Circular dynamics

A body going round a circle at constant speed is accelerating towards the centre at $v^{2} / r$ , and since acceleration requires force, something real has to supply it, which turns a geometric fact into an engineering constraint.

Two results are carried in from earlier lessons: the centripetal acceleration of a body moving at speed $v$ on a circle of radius $r$ has magnitude $v^{2} / r$ and points at the centre, and the net force on a body of mass $m$ equals $m$ times its acceleration.

Inverting the question

Every problem so far has run forwards: given the forces, find the motion. Circular motion is naturally run backwards, because the path is usually known in advance. A car must follow the road, a satellite must stay in its orbit, a conker on a string must go round in a circle because the string is that long. The path fixes the acceleration, and the second law then says what the forces must add up to:

\sum F_{towards centre} = \frac{m v^{2}}{r}

That is the entire content of this lesson. The equation is not a new law and $m v^{2} / r$ is not a new force: it is the right hand side of $\sum \vec{F} = m \vec{a}$ with the known acceleration substituted in. The skill is entirely in the free body diagram, and the discipline is to write only forces with named agents on the left and put $m v^{2} / r$ on the right, never both on the same side.

The pattern is worth noticing before the examples. The required force grows with the square of the speed and falls with the radius, so the dangerous combination is fast and tight. Real forces have limits, and where the demand exceeds the supply, the body does not go round. It leaves.

The flat curve, and the limit of friction

A car of mass $m$ takes a level bend of radius $r$ at speed $v$ . Draw its free body diagram: weight $m g$ down, normal force $N$ up, and friction from the road, horizontally. Nothing else touches it.

Vertically there is no acceleration, so $N = m g$ . Horizontally the only force available is friction, and it must supply the whole centripetal requirement: $f = m v^{2} / r$ . Notice that this is static friction, not kinetic, because a rolling tyre is not sliding across the road: the contact patch is momentarily at rest. Static friction is capped at $μ_{s} N = μ_{s} m g$ , so

\frac{m v^{2}}{r} \leq μ_{s} m g \Rightarrow v_{\max} = \sqrt{μ_{s} g r}

The mass cancels, so a loaded lorry and an empty one have the same cornering limit, at least until tyre behaviour stops being ideal. Above $v_{\max}$ the friction available is less than the friction required, and the car travels on a path of larger radius than the bend, which is to say it runs wide off the outside of the corner. It is not "thrown outward": nothing pushes it out, and what actually happens is that the inward force ran out and the car went comparatively straight.

Example. A bend of radius 60 m has $μ_{s} = 0.80$ in the dry. What is the fastest speed a car can take it?

$v_{\max} = \sqrt{0.80 \times 9.81 \times 60} = \sqrt{470.9} = 21.7$ m s⁻¹, which is 78 km/h. The square root is unforgiving in both directions: halving the grip costs only 29 per cent of the speed, but a corner posted for 78 km/h taken at 95 km/h needs a coefficient of 1.2, which no ordinary tyre and road can deliver.

Now you. The same road is wet, with $μ_{s} = 0.35$ , on a bend of radius 120 m. What is the fastest safe speed?

Answer

$v_{\max} = \sqrt{0.35 \times 9.81 \times 120} = \sqrt{412.0} = 20.3$ m s⁻¹, about 73 km/h. Doubling the radius did not recover what halving the grip took away.

Banking, which removes the need for friction

Tilt the road inward by an angle $θ$ and the normal force, which is perpendicular to the surface, acquires a horizontal component pointing at the centre of the bend. That component can do the whole job.

Assume no friction at all, which is the design case. The forces are $N$ perpendicular to the road and $m g$ down. Resolve into horizontal and vertical, not along the slope, because the acceleration is horizontal. Vertically there is no acceleration, so $N \cos θ = m g$ . Horizontally, $N \sin θ = m v^{2} / r$ . Divide the second by the first and $N$ and $m$ both disappear:

\tan θ = \frac{v^{2}}{r g}

A banked curve has exactly one design speed, at which no friction is needed. Below it the car tends to slide down the bank and friction must hold it up; above it friction must hold it in. Real roads are banked for a chosen speed and rely on friction for the spread of speeds around it.

