetic energy. At the bottom, $MgH = K_{trans} + K_{rot}$. On the second, frictionless incline, the ball will slide, not roll. The rotational motion will continue at a constant angular velocity (as there are no tangential forces or torques), so the rotational kinetic energy is not converted back into potential energy. Only the translational kinetic energy is converted into potential energy, $K_{trans} = MgH'$. Since $K_{trans} < MgH$, it follows that $H' < H$.
5. A solid disk and a spoked wheel have the same mass $M$ and the same outer radius $R$. Both are accelerated from rest to the same final angular velocity $\omega$. Which object has more rotational kinetic energy, and why?
   Answer: The spoked wheel, because more of its mass is located at a larger radius, giving it a larger rotational inertia. Rotational kinetic energy is $K_{rot} = \frac{1}{2}I\omega^2$. Since both objects have the same mass $M$ and are accelerated to the same angular velocity $\omega$, the one with the larger rotational inertia $I$ will have more kinetic energy. Rotational inertia depends on how mass is distributed relative to the axis of rotation. The spoked wheel has most of its mass concentrated at the outer radius, while the solid disk's mass is distributed throughout. Therefore, the spoked wheel has a larger rotational inertia and thus more rotational kinetic energy.
6. Based on the scenario, calculate the flywheel's rotational kinetic energy for $I=0.80\ \text{kg}\cdot\text{m}^2$ and $\omega=50\ \text{rad/s}$.
   Answer: $1000\ \text{J}$ This question tests AP Physics C understanding of rotational kinetic energy in rotating systems. Rotational kinetic energy is calculated using the formula K_rot = (1/2)Iω², where I is the moment of inertia and ω is the angular velocity. In this scenario, we have a flywheel with I = 0.80 kg·m² and ω = 50 rad/s. Choice B is correct because K_rot = (1/2)(0.80)(50)² = (1/2)(0.80)(2500) = 1000 J. Choice D is incorrect because it has the wrong units (N·m instead of J), though numerically it might seem plausible. To help students: Emphasize that rotational kinetic energy has units of joules, not newton-meters (though they are dimensionally equivalent). Practice substituting values carefully and checking units throughout calculations.
7. Two identical uniform rods, each of mass $m$ and length $L$, are joined at right angles to form an L-shaped object. The system rotates about an axis perpendicular to both rods, passing through the junction point. If one rod lies along the x-axis and the other along the y-axis, both extending in the positive directions from the origin, what is the rotational kinetic energy when the system rotates with angular velocity $\omega$?
   Answer: $\frac{1}{3}mL^2\omega^2$ Each rod rotates about an axis through one of its ends. For a uniform rod of mass $m$ and length $L$ rotating about one end, the moment of inertia is $I = \frac{1}{3}mL^2$. Since we have two identical rods in the same configuration, the total moment of inertia is $I_{total} = 2 \times \frac{1}{3}mL^2 = \frac{2}{3}mL^2$. The rotational kinetic energy is $KE = \frac{1}{2}I_{total}\omega^2 = \frac{1}{2} \times \frac{2}{3}mL^2\omega^2 = \frac{1}{3}mL^2\omega^2$. Choice A uses the moment of inertia for one rod only. Choice C incorrectly uses $I = \frac{1}{2}mL^2$ for each rod. Choice D gives the total moment of inertia instead of kinetic energy.
8. A spinning platform has rotational kinetic energy $K$. A student modifies the platform, halving its rotational inertia and doubling its angular velocity. What is the new rotational kinetic energy, $K'$, in terms of $K$?
   Answer: $K' = 2K$ The initial kinetic energy is $K = \frac{1}{2}I\omega^2$. The new inertia is $I' = I/2$ and the new angular velocity is $\omega' = 2\omega$. The new kinetic energy is $K' = \frac{1}{2}I'(\omega')^2 = \frac{1}{2}(\frac{I}{2})(2\omega)^2 = \frac{1}{2}(\frac{I}{2})(4\omega^2) = 2(\frac{1}{2}I\omega^2) = 2K$.
9. A bicycle wheel is mounted on a fixed, frictionless axle so that its center of mass is stationary. The wheel is spun, giving it a certain amount of rotational kinetic energy. Which statement accurately describes the kinetic energy of the wheel?
   Answer: The wheel possesses only rotational kinetic energy since its center of mass is not translating. Total kinetic energy for a rigid body is the sum of translational kinetic energy of the center of mass and rotational kinetic energy about the center of mass. Since the axle is fixed, the center of mass is at rest, so the translational kinetic energy is zero. The wheel is rotating, so it has rotational kinetic energy. Therefore, its total kinetic energy is purely rotational.
