AP Physics C Mechanics Quiz: Rotational Kinetic Energy
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Rotational Kinetic EnergyQuestion 1 of 20

A solid disk and a spoked wheel have the same mass MM and the same outer radius RR. Both are accelerated from rest to the same final angular velocity ω\omega. Which object has more rotational kinetic energy, and why?

The solid disk, because its mass is more uniformly distributed, leading to a more efficient rotation.
The spoked wheel, because more of its mass is located at a larger radius, giving it a larger rotational inertia.
They have the same rotational kinetic energy, because their mass, radius, and angular velocity are identical.
They have the same rotational kinetic energy, because the work required to accelerate them is the same.
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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Rotational Kinetic Energy

Practice Rotational Kinetic Energy 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 Kinetic Energy, 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

A solid disk and a spoked wheel have the same mass MM and the same outer radius RR. Both are accelerated from rest to the same final angular velocity ω\omega. Which object has more rotational kinetic energy, and why?

  1. The solid disk, because its mass is more uniformly distributed, leading to a more efficient rotation.
  2. The spoked wheel, because more of its mass is located at a larger radius, giving it a larger rotational inertia. (correct answer)
  3. They have the same rotational kinetic energy, because their mass, radius, and angular velocity are identical.
  4. They have the same rotational kinetic energy, because the work required to accelerate them is the same.
Explanation: Rotational kinetic energy is Krot=12Iω2K_{rot} = \frac{1}{2}I\omega^2. Since both objects have the same mass MM and are accelerated to the same angular velocity ω\omega, the one with the larger rotational inertia II 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.

Question 2

A uniform rod of length LL and mass MM 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=13ML2I = \frac{1}{3}ML^2. Use gg for the acceleration due to gravity.

  1. MgLMgL
  2. MgL/2MgL/2 (correct answer)
  3. MgL/3MgL/3
  4. MgL/4MgL/4
Explanation: This is a conservation of energy problem. The center of mass of the rod is at L/2L/2 from the pivot. When the rod is released from a horizontal position, its center of mass falls a vertical distance of L/2L/2. The change in gravitational potential energy is ΔUg=Mg(L/2)\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 Krot=MgL/2K_{rot} = MgL/2.

Question 3

A thin hoop of mass MM and radius RR rolls without slipping on a horizontal surface. Its rotational inertia is I=MR2I = MR^2. What fraction of its total kinetic energy is rotational kinetic energy?

  1. 1/41/4
  2. 1/31/3
  3. 1/21/2 (correct answer)
  4. 2/32/3
Explanation: The total kinetic energy is Ktotal=Ktrans+KrotK_{total} = K_{trans} + K_{rot}. For a hoop rolling without slipping (v=Rωv=R\omega), Ktrans=12Mv2K_{trans} = \frac{1}{2}Mv^2 and Krot=12Iω2=12(MR2)(vR)2=12Mv2K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(MR^2)(\frac{v}{R})^2 = \frac{1}{2}Mv^2. So, Ktotal=12Mv2+12Mv2=Mv2K_{total} = \frac{1}{2}Mv^2 + \frac{1}{2}Mv^2 = Mv^2. The fraction that is rotational is KrotKtotal=12Mv2Mv2=12\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{2}Mv^2}{Mv^2} = \frac{1}{2}.

Question 4

A solid ball starts from rest at height HH on an incline and rolls without slipping to the bottom. It then encounters a second, frictionless incline. What is the maximum vertical height H′H' the ball reaches on the second incline?

  1. H′H' is less than HH. (correct answer)
  2. H′H' is equal to HH.
  3. H′H' is greater than HH.
  4. The relationship between H′H' and HH depends on the ball's mass.
Explanation: On the first incline, the initial potential energy MgHMgH is converted into both translational and rotational kinetic energy. At the bottom, MgH=Ktrans+KrotMgH = 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, Ktrans=MgH′K_{trans} = MgH'. Since Ktrans<MgHK_{trans} < MgH, it follows that H′<HH' < H.

Question 5

A point mass mm is moving in a circle of radius RR with a constant speed vv. Its rotational kinetic energy about the center of the circle is KrotK_{rot}. Which expression correctly relates KrotK_{rot} to its translational kinetic energy, Ktrans=12mv2K_{trans} = \frac{1}{2}mv^2?

