In the fields of Maharashtra, a farmer guards his jowar crop from birds with a gophan. It is a sling of rope with a pouch for a stone. He whirls the stone round and round over his head. His hand feels a steady pull along the rope, and the faster he whirls, the harder the rope pulls. Then he lets one end of the rope go. The stone does not fly straight out, away from his head. It shoots off sideways, towards the birds in front of him.
From an acceleration to a force
Take a body that moves on a circle of radius \(r\) at a steady speed \(v\). The chapter on the kinematics of circular motion proved one fact about it. Its velocity keeps turning, so the body accelerates. The acceleration points at the centre of the circle, and its size is \(v^2/r\). We called it the centripetal acceleration. With the angular speed \(\omega = v/r\), the same size is \(\omega^2 r\).
Newton's second law now turns this fact about motion into a fact about forces. The net force on a body equals its mass times its acceleration, and it points the same way. So a body of mass \(m\) in uniform circular motion must have a net force of size \(mv^2/r\) on it, pointing at the centre. Without that net inward force, the velocity could not keep turning.
A body of mass \(m\) moving on a circle of radius \(r\) at speed \(v\) has an acceleration \(v^2/r\) towards the centre. So the vector sum of the real forces on it must have a component \(mv^2/r = m\omega^2 r\) towards the centre. If the speed also changes, the forces have a second component along the tangent, equal to \(m\) times the tangential acceleration.
The requirement grows fast with speed. At twice the speed on the same circle, the inward force must be four times as large. On a circle of half the radius at the same speed, it must be twice as large. The farmer feels exactly this: whirl faster, and the rope pulls much harder on his hand.
A car of mass \(1000\,\text{kg}\) goes round a curve of radius \(100\,\text{m}\) at a steady \(20\,\text{m/s}\). What net force towards the centre of the curve must act on it, in newtons?
A job, not a new force
The net inward force \(mv^2/r\) is called the centripetal force. The name misleads many students, because it sounds like a new kind of force, like gravity or friction. It is not. "Centripetal" means "seeking the centre". The word names a job, not a force. The job is to pull or push the body towards the centre. Some real force, with a real body as its agent, always does it.
The chapter on the kinematics of circular motion named three standard circles. Now we can name the force that does the job in each one. In a stone whirled on a string, the tension of the string pulls the stone towards the hand. In the Moon going round the Earth, the gravity of the Earth pulls the Moon towards the Earth's centre. In the hydrogen atom as Bohr pictured it, the electric attraction of the nucleus pulls the electron towards the nucleus.
The rows of this table have one thing in common. In every case the force that does the job has an agent, and the agent is a body you can point to. Later lessons meet the last two rows. There the road and the wall push the body towards the centre, and the rest of the forces balance.
The free-body diagram never shows it
This leads to the rule for free-body diagrams. Draw only the real forces, each with its agent: the weight, the tension, the normal force, friction. Then add their components along the radius. That sum must equal \(mv^2/r\). Do not draw a separate arrow labelled "centripetal force". It would count the same force twice.
The most common error of this chapter is an extra arrow. A student draws the tension on a whirled stone, and then a second arrow, \(mv^2/r\), towards the centre. The diagram now claims that two inward forces act, and the equation that follows is wrong by exactly that much. The tension is the centripetal force here. Write \(T = mv^2/r\), not \(T + mv^2/r = \ldots\).
A student draws the free-body diagram of the Moon with two arrows towards the Earth: the Earth's gravity and a centripetal force. What is wrong?
A force that always points at the centre is always at right angles to the velocity, which lies along the tangent. A force at right angles to the motion does not speed the body up or slow it down; it only turns the velocity. So the centripetal force does no work on a body in uniform circular motion, and the speed stays the same.
A stone goes once round a circle at a steady speed, held by a string. How much work does the tension do on it in one turn?
In 1666 the plague had closed Cambridge, and the young Isaac Newton was at home. He wondered whether the pull that brings an apple down also holds the Moon on its circle. The Moon is about 60 Earth radii away and goes round once in 27.3 days. Its centripetal acceleration is therefore \(4\pi^2 r/T^2\), about \(0.0027\,\text{m/s}^2\). If gravity weakens as the square of the distance, the pull at the Moon should give \(g/60^2 = 9.8/3600\), also about \(0.0027\,\text{m/s}^2\). The two numbers agreed. The force that does the Moon's centripetal job is the same gravity that drops the apple. Newton published the comparison in the Principia in 1687.
