Rotation

The centre of mass of any body moves like a single particle, which leaves entirely undescribed the tumbling, spinning and rolling that goes on around it.

This lesson needs the kinematics of the first two lessons, the kinetic energy $\frac{1}{2} m v^{2}$ , and the result that the centre of mass of a system is its mass weighted average position. It builds the rotational half of mechanics from those.

Angular variables

A rigid body turning about a fixed axis has one degree of freedom: the angle $θ$ through which it has turned. Every particle in it sweeps the same angle in the same time, which is what "rigid" means, so one variable describes the whole object however complicated its shape.

Measure $θ$ in radians, defined as arc length divided by radius, $θ = s / r$ . The definition is what makes the whole formalism tidy, because it makes the arc length $s = r θ$ with no conversion factor, and a full turn is $2 π$ radians. Degrees would insert $π / 180$ into every equation in this lesson.

Differentiate as before. The angular velocity is $ω = d θ / d t$ in radians per second, and the angular acceleration is $α = d ω / d t$ in radians per second squared. Every statement of the first lesson carries over by replacing each symbol with its angular counterpart, so for constant $α$ ,

ω = ω_{0} + α t θ = θ_{0} + ω_{0} t + \frac{1}{2} α t^{2} ω^{2} = ω_{0}^{2} + 2 α Δ θ

These are not new physics and need no separate derivation: they are the same integrations performed on different letters.

The link back to linear quantities is through the radius. A particle at distance $r$ from the axis moves at speed $v = ω r$ along its circle, has a tangential acceleration $a_{t} = α r$ that changes that speed, and a radial acceleration $a_{r} = v^{2} / r = ω^{2} r$ that keeps it on the circle. The two accelerations are perpendicular, and both are present whenever a spinning body is speeding up. Note that $ω$ is a property of the whole body while $v$ is not: the rim of a wheel moves faster than a point near the hub, which is why the outer edge of a grinding wheel does the cutting.

Example. A wheel starts from rest and is given a constant angular acceleration of 2.5 rad s⁻² for 8.0 s. Find the final angular velocity, the angle turned and the number of revolutions. If the wheel has a radius of 0.40 m, how fast is a point on its rim moving at the end?

$ω = 2.5 \times 8.0 = 20$ rad s⁻¹. The angle is $\frac{1}{2} (2.5) (64) = 80$ radians, which is $80 / 2 π = 12.7$ revolutions. The rim speed is $v = ω r = 20 \times 0.40 = 8.0$ m s⁻¹, and the rim's radial acceleration at that moment is $ω^{2} r = 400 \times 0.40 = 160$ m s⁻², sixteen times $g$ .

Now you. A centrifuge rotor reaches 1200 revolutions per minute from rest in 5.0 s at constant angular acceleration. Find the final angular velocity, the angular acceleration and the number of revolutions made.

Answer

$ω = 1200 \times 2 π / 60 = 125.7$ rad s⁻¹, so $α = 125.7 / 5.0 = 25.1$ rad s⁻². The angle is $\frac{1}{2} (25.1) (25) = 314$ radians, which is exactly 50 revolutions.

Rotational kinetic energy, and where the moment of inertia comes from

A spinning body has kinetic energy even though its centre of mass is at rest, and adding it up is what produces the central quantity of this lesson.

Divide the body into particles of mass $m_{i}$ at distance $r_{i}$ from the axis. Each moves at $v_{i} = ω r_{i}$ , so the total kinetic energy is

K = \sum_{i} \frac{1}{2} m_{i} v_{i}^{2} = \sum_{i} \frac{1}{2} m_{i} ω^{2} r_{i}^{2} = \frac{1}{2} (\sum_{i} m_{i} r_{i}^{2}) ω^{2}

The angular velocity came outside the sum because it is the same for every particle. What is left inside is a property of the body and the chosen axis alone:

I = \sum_{i} m_{i} r_{i}^{2} K = \frac{1}{2} I ω^{2}

$I$ is the moment of inertia, in kg m². Compare $\frac{1}{2} m v^{2}$ and the analogy is exact: $I$ plays the part of mass in rotational motion, and it is the resistance a body offers to being spun up.