The numbers are checkable. Daytona's turns, banked at 31° with a radius of about 300 m, have a design speed of $\sqrt{300 \times 9.81 \times \tan 31^{\circ}} = \sqrt{1768} = 42.1$ m s⁻¹, or 151 km/h, and cars lap far faster than that, which is why they need enormous downforce and enormous tyres: everything above the design speed is paid for by friction. An ordinary motorway curve of radius 200 m intended for 30 m s⁻¹ would need $\tan θ = 900 / (200 \times 9.81) = 0.459$ , an angle of 24.6°, which is far steeper than any road is built, and so real motorway curves are much larger in radius instead.

The conical pendulum

Hang a mass on a string of length $L$ and swing it so that it travels in a horizontal circle with the string making a constant angle $θ$ to the vertical. The string sweeps out a cone, and the analysis is the banked curve with tension in place of the normal force.

The radius of the circle is $r = L \sin θ$ . Vertically, $T \cos θ = m g$ ; horizontally, $T \sin θ = m v^{2} / r$ . Dividing gives $\tan θ = v^{2} / (r g)$ again, and substituting $r = L \sin θ$ and $v = 2 π r / τ$ for a period $τ$ gives the tidy result

τ = 2 π \sqrt{\frac{L \cos θ}{g}}

The tension is $T = m g / \cos θ$ , which exceeds the weight always and diverges as $θ$ approaches 90°: a string can never be pulled horizontal by a mass on its end, however fast it is swung, because a horizontal string has no vertical component to hold the weight up.

That period formula is also the first appearance of a result the course returns to. As $θ$ becomes small, $\cos θ \to 1$ and $τ \to 2 π \sqrt{L / g}$ , which is the period of an ordinary pendulum swinging back and forth. The two motions are the same motion seen from different sides, and the connection is made properly in the lesson on oscillations.

Example. A 0.25 kg bob on a 1.2 m string swings in a horizontal circle at 30° to the vertical. Find the radius, the speed, the period and the tension.

The radius is $1.2 \sin 30^{\circ} = 0.600$ m. From $\tan θ = v^{2} / (r g)$ , $v = \sqrt{9.81 \times 0.600 \times \tan 30^{\circ}} = \sqrt{3.399} = 1.84$ m s⁻¹. The period is $2 π \sqrt{1.2 \cos 30^{\circ} / 9.81} = 2 π \sqrt{0.1059} = 2.05$ s, which the check $τ = 2 π r / v = 3.770 / 1.843$ confirms. The tension is $m g / \cos 30^{\circ} = (0.25) (9.81) / 0.866 = 2.83$ N, some 15 per cent more than the bob's weight of 2.45 N.

Now you. A bob on a 0.80 m string swings in a horizontal circle at 40° to the vertical. Find the radius, the speed and the period.

Answer

$r = 0.80 \sin 40^{\circ} = 0.514$ m, $v = \sqrt{9.81 \times 0.514 \times \tan 40^{\circ}} = 2.06$ m s⁻¹, and $τ = 2 π \sqrt{0.80 \cos 40^{\circ} / 9.81} = 1.57$ s.

The vertical loop

Now let the circle be vertical, so gravity is sometimes towards the centre and sometimes away from it. The speed is no longer constant, but at any instant the components of force towards the centre must still sum to $m v^{2} / r$ .

At the top of the loop, both the weight and the track's push point downward, which is to say towards the centre:

N + m g = \frac{m v^{2}}{r}

so $N = m v^{2} / r - m g$ . As the speed falls, $N$ falls, and it reaches zero when $v^{2} = g r$ . Below that speed the equation demands a negative $N$ , meaning the track would have to pull the car inward, which a track that only pushes cannot do. So the minimum speed at the top of a loop is

v_{top} = \sqrt{g r}

and it is independent of mass. For a loop of radius 8.0 m that is $\sqrt{9.81 \times 8.0} = 8.86$ m s⁻¹. At the bottom of the same loop, energy conservation, which the next lesson derives properly, gives $v_{bot}^{2} = v_{top}^{2} + 4 g r = 5 g r$ , so $v_{bot} = 19.8$ m s⁻¹, and there the normal force is $N = m v^{2} / r + m g = 6 m g$ : the rider is pressed into the seat at six times their weight. That is why real roller coaster loops are not circles but clothoids, tightening as they rise, so that the radius is small where the speed is low and large where the speed is high, and the load on the rider stays near 4 $g$ throughout instead of spiking at the bottom.

