In a village near Nashik, children race old bicycle rims along a dusty lane, driving each one with a short stick. Bringing a rim to a running pace takes a surprising number of taps. A wooden block of the same mass, slid along the same lane at the same speed, would need less energy. Where does the extra energy of the rolling rim go?
The ride and the spin
The chapter on Work, Kinetic Energy and Power defined the kinetic energy of a particle as \(\tfrac12 mv^2\). The chapter on Rigid Bodies found the kinetic energy of a body turning about a fixed axis, \(\tfrac12 I\omega^2\). A rolling wheel does both at once: it moves along the road and it turns. What is its kinetic energy?
Split the velocity of every particle into two parts, as in lesson 1, using the centre of mass as the reference point. Each particle has the velocity \(\vec v_{cm}\) of the centre of mass plus a velocity \(\vec v\,'\) relative to it. Then sum \(\tfrac12 mv^2\) over all the particles.
\[ K = \sum \tfrac12 m\,(\vec v_{cm} + \vec v\,')\cdot(\vec v_{cm} + \vec v\,') \]
Why. the kinetic energy of the body is the sum of \(\tfrac12 mv^2\) over its particles, with \(\vec v = \vec v_{cm} + \vec v\,'\).
\[ K = \tfrac12 Mv_{cm}^2 + \vec v_{cm}\cdot\sum m\vec v\,' + \sum \tfrac12 m v'^2 \]
Why. expand the dot product; \(\vec v_{cm}\) is the same for every particle, so it is a common factor of each sum, and \(\sum m = M\).
\[ \sum m\vec v\,' = 0 \]
Why. this is the total momentum measured from the centre of mass. By the definition of the centre of mass, \(\sum m\vec r\,' = 0\). Its rate of change is zero too, so the middle term vanishes exactly.
\[ \sum \tfrac12 m v'^2 = \tfrac12\Big(\sum m r'^2\Big)\omega^2 = \tfrac12 I_{cm}\omega^2 \]
Read it. relative to its centre of mass a rigid body can only turn. So each particle has \(v' = \omega r'\), where \(r'\) is its distance from the axis.
The kinetic energy of a rigid body moving in a plane has two parts. One is the energy of its whole mass moving with the centre of mass. The other is the energy of its rotation about the centre of mass: \(K = \tfrac12 Mv_{cm}^2 + \tfrac12 I_{cm}\omega^2\).
The first term is the translational kinetic energy, the energy of the ride. It is what the body would have if all its mass moved with the centre of mass. The second is the rotational kinetic energy, the energy of the spin. It is what the body would have if its centre of mass were held still while it turned. Nothing is dropped and nothing is approximated. The split is exact for any motion.
A frisbee of mass \(0.2\,\text{kg}\) and radius \(0.1\,\text{m}\) is thrown on a beach in Goa. It moves at \(8\,\text{m/s}\) and spins at \(50\,\text{rad/s}\) about its own axis. Treat it as a uniform disc and find its kinetic energy.
The ride: \(\tfrac12 Mv^2 = \tfrac12 \times 0.2 \times 8^2 = 6.4\,\text{J}\).
The spin: \(I_{cm} = \tfrac12 MR^2 = \tfrac12 \times 0.2 \times 0.1^2 = 0.001\,\text{kg m}^2\), so \(\tfrac12 I_{cm}\omega^2 = \tfrac12 \times 0.001 \times 50^2 = 1.25\,\text{J}\).
The total is \(6.4 + 1.25 = 7.65\,\text{J}\). The spin holds only about \(16\,\%\) of it, because a light, thin disc has a small moment of inertia.
A frisbee does not roll, so its spin and its speed are free of each other. The thrower can give it more spin or less. A rolling body has no such freedom. The rolling condition ties \(\omega\) to \(v\), and that fixes how the energy divides, as the third section of this lesson shows.
A wheel of mass \(M\) rolls along a road with its centre moving at \(v\). Its kinetic energy is
Why the centre of mass
The middle term of the derivation vanished for one reason: the momentum measured from the centre of mass is zero. That is true of the centre of mass and of no other point in general. Take the velocities relative to some other point X of the body, and the matching sum \(\sum m\vec v\,'\) is not zero. A cross term survives, and \(\tfrac12 Mv_X^2 + \tfrac12 I_X\omega^2\) is then not the kinetic energy.
The numbers in Fig 6.1 are easy to check. About the centre, \(\tfrac12 \times 2 \times 3^2 = 9\,\text{J}\) and \(\tfrac12 \times \tfrac12 MR^2 \times (v/R)^2 = \tfrac14 Mv^2 = 4.5\,\text{J}\): \(13.5\,\text{J}\) in all. About the top, the point moves at \(6\,\text{m/s}\), giving \(36\,\text{J}\). With \(I_T = \tfrac32 MR^2\) the spin term is \(13.5\,\text{J}\), and the bar reaches \(49.5\,\text{J}\). That is wrong.
About the contact point P the formula works again, for a different reason. P is at rest at that instant, so \(\tfrac12 Mv_P^2 = 0\). The whole disc is turning about P, as lesson 4 showed, and the energy of a pure rotation about a point at rest is \(\tfrac12 I_P\omega^2\). The last section of this lesson returns to it.
Split the kinetic energy about the centre of mass. Or write it as a pure rotation about a point at rest at that instant. Never split it about any other point, such as the top of a wheel or the end of a thrown bat. The two terms there do not add to the kinetic energy.