  1. Krot=KtransK_{rot} = K_{trans} (correct answer)
  2. Krot=R⋅KtransK_{rot} = R \cdot K_{trans}
  3. Krot=Ktrans/RK_{rot} = K_{trans} / R
  4. Krot=0K_{rot} = 0 since it is a point mass.
Explanation: For a point mass mm at radius RR, the rotational inertia is I=mR2I = mR^2. The angular velocity is ω=v/R\omega = v/R. The rotational kinetic energy is Krot=12Iω2=12(mR2)(vR)2=12mR2(v2R2)=12mv2K_{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 Krot=KtransK_{rot} = K_{trans}.

Question 6

The rotational kinetic energy of a rigid body is calculated using the expression 12Iω2\frac{1}{2}I\omega^2. Given that the SI unit for rotational inertia II is kg⋅m² and for angular velocity ω\omega is rad/s, what are the fundamental SI units for rotational kinetic energy? Note that the radian is a dimensionless unit.

  1. kg⋅m⋅s−1\text{kg} \cdot \text{m} \cdot \text{s}^{-1}
  2. kg⋅m2⋅s−2\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2} (correct answer)
  3. kg⋅m2⋅s−1\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-1}
  4. kg⋅m−1⋅s−2\text{kg} \cdot \text{m}^{-1} \cdot \text{s}^{-2}
Explanation: The units can be found by substituting the units of II and ω\omega into the expression. The unit for II is kg⋅m2\text{kg} \cdot \text{m}^2. The unit for ω\omega is rad/s\text{rad/s}. Since the radian is dimensionless, the unit for ω\omega is effectively s−1\text{s}^{-1}. Squaring this gives s−2\text{s}^{-2}. Multiplying the units gives (kg⋅m2)⋅(s−2)=kg⋅m2⋅s−2(\text{kg} \cdot \text{m}^2) \cdot (\text{s}^{-2}) = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}. This is the unit for energy, the Joule.

Question 7

A rigid body is both translating with its center of mass moving at speed vcmv_{cm} and rotating with angular speed ω\omega. Which of the following statements provides the most fundamental definition of its total kinetic energy?

  1. The sum of the translational kinetic energy 12Mvcm2\frac{1}{2}Mv_{cm}^2 and the rotational kinetic energy 12Iω2\frac{1}{2}I\omega^2.
  2. The kinetic energy of a point mass with the same total mass moving at the speed of the fastest point on the body.
  3. The sum of the kinetic energies, 12mivi2\frac{1}{2}m_i v_i^2, of all the individual particles that constitute the rigid body. (correct answer)
  4. The work required to bring the body from rest to its current state of motion, assuming no dissipative forces.
Explanation: The most fundamental definition of the kinetic energy of any system of particles, including a rigid body, is the scalar sum of the kinetic energies of all its constituent particles. The formula Ktotal=12Mvcm2+12Icmω2K_{total} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2 is a convenient result derived from this fundamental definition for the special case of a rigid body. The work-energy theorem (D) relates work to the change in kinetic energy but is not the definition of kinetic energy itself.

Question 8

Based on the scenario, calculate the flywheel's rotational kinetic energy for I=0.80 kg\cdotpm2I=0.80\ \text{kg·m}^2 and ω=50 rad/s\omega=50\ \text{rad/s}.

  1. 800 J800\ \text{J}
  2. 1000 J1000\ \text{J} (correct answer)
  3. 1250 J1250\ \text{J}
  4. 2000 N\cdotpm2000\ \text{N·m}
Explanation: 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.

Question 9

A thin-walled hollow sphere of mass MM and radius RR rolls without slipping. Its total kinetic energy is KtotalK_{total}. What is the translational speed vv of its center of mass? The rotational inertia of a thin-walled hollow sphere is I=23MR2I=\frac{2}{3}MR^2.

  1. KtotalM\sqrt{\frac{K_{total}}{M}}
  2. 3Ktotal2M\sqrt{\frac{3K_{total}}{2M}}
  3. 6Ktotal5M\sqrt{\frac{6K_{total}}{5M}} (correct answer)
  4. 5Ktotal3M\sqrt{\frac{5K_{total}}{3M}}
Explanation: The total kinetic energy is Ktotal=Ktrans+Krot=12Mv2+12Iω2K_{total} = K_{trans} + K_{rot} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2. For rolling without slipping, ω=v/R\omega = v/R. So, Ktotal=12Mv2+12(23MR2)(vR)2=12Mv2+13Mv2=56Mv2K_{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 vv gives v2=6Ktotal5Mv^2 = \frac{6K_{total}}{5M}, so v=6Ktotal5Mv = \sqrt{\frac{6K_{total}}{5M}}.

