A flywheel on a fixed axle turns in place, so its center of mass does not move and its total momentum is zero. The flywheel still has kinetic energy, because it takes work to stop it, and a heavy flywheel can keep a machine running for a while after the motor is switched off. Unit 5 used \(\omega\), \(I\) and \(\tau_{\text{net}} = I\alpha\) to describe the turning. This topic finds how much kinetic energy a turning rigid body has.

Rotational kinetic energy is not a new kind of energy. It is the kinetic energy \(\tfrac{1}{2}mv^2\) of each particle, added up over every particle of the body. The particles farther from the axis move faster, so the rotational inertia \(I\) appears in the result. After that, you will split the kinetic energy of a body that moves and spins into two terms, and you will use both terms in conservation of energy, the way you did in Unit 3.

1. Rotational kinetic energy is the kinetic energy of every particle added together

A rigid body turning about a fixed axis does not have one speed. Every particle has the same angular velocity \(\omega\). However, a particle at distance \(r_i\) from the axis moves at \(v_i = r_i\omega\). The particles near the rim move fastest, and a particle on the axis does not move at all.

To find the kinetic energy, you add \(\tfrac{1}{2}m_iv_i^2\) over every particle. Since \(\omega\) is the same for all of them, it comes out of the sum. What is left inside the sum is \(\sum m_ir_i^2\), which is the rotational inertia of the body. Check the units: kg m\(^2\) times (rad/s)\(^2\) is kg m\(^2\)/s\(^2\), which is a joule.

\(K_{\text{rot}} = \sum \tfrac{1}{2}m_i(r_i\omega)^2 = \tfrac{1}{2}I\omega^2\)the counterpart of \(K = \tfrac{1}{2}mv^2\)

For a body turning about a fixed axis, \(\tfrac{1}{2}I\omega^2\), with \(I\) about that axis, is the ordinary kinetic energy of all its particles added up. This sum is the body’s total kinetic energy, and no other term is needed. For a hoop, the two expressions are equal term by term: \(\tfrac{1}{2}(MR^2)\omega^2 = \tfrac{1}{2}M(R\omega)^2\). A common mistake is to use \(\tfrac{1}{2}Mv_{cm}^2\) for a wheel on a fixed axle. That is wrong because the center of mass is at rest, so the translational term is zero while every other particle is moving. Another mistake is to give every particle the rim speed and write \(\tfrac{1}{2}M(R\omega)^2\). That expression is correct for a hoop, since all of the mass of a hoop is at one radius. For a uniform disk it gives twice the true value, because the particles of the disk have speeds from \(0\) up to \(R\omega\).

Rotational kinetic energy is a scalar. It has a size but no direction. Since \(\omega\) is squared, the energy can never be negative. A wheel turning clockwise has the same \(K_{\text{rot}}\) as the same wheel turning counterclockwise at the same rate. When a system has several turning parts, you add their kinetic energies as plain numbers, whichever way each part turns.

2. The mass farthest from the axis carries the most energy

Each particle’s share of the kinetic energy is \(\tfrac{1}{2}m_ir_i^2\omega^2\). This share grows as the square of the particle’s distance from the axis. Mass near the rim carries far more energy than the same mass near the center. A hoop and a uniform disk with the same \(M\), \(R\) and \(\omega\) differ by a factor of two. The reason is that \(I\) is \(MR^2\) for the hoop and \(\tfrac{1}{2}MR^2\) for the disk.

You can also find the kinetic energy by integrating directly, without first finding \(I\). A thin ring of a uniform disk at radius \(r\), with width \(dr\), has mass \(dm = (2M/R^2)\,r\,dr\). Integrate its energy from the center out to \(r\), and you find that the share of the energy inside radius \(r\) is \((r/R)^4\). The inner half of the radius therefore holds only \(1/16\) of the energy, and the outer half holds \(15/16\).

\(dK = \tfrac{1}{2}\omega^2r^2\,dm\)integrated over the body, this gives \(\tfrac{1}{2}I\omega^2\)

For a uniform rod turning about one end, the same integral gives \(K = \tfrac{1}{6}ML^2\omega^2\). That is \(\tfrac{1}{2}I\omega^2\) with \(I = \tfrac{1}{3}ML^2\). Finding the energy both ways gives the same result, which is a check on your work.

