Assume that there is no friction loss due to the rotating wheels.

Because gravity is a conservative force, the Principle of Conservation of Energy can be applied,

     T₁ + V₁ = T₂ + V₂

Moment of Inertia of Wheels

Angular Rotation and Linear Velocity

The car body has only translational kinetic energy since the rotation due to the hill is small and can be ignored.

The wheels of both cars have both linear velocity and angular velocity so the total kinetic energy of the car and its wheels are,

     T₁ = 0

     T₂ = 1/2 m v_G² + 1/2 I_G ω²

        = 1/2 (m_body + 4m_wheels) v² + 1/2 4I_wheel ω²

Angular velocity ω and linear velocity v are related (assumes no slipping occurs) as

     v = ωr

If the datum line is drawn to coincide with the center of gravity of the wheel when it is at the bottom of the hill, then the potential energy of the cars at position 2 is zero.

The potential energy of the cars at the start is based on the height of the hill:

     V₁ = mgy = (m_body + m_wheels) g (13.7 ft)

Now that all the energy terms have been identified, the conservation of energy can be used to solve for the velocity of the car with disk wheels.

   T₁ + V₁ = T₂ + V₂
   
    v² = 23.98 g

    v_box = 27.8 ft/s

The only difference between the cars is the moment of inertia of the wheels. The conservation of energy for the sports car is

    T₁ + V₁ = T₂ + V₂

    v² = 21.31g

    v_sport = 26.2 ft/s

The box car is faster than the sports car because of the wheel's lower moment of inertia. Recall, the wheels were assumed to have the same mass.