Question 10

A flywheel with rotational inertia II rotates with angular velocity ω\omega, possessing rotational kinetic energy KrotK_{rot}. If the flywheel's angular velocity is increased to 4ω4\omega while its rotational inertia remains constant, what is its new rotational kinetic energy?

  1. 2Krot2K_{rot}
  2. 4Krot4K_{rot}
  3. 8Krot8K_{rot}
  4. 16Krot16K_{rot} (correct answer)
Explanation: Rotational kinetic energy is given by the formula Krot=12Iω2K_{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 42=164^2 = 16. The new kinetic energy will be 16Krot16K_{rot}.

Question 11

A uniform solid cylinder of mass MM and radius RR rolls without slipping on a horizontal surface. The center of mass of the cylinder has a speed vv. The rotational inertia of a solid cylinder about its central axis is I=12MR2I = \frac{1}{2}MR^2. What is the total kinetic energy of the cylinder?

  1. 12Mv2\frac{1}{2}Mv^2
  2. 34Mv2\frac{3}{4}Mv^2 (correct answer)
  3. Mv2Mv^2
  4. 54Mv2\frac{5}{4}Mv^2
Explanation: The total kinetic energy is the sum of the translational and rotational kinetic energies, Ktotal=Ktrans+KrotK_{total} = K_{trans} + K_{rot}. The translational kinetic energy is Ktrans=12Mv2K_{trans} = \frac{1}{2}Mv^2. The rotational kinetic energy is Krot=12Iω2K_{rot} = \frac{1}{2}I\omega^2. For rolling without slipping, v=Rωv = R\omega, so ω=v/R\omega = v/R. Substituting for II and ω\omega, we get Krot=12(12MR2)(vR)2=14Mv2K_{rot} = \frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2 = \frac{1}{4}Mv^2. The total kinetic energy is Ktotal=12Mv2+14Mv2=34Mv2K_{total} = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2 = \frac{3}{4}Mv^2.

Question 12

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?

  1. Both have the same total kinetic energy, but the hoop has greater translational kinetic energy.
  2. Both have the same total kinetic energy, but the disk has greater translational kinetic energy. (correct answer)
  3. The disk has greater total kinetic energy, and also greater translational kinetic energy.
  4. The hoop has greater total kinetic energy, but the disk has greater translational kinetic energy.
Explanation: By conservation of energy, both objects convert the same amount of gravitational potential energy (mghmgh) into total kinetic energy. Thus, their total kinetic energies are equal. The hoop has a larger rotational inertia (MR2MR^2) than the disk (12MR2\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.

Question 13

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?

  1. The wheel possesses only translational kinetic energy since every point on the rim is moving.
  2. The wheel possesses only rotational kinetic energy since its center of mass is not translating. (correct answer)
  3. The wheel has zero total kinetic energy because the velocity of its center of mass is zero.
  4. The wheel has equal amounts of translational and rotational kinetic energy.
Explanation: 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.

Question 14

A solid sphere with rotational inertia I=25MR2I = \frac{2}{5}MR^2 and a hollow sphere with rotational inertia I=23MR2I = \frac{2}{3}MR^2 have the same mass MM and radius RR. If they rotate with the same angular velocity ω\omega, what is the ratio of the hollow sphere's rotational kinetic energy to the solid sphere's rotational kinetic energy?

  1. 3/53/5
  2. 11
  3. 5/35/3 (correct answer)
  4. 25/925/9
Explanation: The ratio of their kinetic energies is KhollowKsolid=12Ihollowω212Isolidω2=IhollowIsolid\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 23MR225MR2=2/32/5=23×52=53\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}.

Question 15

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?

  1. 4
  2. 8
  3. 16 (correct answer)
  4. 32
Explanation: Mass MM is proportional to volume, so M∝R2tM \propto R^2t. Since thickness tt is the same, M∝R2M \propto R^2. Thus, M1=4M2M_1 = 4M_2. Rotational inertia for a disk is I=12MR2I = \frac{1}{2}MR^2. The ratio of inertias is I1I2=12M1R1212M2R22=(4M2)(2R2)2M2R22=4M2⋅4R22M2R22=16\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 Krot=12Iω2K_{rot} = \frac{1}{2}I\omega^2 and ω\omega is the same for both, the ratio of kinetic energies is equal to the ratio of their rotational inertias, which is 16.

Question 16

A solid sphere rolls without slipping on a horizontal surface. The sphere's rotational kinetic energy is KrotK_{rot}. What is its translational kinetic energy, KtransK_{trans}, in terms of KrotK_{rot}? The rotational inertia of a solid sphere is I=25MR2I = \frac{2}{5}MR^2.