A body rolls without slipping. Let \(v_X\) be the speed of a point X and \(I_X\) the moment of inertia about X. About which points does \(\tfrac12 Mv_X^2 + \tfrac12 I_X\omega^2\) give its kinetic energy?
The split in rolling
For a rolling body, put \(\omega = v/R\) into the split. Write the moment of inertia through the radius of gyration \(k\), as \(I_{cm} = Mk^2\). Then the spin term is \(\tfrac12 Mk^2 v^2/R^2\), and the two terms share the factor \(\tfrac12 Mv^2\).
\[ K = \tfrac12 Mv^2 + \tfrac12 Mk^2\,\frac{v^2}{R^2} = \tfrac12 Mv^2\left(1 + \frac{k^2}{R^2}\right) \]
The ratio \(k^2/R^2\) depends only on the shape, not on the mass or the size. So two rolling bodies of the same shape share their energy between ride and spin in the same proportion.
The ring has all its mass at the rim, so \(k = R\), and its spin carries as much energy as its ride. The solid sphere keeps much of its mass near the centre, so \(k^2/R^2\) is only \(2/5\), and its spin takes the smallest share, \(2/7\). The share of the spin is \((k^2/R^2)/(1 + k^2/R^2)\) in every case.
A gardener in Bengaluru pushes a stone roller across a lawn. Treat it as a solid cylinder of mass \(60\,\text{kg}\), rolling without slipping at \(1\,\text{m/s}\). Find its kinetic energy and the share of the spin.
A solid cylinder has \(I_{cm} = \tfrac12 MR^2\), so \(k^2/R^2 = \tfrac12\).
\(K = \tfrac12 Mv^2(1 + \tfrac12) = \tfrac34 Mv^2\).
The ride is \(\tfrac12 \times 60 \times 1^2 = 30\,\text{J}\), the spin \(15\,\text{J}\), and \(K = 45\,\text{J}\).
The roller carries \(45\,\text{J}\), and one third of it, \(15\,\text{J}\), is in the spin. To bring it to \(1\,\text{m/s}\) the gardener must supply the whole \(45\,\text{J}\), half as much again as a sledge of the same mass needs.
A solid sphere of mass \(5\,\text{kg}\) rolls without slipping at \(2\,\text{m/s}\). What is its kinetic energy, in joules?
A ring, a disc, a thin hollow sphere and a solid sphere have the same mass and roll at the same speed. Rank them by kinetic energy, largest first.
A car of mass \(1000\,\text{kg}\) moves along a highway. Each of its four wheels has a mass of about \(15\,\text{kg}\) and can be treated as a uniform disc. Estimate what percentage of the car's kinetic energy is in the spin of its wheels.
One term about the contact point
Lesson 4 showed that a rolling body turns about its contact point P at each instant. The energy of a body turning about an axis is \(\tfrac12 I\omega^2\) about that axis. So the whole kinetic energy of a rolling body can be written as a single term about P.
\[ K = \tfrac12 I_P\,\omega^2, \qquad I_P = I_{cm} + MR^2 \]
The second equation is the parallel-axis theorem: P is a distance \(R\) from the centre of mass.
Expand the single term, using \(\omega R = v\):
\[ \tfrac12 (I_{cm} + MR^2)\,\omega^2 = \tfrac12 I_{cm}\omega^2 + \tfrac12 MR^2\omega^2 = \tfrac12 I_{cm}\omega^2 + \tfrac12 Mv^2 \]
The one-term answer becomes the two-term answer. Nothing new was used. The \(MR^2\) inside \(I_P\) carries exactly the energy of the ride. For a disc, \(I_P = \tfrac32 MR^2\), and \(\tfrac12 \times \tfrac32 MR^2 \times v^2/R^2 = \tfrac34 Mv^2\), the value in the table.
Use one route or the other, never both at once. \(\tfrac12 I_P\omega^2\) is already the whole kinetic energy of a rolling body. Adding \(\tfrac12 Mv^2\) to it counts the ride twice. And remember that the contact-point route needs a point at rest; for a body that skids, only the split about the centre of mass works.
A ring of mass \(M\) and radius \(R\) rolls without slipping at speed \(v\). Using the contact point, its kinetic energy is
A student says that \(K = \tfrac12 I_P\omega^2\) for a rolling body leaves out the energy of translation. Explain why the student is wrong.
The moment of inertia about the contact point is \(I_P = I_{cm} + MR^2\) by the parallel-axis theorem. Expanding \(\tfrac12 I_P\omega^2\) with \(\omega R = v\) gives \(\tfrac12 I_{cm}\omega^2 + \tfrac12 Mv^2\). The \(MR^2\) part carries exactly the energy of translation, so nothing is left out.
Return to the children racing rims near Nashik. A bicycle rim is very nearly a ring, with all its mass at the edge. So its spin carries exactly as much energy as its run along the lane. Rolling it at a given speed takes twice the energy of sliding a block of the same mass. Half of every tap goes into the spin.
- Writing only \(\tfrac12 Mv^2\) for a rolling body.A rolling body also spins. Add \(\tfrac12 I_{cm}\omega^2\), or use \(\tfrac12 Mv^2(1 + k^2/R^2)\).
- Splitting the energy about a point other than the centre of mass.Use the centre of mass for the split, or the contact point for a single term. Any other point leaves a cross term.
- Adding \(\tfrac12 Mv^2\) to \(\tfrac12 I_P\omega^2\).\(\tfrac12 I_P\omega^2\) is already the total. Its \(MR^2\) part is the energy of the ride.