The angular velocity has a larger effect on the kinetic energy than the mass has. Doubling \(M\) doubles \(K\), but doubling \(\omega\) multiplies it by four. For this reason, a flywheel built to store energy turns at a high angular velocity and has most of its mass in a heavy rim. A uniform disk flywheel with \(M = 20\) kg and \(R = 0.30\) m has \(I = 0.90\) kg m\(^2\). At \(150\) rad/s, it stores about \(1.0 \times 10^4\) J.

3. A body that moves and spins has two kinetic energy terms

A thrown flying disc moves and spins at the same time. To handle that, split each particle’s velocity into the velocity of the center of mass plus its velocity relative to the center of mass, \(\vec{v}_i = \vec{v}_{cm} + \vec{v}\,’_i\). Square this and add it up over the body, and you get three terms: \(\tfrac{1}{2}Mv_{cm}^2\), the spin term \(\tfrac{1}{2}I_{cm}\omega^2\), and a cross term \(\vec{v}_{cm}\cdot\sum m_i\vec{v}\,’_i\).

The cross term is zero, since motion relative to the center of mass carries no net momentum. Then you are left with only two terms. The zero cross term is also the reason the rotational term must use \(I\) about the center of mass. If you pair \(\tfrac{1}{2}Mv_{cm}^2\) with \(I\) about some other point, you count an extra \(\tfrac{1}{2}Md^2\omega^2\) that the body does not have.

\(K = \tfrac{1}{2}Mv_{cm}^2 + \tfrac{1}{2}I_{cm}\omega^2\)for any rigid body, rolling or not

The two descriptions agree whenever both apply. Take a uniform rod turning about a fixed pivot at one end. Its total kinetic energy is \(\tfrac{1}{2}I_{\text{end}}\omega^2 = \tfrac{1}{6}ML^2\omega^2\), as in Section 2. You can also split this energy into two terms. The center moves at \(v_{cm} = \tfrac{1}{2}L\omega\), so \(\tfrac{1}{2}Mv_{cm}^2 = \tfrac{1}{8}ML^2\omega^2\). The spin about the center adds \(\tfrac{1}{2}(\tfrac{1}{12}ML^2)\omega^2 = \tfrac{1}{24}ML^2\omega^2\). The two add to \(\tfrac{1}{6}ML^2\omega^2\). Use one description or the other. Adding the terms from both counts part of the energy twice.

It helps to find each term on its own before you add them. Model a flying disc of mass \(0.175\) kg and radius \(0.135\) m as a uniform disk, thrown at \(8.0\) m/s and spinning at \(50\) rad/s. For this disc, \(I_{cm} = 1.59 \times 10^{-3}\) kg m\(^2\), \(K_{\text{trans}} = 5.6\) J and \(K_{\text{rot}} = 2.0\) J. About a quarter of the disc’s kinetic energy is rotational. The motion of your arm gives the disc most of its energy as translational kinetic energy, and the motion of your wrist at release gives it the rotational part.

4. Energy conservation with a massive pulley

In Topic 5.6 you used forces and torques to find the motion of a block hanging from a cord wound on a pulley. This time you will use energy. As the block falls a height \(h\), the potential energy it loses is shared between its own kinetic energy and the pulley’s rotational kinetic energy. The cord links the two speeds through \(v = R\omega\), so you can write everything in terms of \(v\).

\(mgh = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2, \qquad v = R\omega\)the energy the block loses, shared with the pulley

Solving gives \(v^2 = 2gh/(1 + I/mR^2)\). That is \(v^2 = 2ah\) with \(a = g/(1 + I/mR^2)\), the same acceleration you found for this system in Topic 5.6. With energy, the result takes one line instead of three equations. The block ends up slower than in free fall, since part of the energy went into turning the pulley.

However, energy does not give you the tension. The tension is internal to the block and pulley system. It does negative work on the block and equal positive work on the pulley, so it drops out of the energy equation. To find it, apply \(mg-T = ma\) to the block alone.

The same idea works for any rigid body that turns as it falls. Consider a uniform rod of length \(L\), pivoted at one end and released from horizontal. Its center of mass falls only \(L/2\), so \(Mg(L/2) = \tfrac{1}{6}ML^2\omega^2\) at the bottom. Its free end then moves at \(\sqrt{3gL}\). That is faster than a stone dropped through \(L\), which reaches \(\sqrt{2gL}\). This does not violate conservation of energy. The parts of the rod near the pivot move slower than they would in free fall, and the rigid rod transfers energy from those parts to its free end.