The method, on the simplest circle
A small stone lies on a smooth horizontal table. A light string ties it to a peg fixed in the table. The stone is set moving, and it goes round the peg on a circle at a steady speed. This is the simplest circle there is, and it shows the method that the rest of the chapter uses.
Follow four steps. First, find the circle and its centre: here the centre is the peg. Second, draw the free-body diagram with the real forces only. Third, take one axis along the radius, pointing towards the centre, and a second axis perpendicular to the plane of the circle. Fourth, write the net force along the radius as \(mv^2/r\), and the net force along the other axis as zero.
Apply the steps. Three forces act on the stone. They are its weight \(mg\) down, the normal force \(N\) of the table up, and the tension \(T\) towards the peg. The stone stays on the table, so it has no vertical acceleration, and \(N = mg\). The tension is the only horizontal force left. So the tension alone does the centripetal job:
\[ T = \frac{mv^2}{r} = m\omega^2 r \]
A real string breaks when the tension passes some largest value \(T_{\max}\). Putting \(T = T_{\max}\) gives the fastest the stone can go: \(v_{\max} = \sqrt{T_{\max}\,r/m}\). Notice the direction of the effect. A heavier stone or a longer string with the same speed needs a larger tension, so the heavier stone breaks the string sooner.
A stone of mass \(0.2\,\text{kg}\) is whirled on a smooth table at the end of a string \(0.5\,\text{m}\) long. The string breaks when the tension exceeds \(40\,\text{N}\). (a) Find the greatest speed of the stone. (b) Find the greatest angular speed.
(a) The tension is the whole centripetal force, so \(T_{\max} = mv_{\max}^2/r\). Then \(v_{\max} = \sqrt{T_{\max}\,r/m} = \sqrt{40 \times 0.5 / 0.2} = \sqrt{100} = 10\,\text{m/s}\).
(b) \(\omega_{\max} = v_{\max}/r = 10/0.5 = 20\,\text{rad/s}\), a little more than three turns every second.
A \(0.1\,\text{kg}\) stone goes round a smooth table on a string \(0.4\,\text{m}\) long, making \(2\) revolutions every second. Find the tension in the string, in newtons.
When the job stops
Now cut the string while the stone is at the lowest point of its circle, seen from above. What does the stone do next? Many students expect it to fly outwards, away from the peg, as if something had been holding it back from escaping. Make your prediction before you play the figure.
A stone goes round a peg on a smooth table. At the lowest point of the circle, where it moves to the right, the string is cut. Which path does the stone follow?
In the figure the stone goes round one and a half times while the string pulls it towards the peg. The moment the string is cut, the stone moves off along the tangent in a straight line at the same speed. It does not move outwards along the radius. The stone was never trying to escape outwards. It was always trying to go straight, and the string kept turning it.
Now return to the farmer. The rope of his sling pulls the stone towards his hand. That tension is the centripetal force, and it grows as the square of the speed. When he lets one end go, the tension vanishes. The stone keeps the velocity it had at that instant, which points along the tangent to its circle. So the farmer lets go at the moment the stone moves towards the birds.
A hammer thrower in an athletics stadium spins round three or four times, pulling hard on the wire that holds the heavy ball. The wire's tension, several times the ball's weight, is its centripetal force. At release the ball flies along the tangent to its last circle. So the throw is aimed by the moment of release, not by the direction the thrower faces. The discus thrower uses the same rule.
A friend says, "When the string breaks, the stone flies outwards. The outward force on it is no longer balanced." Explain what is wrong.
Seen from the table, no outward force acts on the stone at any time. The only horizontal force is the tension, and it points inwards. While the string holds, the tension turns the velocity; the net force is not zero, it is \(mv^2/r\) inwards. When the string breaks, the net force becomes zero, and by the first law the stone keeps its velocity at that instant. That velocity is along the tangent, so the stone moves along the tangent, not outwards along the radius.
- Drawing a "centripetal force" arrow in addition to the real forces.Draw only forces with agents. The inward sum of those forces is the centripetal force, so write it equal to \(mv^2/r\).
- Using the speed in place of its square.The requirement is \(mv^2/r\). Doubling the speed makes it four times as large.
- Saying a released body flies outwards along the radius.With the inward force gone, the body keeps its velocity, which is along the tangent.