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.

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 $$K_{rot} = MgL/2$$.

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 $$\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{2}Mv^2}{Mv^2} = \frac{1}{2}$$.

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 $$H' < H$$.

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 $$K_{rot} = K_{trans}$$.

The units can be found by substituting the units of $$I$$ and $$\omega$$ into the expression. The unit for $$I$$ is $$\text{kg} \cdot \text{m}^2$$. The unit for $$\omega$$ is $$\text{rad/s}$$. Since the radian is dimensionless, the unit for $$\omega$$ is effectively $$\text{s}^{-1}$$. Squaring this gives $$\text{s}^{-2}$$. Multiplying the units gives $$(\text{kg} \cdot \text{m}^2) \cdot(\text{s}^{-2}) = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}$$. This is the unit for energy, the Joule.

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 $$K_{total} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$$ 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.

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.

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.

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.