The crucial difference from mass is the $r^{2}$ . Moment of inertia depends on how the mass is distributed, not merely on how much there is, and mass far from the axis counts enormously more. Two wheels of the same mass, one a solid disc and one a hoop with all its mass at the rim, differ by a factor of two in $I$ . It also means a body has no single moment of inertia: the axis must be named, and the same rod has three different values about three natural axes.

Computing it

For a continuous body the sum becomes $I = \int r^{2} d m$ , and the skill is expressing $d m$ and choosing the element so that every part of it is at the same $r$ .

Take a uniform rod of mass $M$ and length $L$ about a perpendicular axis through its centre. A slice of length $d x$ at distance $x$ has $d m = (M / L) d x$ , and every point of it is at distance $| x |$ from the axis. So

I = \int_{- L / 2}^{L / 2} x^{2} \frac{M}{L} d x = \frac{M}{L} {[\frac{x^{3}}{3}]}_{- L / 2}^{L / 2} = \frac{M}{L} \cdot \frac{2}{3} \cdot \frac{L^{3}}{8} = \frac{M L^{2}}{12}

For a hoop of radius $R$ every particle is at $r = R$ , so $I = M R^{2}$ with no integration at all. For a solid disc about its axis, take a ring of radius $r$ and thickness $d r$ , whose mass is $d m = (M / π R^{2}) (2 π r d r)$ , and

I = \int_{0}^{R} r^{2} \frac{2 M r}{R^{2}} d r = \frac{2 M}{R^{2}} \cdot \frac{R^{4}}{4} = \frac{1}{2} M R^{2}

The results worth remembering are the hoop at $M R^{2}$ , the disc or cylinder at $\frac{1}{2} M R^{2}$ , the solid sphere at $\frac{2}{5} M R^{2}$ , the thin spherical shell at $\frac{2}{3} M R^{2}$ , and the rod at $\frac{1}{12} M L^{2}$ about its centre. Each is $M R^{2}$ multiplied by a pure number below one, and that number measures how far out the mass sits. The Earth's is 0.3307, lower than the 0.4 of a uniform sphere, and that single number is one of the main pieces of evidence that the Earth has a dense iron core.

The parallel axis theorem

Moment of inertia is defined about an axis, and it is usually easiest to compute about one through the centre of mass. The parallel axis theorem gets every other parallel axis for free.

Put the centre of mass at the origin and consider an axis through it, and a second axis parallel to it at distance $d$ . For a particle at $(x_{i}, y_{i})$ , its distance from the second axis, taken as displaced along $x$ , satisfies $r^{2} = (x_{i} - d)^{2} + y_{i}^{2} = r_{i}^{2} - 2 d x_{i} + d^{2}$ . Sum with the masses:

I^{'} = \sum m_{i} r_{i}^{2} - 2 d \sum m_{i} x_{i} + d^{2} \sum m_{i}

The middle sum is $M X_{cm}$ , which is zero because the origin was put at the centre of mass. So

I^{'} = I_{cm} + M d^{2}

Two things follow immediately. The moment of inertia is smallest about an axis through the centre of mass, so a body is always easiest to spin about its own centre. And the correction is the whole mass treated as a point at distance $d$ , which is the centre of mass theorem showing up again. Applying it to a rod about its end, $I = \frac{1}{12} M L^{2} + M (L / 2)^{2} = \frac{1}{3} M L^{2}$ , four times the central value, which is why a bat is harder to swing from the handle than from the middle.

Example. A flywheel is a uniform steel disc of mass 200 kg and radius 0.50 m, spinning at 3000 revolutions per minute. What is its moment of inertia and its stored kinetic energy?

$I = \frac{1}{2} (200) (0.25) = 25.0$ kg m². The angular velocity is $3000 \times 2 π / 60 = 314.2$ rad s⁻¹, so $K = \frac{1}{2} (25.0) ({314.2}^{2}) = 1.23 \times 10^{6}$ J. That is 1.23 MJ, more than twice the 588 kJ of kinetic energy in a 1500 kg car at 28 m s⁻¹, stored in a wheel that would fit in a wheelbarrow. The energy goes as $ω^{2}$ , which is why flywheel storage is a matter of spinning fast rather than building heavy, and why the failure mode of a burst flywheel is taken so seriously.