  1. 25Krot\frac{2}{5}K_{rot}
  2. 23Krot\frac{2}{3}K_{rot}
  3. 32Krot\frac{3}{2}K_{rot}
  4. 52Krot\frac{5}{2}K_{rot} (correct answer)
Explanation: Krot=12Iω2=12(25MR2)(vR)2=15Mv2K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2 = \frac{1}{5}Mv^2. The translational kinetic energy is Ktrans=12Mv2K_{trans} = \frac{1}{2}Mv^2. From the expression for rotational energy, Mv2=5KrotMv^2 = 5K_{rot}. Substituting this into the translational energy expression gives Ktrans=12(5Krot)=52KrotK_{trans} = \frac{1}{2}(5K_{rot}) = \frac{5}{2}K_{rot}.

Question 17

A solid cylinder starts from rest at height HH 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′H' the cylinder reaches on ramp B?

  1. H′H' is less than HH.
  2. H′H' is equal to HH. (correct answer)
  3. H′H' is greater than HH.
  4. The relationship depends on the angles of the ramps.
Explanation: 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 MgHMgH 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′MgH' on the second ramp. Therefore, MgH=MgH′MgH = MgH', which means H′=HH' = H.

Question 18

A spinning platform has rotational kinetic energy KK. A student modifies the platform, halving its rotational inertia and doubling its angular velocity. What is the new rotational kinetic energy, K′K', in terms of KK?

  1. K′=K/2K' = K/2
  2. K′=KK' = K
  3. K′=2KK' = 2K (correct answer)
  4. K′=4KK' = 4K
Explanation: The initial kinetic energy is K=12Iω2K = \frac{1}{2}I\omega^2. The new inertia is I′=I/2I' = I/2 and the new angular velocity is ω′=2ω\omega' = 2\omega. The new kinetic energy is K′=12I′(ω′)2=12(I2)(2ω)2=12(I2)(4ω2)=2(12Iω2)=2KK' = \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.

Question 19

A uniform solid cylinder of mass MM and radius RR rolls without slipping. Its total kinetic energy is EE. What is its angular velocity ω\omega? The rotational inertia of a solid cylinder is I=12MR2I = \frac{1}{2}MR^2.

  1. 1REM\frac{1}{R}\sqrt{\frac{E}{M}}
  2. 1R2EM\frac{1}{R}\sqrt{\frac{2E}{M}}
  3. 1R4E3M\frac{1}{R}\sqrt{\frac{4E}{3M}} (correct answer)
  4. 1R3E4M\frac{1}{R}\sqrt{\frac{3E}{4M}}
Explanation: The total kinetic energy EE is the sum of translational and rotational kinetic energies. E=Ktrans+Krot=12Mv2+12Iω2E = K_{trans} + K_{rot} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2. For rolling without slipping, v=Rωv = R\omega. Substituting this and I=12MR2I = \frac{1}{2}MR^2 gives E=12M(Rω)2+12(12MR2)ω2=12MR2ω2+14MR2ω2=34MR2ω2E = \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 ω\omega gives ω2=4E3MR2\omega^2 = \frac{4E}{3MR^2}, so ω=1R4E3M\omega = \frac{1}{R}\sqrt{\frac{4E}{3M}}.

Question 20

A bowling ball (solid sphere) and a volleyball (hollow sphere) have the same mass and radius. They are rolled on a horizontal surface so that their centers of mass have the same translational speed. Which ball has the greater total kinetic energy?

  1. The bowling ball, because it is solid and has a smaller rotational inertia for the same mass.
  2. The volleyball, because it is hollow and has a larger rotational inertia for the same mass. (correct answer)
  3. They both have the same total kinetic energy because their mass and speed are the same.
  4. They both have the same total kinetic energy because their translational kinetic energies are the same.
Explanation: Both balls have the same translational kinetic energy (12Mv2\frac{1}{2}Mv^2) since their masses and speeds are the same. The total kinetic energy is Ktotal=Ktrans+KrotK_{total} = K_{trans} + K_{rot}. The rotational kinetic energy is Krot=12Iω2K_{rot} = \frac{1}{2}I\omega^2. Since vv is the same, ω=v/R\omega = v/R is also the same for both. The volleyball, being a hollow sphere, has a larger rotational inertia (I=23MR2I = \frac{2}{3}MR^2) than the solid bowling ball (I=25MR2I = \frac{2}{5}MR^2). Therefore, the volleyball has greater rotational kinetic energy, and thus greater total kinetic energy.