10. Based on the scenario, if a disk's angular velocity doubles, what happens to its rotational kinetic energy?
    Answer: It quadruples. This question tests AP Physics C understanding of rotational kinetic energy in rotating systems. Rotational kinetic energy is calculated using the formula K_rot = (1/2)Iω², where I is the moment of inertia and ω is the angular velocity. In this scenario, we examine what happens when angular velocity doubles while moment of inertia remains constant. Choice C is correct because when ω doubles, ω² becomes four times larger, making K_rot quadruple since K_rot is proportional to ω². Choice B is incorrect because it assumes a linear relationship between K_rot and ω, missing the squared term. To help students: Emphasize the quadratic relationship between rotational kinetic energy and angular velocity. Use graphical representations to show how K_rot varies with ω².
11. A flywheel with rotational inertia $I$ rotates with angular velocity $\omega$, possessing rotational kinetic energy $K_{rot}$. If the flywheel's angular velocity is increased to $4\omega$ while its rotational inertia remains constant, what is its new rotational kinetic energy?
    Answer: $16K_{rot}$ Rotational kinetic energy is given by the formula $K_{rot} = \frac{1}{2}I\omega^2$. Since the kinetic energy is proportional to the square of the angular velocity, increasing the angular velocity by a factor of 4 will increase the kinetic energy by a factor of $4^2 = 16$. The new kinetic energy will be $16K_{rot}$.
12. A uniform rod of length $L$ and mass $M$ is pivoted at one end and released from a horizontal position. What is its rotational kinetic energy when it reaches the vertical position? The rotational inertia of a rod about its end is $I = \frac{1}{3}ML^2$. Use $g$ for the acceleration due to gravity.
    Answer: $MgL/2$ This is a conservation of energy problem. The center of mass of the rod is at $L/2$ from the pivot. When the rod is released from a horizontal position, its center of mass falls a vertical distance of $L/2$. The change in gravitational potential energy is $\Delta U_g = Mg(L/2)$. This potential energy is converted into rotational kinetic energy. Therefore, the rotational kinetic energy at the vertical position is $K_{rot} = MgL/2$.
13. Two uniform disks, Disk 1 and Disk 2, are made from the same material and have the same thickness. The radius of Disk 1 is twice the radius of Disk 2. If both disks rotate with the same angular speed about their central axes, what is the ratio of the rotational kinetic energy of Disk 1 to that of Disk 2?
    Answer: 16 Mass $M$ is proportional to volume, so $M \propto R^2t$. Since thickness $t$ is the same, $M \propto R^2$. Thus, $M_1 = 4M_2$. Rotational inertia for a disk is $I = \frac{1}{2}MR^2$. The ratio of inertias is $\frac{I_1}{I_2} = \frac{\frac{1}{2}M_1R_1^2}{\frac{1}{2}M_2R_2^2} = \frac{(4M_2)(2R_2)^2}{M_2R_2^2} = \frac{4M_2 \cdot 4R_2^2}{M_2R_2^2} = 16$. Since $K_{rot} = \frac{1}{2}I\omega^2$ and $\omega$ is the same for both, the ratio of kinetic energies is equal to the ratio of their rotational inertias, which is 16.
14. A hollow sphere of mass $M$ and radius $R$ rolls without slipping down an inclined plane of angle $\alpha$. At the bottom of the incline, it has translational speed $v$. The sphere then rolls up a frictionless curved track. What is the maximum height $h$ it reaches above the bottom of the track?
    Answer: $\frac{5v^2}{6g}$ At the bottom, the sphere has both translational and rotational kinetic energy. For a hollow sphere, $I = \frac{2}{3}MR^2$. Since it's rolling, $\omega = v/R$. Total KE at bottom: $KE = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}Mv^2 + \frac{1}{2}\cdot\frac{2}{3}MR^2\cdot\frac{v^2}{R^2} = \frac{1}{2}Mv^2 + \frac{1}{3}Mv^2 = \frac{5}{6}Mv^2$. On the frictionless track, the sphere slides (no rolling), so only translational energy is conserved since rotational energy cannot change without torque. Actually, on a frictionless surface, both translational and rotational energy are conserved separately. So total energy $\frac{5}{6}Mv^2 = Mgh$, giving $h = \frac{5v^2}{6g}$. Choice A assumes only translational energy. Choice C uses solid sphere moment of inertia. Choice D uses incorrect energy fraction.
