Topic 6.1 Rotational Kinetic Energy: define rotational kinetic energy as $\tfrac{1}{2}I\omega^2$, combine it with translational kinetic energy for a moving, spinning body, and use it in energy conservation.

What this topic is asking

The College Board (Topic 6.1) wants you to define rotational kinetic energy as $\tfrac{1}{2}I\omega^2$, to combine it with translational kinetic energy for a body that both moves and spins, and to use it in energy conservation. This opens Unit 6 by extending the energy framework to rotation: a rolling or spinning body stores kinetic energy in its rotation, and accounting for it is the key to rolling problems.

Rotational kinetic energy

The form is the exact rotational analogue of translational kinetic energy: $\tfrac{1}{2}mv^2$ becomes $\tfrac{1}{2}I\omega^2$, with mass replaced by rotational inertia and linear speed by angular speed. It can be derived by summing $\tfrac{1}{2}m_i v_i^2$ over all the mass elements, using $v_i = r_i\omega$, which gives $\tfrac{1}{2}(\sum m_i r_i^2)\omega^2 = \tfrac{1}{2}I\omega^2$. A spinning flywheel stores energy this way, which is why flywheels are used as energy reservoirs.

Total kinetic energy of a moving, spinning body

A body can do both at once: a rolling wheel translates while it spins. Its total kinetic energy is the sum of the two parts:

The first term is the translational kinetic energy of the center of mass; the second is the rotational kinetic energy about the center of mass. This neat split (sometimes called the König decomposition) holds for any rigid body. For a body rolling without slipping, the two are linked by $v_{cm} = R\omega$, so you can write the total energy in terms of a single variable and use it directly in energy conservation.

Energy conservation with rotation

When you apply conservation of energy to a rotating system, simply include the rotational kinetic energy term. For a body released from rest at height $h$ and rolling down without slipping (rolling friction does no work because the contact point is instantaneously at rest), energy conservation reads

Substituting $\omega = v_{cm}/R$ and the body's $I_{cm}$ gives the speed at the bottom. The result depends on the shape through $I_{cm}$: a hoop ($I = MR^2$) is slower than a disk ($\tfrac{1}{2}MR^2$), which is slower than a sphere ($\tfrac{2}{5}MR^2$), because more rotational inertia diverts more energy into spinning. The mass always cancels.

Try this

Q1. A flywheel with $I = 0.50$ kg m squared spins at $20$ rad/s. Calculate its rotational kinetic energy. [2 points]

• Cue. $K_{rot} = \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}(0.50)(20)^2 = 100$ J.

Q2. Explain why a solid disk rolls down a ramp faster than a hoop of the same mass and radius. [2 points]

• Cue. The disk has smaller rotational inertia, so less of the energy goes into rotation and more into translation, giving a larger center-of-mass speed.