Now you. A solid sphere of mass 5.0 kg and radius 0.20 m rolls with its centre moving at 3.0 m s⁻¹. What is its moment of inertia about its centre, and its rotational kinetic energy?

Answer

$I = \frac{2}{5} (5.0) (0.04) = 0.080$ kg m². Rolling without slipping gives $ω = v / R = 3.0 / 0.20 = 15$ rad s⁻¹, so $K_{rot} = \frac{1}{2} (0.080) (225) = 9.0$ J, against a translational $\frac{1}{2} (5.0) (9.0) = 22.5$ J.

Rolling, which is both motions at once

A wheel rolling along the ground without slipping is turning about its axle and travelling along the road simultaneously, and the condition that links the two is that the contact point does not slide. In one full turn the wheel advances one circumference, so $v_{cm} = ω R$ , and differentiating, $a_{cm} = α R$ .

The kinetic energy is then the sum of the two parts, and the theorem behind that split is worth stating: the kinetic energy of any body equals the kinetic energy of its total mass moving with the centre of mass, plus the kinetic energy of the motion about the centre of mass.

K = \frac{1}{2} M v_{cm}^{2} + \frac{1}{2} I_{cm} ω^{2}

For rolling, substitute $ω = v / R$ and write $I_{cm} = k M R^{2}$ , where $k$ is the pure number tabulated above. Then $K = \frac{1}{2} M v^{2} (1 + k)$ , and the fraction of the energy that is rotational is $k / (1 + k)$ : a third for a solid disc, two sevenths for a solid sphere, a half for a hoop.

Now roll several bodies down the same slope from the same height. Energy conservation gives $M g h = \frac{1}{2} M v^{2} (1 + k)$ , so

v = \sqrt{\frac{2 g h}{1 + k}}

and the mass and radius both cancel. The winner is the body with the smallest $k$ , which is the one with its mass closest to the axis, and the result is independent of size: a marble beats a bowling ball only if the bowling ball is hollow. From a height of 2.0 m, a solid sphere arrives at 5.29 m s⁻¹, a solid disc at 5.11, a hoop at 4.43, and a frictionless sliding block, which stores nothing rotationally, at 6.26.

Example. A solid cylinder rolls from rest down a 25° incline without slipping. What is the acceleration of its centre?

With $k = \frac{1}{2}$ , energy gives $v^{2} = 2 g h / 1.5$ and, since $h = x \sin θ$ along the slope, $v^{2} = (4 / 3) g x \sin θ$ . Comparing with $v^{2} = 2 a x$ gives $a = \frac{2}{3} g \sin θ = \frac{2}{3} (9.81) (0.4226) = 2.76$ m s⁻², against the 4.15 m s⁻² of a block sliding on a frictionless slope. A third of the available energy is being diverted into spin.

Now you. A solid sphere rolls from rest down the same 25° incline. What is the acceleration of its centre?

Answer

With $k = \frac{2}{5}$ , the same argument gives $a = g \sin θ / (1 + k) = 9.81 \times 0.4226 / 1.4 = 2.96$ m s⁻², slightly more than the cylinder, which is why the sphere wins the race.

What has been left out

Every result here has come from energy or from kinematics, and none of it says what makes $ω$ change. The rolling problems were solved by conservation, which works but conceals the mechanism: something must be exerting a twisting influence on the sphere, and it is the friction at the contact point, which does no work yet is indispensable, since a sphere on a frictionless slope would slide down without turning at all.

There is also an unpaid debt. Rotation was set up about a fixed axis, and the analogy with linear motion was drawn by substituting symbols rather than by deriving anything. Establishing that $α$ is caused by a quantity built from force and geometry, and that this quantity has its own conservation law, is the next lesson. That conservation law is what lets a skater double their spin rate by pulling their arms in, and, applied to a planet, it turns out to be a law Kepler had written down seventy years before Newton.