15. Based on the scenario, determine $I$ for $K_\text{rot}=125\ \text{J}$ when $\omega=10\ \text{rad/s}$.
    Answer: $2.50\ \text{kg}\cdot\text{m}^2$ This question tests AP Physics C understanding of rotational kinetic energy in rotating systems. Rotational kinetic energy is calculated using the formula K_rot = (1/2)Iω², which we rearrange to solve for I: I = 2K_rot/ω². In this scenario, we need to find I when K_rot = 125 J and ω = 10 rad/s. Choice C is correct because I = 2(125)/(10)² = 250/100 = 2.50 kg·m². Choice B is incorrect as it represents half the correct value, likely from forgetting to multiply by 2 when rearranging the formula. To help students: When solving for I, remember that the rearranged formula includes a factor of 2 in the numerator. Practice dimensional analysis to verify your algebraic manipulations.
16. A solid sphere with rotational inertia $I = \frac{2}{5}MR^2$ and a hollow sphere with rotational inertia $I = \frac{2}{3}MR^2$ have the same mass $M$ and radius $R$. If they rotate with the same angular velocity $\omega$, what is the ratio of the hollow sphere's rotational kinetic energy to the solid sphere's rotational kinetic energy?
    Answer: $5/3$ The ratio of their kinetic energies is $\frac{K_{hollow}}{K_{solid}} = \frac{\frac{1}{2}I_{hollow}\omega^2}{\frac{1}{2}I_{solid}\omega^2} = \frac{I_{hollow}}{I_{solid}}$. Substituting the given rotational inertias, the ratio is $\frac{\frac{2}{3}MR^2}{\frac{2}{5}MR^2} = \frac{2/3}{2/5} = \frac{2}{3} \times \frac{5}{2} = \frac{5}{3}$.
17. A point mass $m$ is moving in a circle of radius $R$ with a constant speed $v$. Its rotational kinetic energy about the center of the circle is $K_{rot}$. Which expression correctly relates $K_{rot}$ to its translational kinetic energy, $K_{trans} = \frac{1}{2}mv^2$?
    Answer: $K_{rot} = K_{trans}$ For a point mass $m$ at radius $R$, the rotational inertia is $I = mR^2$. The angular velocity is $\omega = v/R$. The rotational kinetic energy is $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(mR^2)(\frac{v}{R})^2 = \frac{1}{2}mR^2(\frac{v^2}{R^2}) = \frac{1}{2}mv^2$. This is identical to the expression for translational kinetic energy, so $K_{rot} = K_{trans}$.
18. A uniform solid cylinder of mass $M$ and radius $R$ rolls without slipping. Its total kinetic energy is $E$. What is its angular velocity $\omega$? The rotational inertia of a solid cylinder is $I = \frac{1}{2}MR^2$.
    Answer: $\frac{1}{R}\sqrt{\frac{4E}{3M}}$ The total kinetic energy $E$ is the sum of translational and rotational kinetic energies. $E = K_{trans} + K_{rot} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$. For rolling without slipping, $v = R\omega$. Substituting this and $I = \frac{1}{2}MR^2$ gives $E = \frac{1}{2}M(R\omega)^2 + \frac{1}{2}(\frac{1}{2}MR^2)\omega^2 = \frac{1}{2}MR^2\omega^2 + \frac{1}{4}MR^2\omega^2 = \frac{3}{4}MR^2\omega^2$. Solving for $\omega$ gives $\omega^2 = \frac{4E}{3MR^2}$, so $\omega = \frac{1}{R}\sqrt{\frac{4E}{3M}}$.
19. A thin-walled hollow sphere of mass $M$ and radius $R$ rolls without slipping. Its total kinetic energy is $K_{total}$. What is the translational speed $v$ of its center of mass? The rotational inertia of a thin-walled hollow sphere is $I=\frac{2}{3}MR^2$.
    Answer: $\sqrt{\frac{6K_{total}}{5M}}$ The total kinetic energy is $K_{total} = K_{trans} + K_{rot} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$. For rolling without slipping, $\omega = v/R$. So, $K_{total} = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{2}{3}MR^2)(\frac{v}{R})^2 = \frac{1}{2}Mv^2 + \frac{1}{3}Mv^2 = \frac{5}{6}Mv^2$. Solving for $v$ gives $v^2 = \frac{6K_{total}}{5M}$, so $v = \sqrt{\frac{6K_{total}}{5M}}$.
20. A thin hoop of mass $M$ and radius $R$ rolls without slipping on a horizontal surface. Its rotational inertia is $I = MR^2$. What fraction of its total kinetic energy is rotational kinetic energy?
    Answer: $1/2$ The total kinetic energy is $K_{total} = K_{trans} + K_{rot}$. For a hoop rolling without slipping ($v=R\omega$), $K_{trans} = \frac{1}{2}Mv^2$ and $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(MR^2)(\frac{v}{R})^2 = \frac{1}{2}Mv^2$. So, $K_{total} = \frac{1}{2}Mv^2 + \frac{1}{2}Mv^2 = Mv^2$. The fraction that is rotational is $\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{2}Mv^2}{Mv^2} = \frac{1}{2}$.

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