The same equation describes a bucket of water swung in a vertical circle. The water stays in because at the top the bucket's base is pushing it downward, adding to gravity to supply $m v^{2} / r$ . For an arm and bucket of radius 1.0 m, the minimum speed is $\sqrt{9.81} = 3.13$ m s⁻¹, a period of 2.0 s. Slower than that and the water leaves the bucket in a parabola, exactly as the lesson on projectiles said it would.

Example. A stone of mass 0.40 kg is whirled on a string in a vertical circle of radius 0.90 m. What is the minimum speed at the top, and what is the tension at the top if it moves at 5.0 m s⁻¹ there?

The minimum is $\sqrt{9.81 \times 0.90} = 2.97$ m s⁻¹. At 5.0 m s⁻¹, the required centripetal force is $m v^{2} / r = (0.40) (25) / 0.90 = 11.1$ N, of which gravity supplies $m g = 3.92$ N, so the string supplies $T = 11.1 - 3.92 = 7.19$ N.

Now you. The same stone is at the bottom of the circle moving at 6.0 m s⁻¹. What is the tension there?

Answer

At the bottom the centre is upward, so the tension acts towards it and gravity away from it: $T - m g = m v^{2} / r$ . That gives $T = (0.40) (36) / 0.90 + 3.92 = 16.0 + 3.92 = 19.9$ N, nearly three times the tension at the top.

Centrifugal force, and what it really is

A passenger in a cornering car feels thrown against the door, and calls the sensation centrifugal force. In the road's frame there is no such force. What acts on the passenger is the door pushing them inward, supplying their share of $m v^{2} / r$ ; without the door they would carry on in a straight line, which relative to the turning car means moving outward. The feeling is the door, and the outward tendency is the absence of a force rather than the presence of one.

That said, the term is not simply an error. In a frame that is itself rotating, the first law fails, and the failure can be repaired by inventing forces. Add an outward $m ω^{2} r$ on every body, plus a velocity dependent Coriolis term, and the second law works again inside the rotating frame. These are fictitious or inertial forces: they have no agent, no third law partner, and they vanish on returning to an inertial frame. They are also indispensable in practice, because meteorology, oceanography and long range gunnery are all done in the Earth's rotating frame and would be unmanageable otherwise.

The Earth supplies a measurable instance. The centrifugal effect of its spin at the equator is $ω^{2} R = 0.0339$ m s⁻², so a body there weighs about 0.35 per cent less than the same body at the pole, part of the reason $g$ measures 9.78 m s⁻² at the equator and 9.83 at the poles, the rest coming from the equatorial bulge that the same spin produced.

When the speed is not constant

Uniform circular motion is the special case where only the direction changes. In general a body on a curved path has both: an acceleration $v^{2} / r$ towards the centre, changing direction, and an acceleration $d v / d t$ along the path, changing speed. The two are perpendicular, so the total has magnitude $\sqrt{(v^{2} / r)^{2} + (d v / d t)^{2}}$ , and a car accelerating out of a corner is using both at once. Since the tyre's total grip is limited by $μ_{s} g$ however it is spent, spending some on speeding up leaves less for turning, which is the entire content of the friction circle that racing drivers work with.

Every problem in this lesson has been solved by knowing the force at each instant of time. That has been the method since the second law appeared, and it is about to become inadequate. Ask how fast a roller coaster is going at the top of a loop given its speed at the bottom, and the time it took is neither known nor wanted. Answering it requires integrating force over distance instead of over time, which produces a quantity that is conserved, and that is the next lesson.