Why a Solid Ball Rolls Down a Slope Faster Than a Hollow Ring

A heavy flywheel resists being started or stopped, a merry-go-round speeds up when riders move inwards, and a solid ball beats a ring down a slope. All of this depends on how mass is spread around an axis — the moment of inertia.

This lesson covers moment of inertia and the axis theorems, the rotational form of Newton's second law, conservation of angular momentum, and rolling motion.

**Moment of inertia, $I=\sum m_i{r}_i^2$, measures how hard a body is to rotate about an axis, and the parallel axis theorem, $I=I_{cm}+M{d}^2$, and perpendicular axis theorem, $I_z=I_x+I_y$, give it about new axes.

Standard values** for mass M:

- Ring about its axis: $M{R}^2$; disc about its axis: $\frac{1}{2}M{R}^2$
- Solid sphere about a diameter: $\frac{2}{5}M{R}^2$; thin rod about its centre: $\frac{1}{12}M{L}^2$
- Radius of gyration k is defined by $I=M{k}^2$

The theorems:

- Parallel axis: $I=I_{cm}+M{d}^2$, for an axis parallel to one through the centre of mass, a distance d away
- Perpendicular axis: $I_z=I_x+I_y$, for a flat body, about an axis perpendicular to its plane

Worked example 1 — a rod about one end. For a 1.2 m rod of mass 0.50 kg:

$$I=\frac{1}{12}M{L}^2+M{\left(\frac{L}{2}\right)}^2=\frac{1}{3}M{L}^2=\frac{1}{3}\times 0.50\times 1.2^2=0.24{\text{ kg m}}^2$$

Worked example 2 — a disc about a diameter. By symmetry $I_x=I_y$, and $I_z=\frac{1}{2}M{R}^2$, so $I_{\text{diameter}}=\frac{1}{4}M{R}^2$. For a 2.0 kg disc of radius 0.30 m, this is $\frac{1}{4}\times 2.0\times 0.30^2=0.045$ kg m${}^2$.

An everyday example. A potter's wheel in Khurja keeps turning between kicks because its heavy rim gives it a large moment of inertia.

The substance. Moment of inertia depends on the axis, not just the mass — a rod is four times harder to spin about one end than about its centre.

**The net torque on a rigid body about a fixed axis equals its moment of inertia times its angular acceleration, $\tau =I\alpha$, the rotational counterpart of $F=ma$.

Rotational analogues.** Angle $\theta$, angular velocity $\omega$, angular acceleration $\alpha$, moment of inertia I and torque $\tau$ play the roles of s, v, a, m and F, so for constant $\alpha$:

$$\omega ={\omega }_0+\alpha t\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2{K}_{\text{rot}}=\frac{1}{2}I{\omega }^2$$

Worked example — a block and pulley. A 2.0 kg block hangs from a string wound on a pulley, a disc of mass 4.0 kg. For the block $mg-T=ma$; for the pulley $TR=\frac{1}{2}M{R}^2\times \frac{a}{R}$, so $T=\frac{1}{2}Ma$:

$$a=\frac{mg}{m+\frac{1}{2}M}=\frac{2.0\times 9.8}{4.0}=4.9{\text{ m s}}^{-2}T=\frac{1}{2}\times 4.0\times 4.9=9.8\text{ N}$$

An everyday example. A ceiling fan takes a few seconds to reach full speed because the motor's torque must overcome the large moment of inertia of its long blades.

The substance. A heavier pulley slows the falling block — part of the pull of gravity goes into giving the pulley angular acceleration.

**When no external torque acts, total angular momentum $L=I\omega$ stays constant, so reducing the moment of inertia makes a body spin faster and increasing it makes it spin slower.**

$$I_1{\omega }_1=I_2{\omega }_2\text{when }{\tau }_{\text{ext}}=0$$

Worked example — a rotating stool. A student on a rotating stool holds weights at arm's length, with total $I=6.0$ kg m${}^2$, turning at 1.0 rev s${}^{-1}$. Pulling the weights in reduces I to 2.0 kg m${}^2$:

$${\omega }_2=\frac{6.0\times 1.0}{2.0}=3.0{\text{ rev s}}^{-1}$$

Since $K=\frac{L^2}{2I}$, the kinetic energy triples; the extra energy comes from the work done pulling the weights inwards.

An everyday example. Children on a playground merry-go-round who move towards the centre make it spin faster, because the system's moment of inertia falls.

The substance. Angular momentum is conserved but kinetic energy need not be — pulling weights in adds energy, while a ring dropped on a turntable loses some to friction.

**A body rolling without slipping combines translation of its centre of mass with rotation about it, with $v=\omega R$, so its kinetic energy is $\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2=\frac{1}{2}M{v}^2\left(1+\frac{k^2}{R^2}\right)$.

Rolling without slipping:**

- The point touching the ground is momentarily at rest, and the top point moves at $2v$
- Rolling down an incline of height h from rest: $v=\sqrt{\frac{2gh}{1+k^2/R^2}}$
- A solid sphere ($k^2/R^2=\frac{2}{5}$) arrives first, then a disc ($\frac{1}{2}$), then a ring (1)

Worked example. A solid sphere and a ring roll from rest down a slope 1.4 m high:

$$v_{\text{sphere}}=\sqrt{\frac{2\times 9.8\times 1.4}{1.4}}=4.43{\text{ m s}}^{-1}{v}_{\text{ring}}=\sqrt{\frac{2\times 9.8\times 1.4}{2}}=3.70{\text{ m s}}^{-1}$$

Mass and radius cancel, so any solid sphere beats any ring. For a ring, half the kinetic energy is rotational; for a solid sphere, only $\frac{2}{7}$ of it is.

An everyday example. Children racing a steel ring and a marble down a sloping lane will always see the marble win, because less of its energy is locked up in spinning.

The substance. Friction lets a body roll but takes no energy from it — the contact point is at rest, so static friction does no work.