A car goes round a large roundabout at a steady \(36\,\text{km/h}\). The speedometer needle does not move at all. Yet the passengers lean against the doors, and a bottle on the seat slides towards the outside. Something about the motion is changing, though the speed is not.
Steady speed, changing velocity
A body that goes round a circle at a constant speed is in uniform circular motion. The word "uniform" refers to the speed. The body covers equal arcs in equal times, and its angular velocity \(\omega\) is constant. Its period \(T = 2\pi/\omega\) is fixed.
The velocity, however, is not constant. The velocity always lies along the tangent to the circle, and the tangent points in a new direction at every point. A vector that changes its direction is a changing vector, even if its length stays the same. So a body in uniform circular motion is accelerated. From Lesson 1, its tangential acceleration is zero, because the speed does not change. The whole acceleration is normal to the path.
A body is in uniform circular motion. Which of these quantities stays constant?
The triangle of velocities
Take a body moving at speed \(v\) on a circle of radius \(r\). In a short time \(\Delta t\) it moves from \(P_1\) to \(P_2\) and turns through the angle \(\Delta\theta\) at the centre. Its velocity changes from \(\vec v_1\) to \(\vec v_2\). Both have the size \(v\), and each is at right angles to its own radius. So the angle between \(\vec v_1\) and \(\vec v_2\) is also \(\Delta\theta\).
Draw \(\vec v_1\) and \(\vec v_2\) from one point, tail to tail. The arrow from the tip of \(\vec v_1\) to the tip of \(\vec v_2\) is the change in velocity, \(\Delta\vec v = \vec v_2 - \vec v_1\). The triangle has two equal sides of length \(v\) with the angle \(\Delta\theta\) between them. The average acceleration \(\Delta\vec v/\Delta t\) points the same way as \(\Delta\vec v\).
\[ |\Delta\vec v| = 2v\sin\dfrac{\Delta\theta}{2} \]
Why. the triangle of velocities has two sides of length \(v\) with the angle \(\Delta\theta\) between them. Its third side, split in half by the bisector of the angle, is \(2v\sin(\Delta\theta/2)\).
\[ |\Delta\vec v| \approx v\,\Delta\theta \]
Why. for a small angle in radians, \(\sin(\Delta\theta/2) \approx \Delta\theta/2\). The chord of the triangle becomes equal to the arc.
\[ a_c = \lim_{\Delta t \to 0}\dfrac{|\Delta\vec v|}{\Delta t} = v\,\lim_{\Delta t \to 0}\dfrac{\Delta\theta}{\Delta t} = v\,\omega \]
Why. the size of the acceleration is the limit of the change in velocity over the time, and \(\Delta\theta/\Delta t\) tends to the angular velocity \(\omega\).
\[ a_c = v\omega = \dfrac{v^2}{r} = \omega^2 r \]
Why. put \(\omega = v/r\) or \(v = \omega r\) from Lesson 2. The three forms are the same quantity.
\[ \text{direction: along the radius, towards the centre} \]
Read it. in the limit the change in velocity is at right angles to the velocity, on the inner side. At right angles to the tangent means along the radius.
The acceleration of a body in uniform circular motion therefore always points to the centre of the circle. It is called the centripetal acceleration, from Latin words meaning "seeking the centre". Its size is fixed, because \(v\) and \(r\) are fixed. Its direction turns with the body, always along the radius.
The two forms tell different stories. At a fixed speed, \(a_c = v^2/r\) is larger on a tighter circle, so a sharp bend needs more acceleration than a gentle one. At a fixed angular velocity, \(a_c = \omega^2 r\) is larger farther from the centre. On a turning wheel, the rim has the largest centripetal acceleration, and the axle has none.
A stone tied to a string \(0.5\,\text{m}\) long is whirled in a horizontal circle, making one revolution every second. Find its angular velocity, its speed and its centripetal acceleration.
One revolution per second is \(f = 1\,\text{Hz}\), so \(\omega = 2\pi f = 2\pi \approx 6.28\,\text{rad/s}\).
\(v = \omega r = 2\pi \times 0.5 = \pi \approx 3.14\,\text{m/s}\).
\(a_c = \omega^2 r = 4\pi^2 \times 0.5 = 2\pi^2 \approx 19.7\,\text{m/s}^2\), directed along the string towards the hand. Check with the other form: \(v^2/r = \pi^2/0.5 = 2\pi^2\).
This is about twice \(g\), from a gentle one turn each second.
The car on the roundabout moves at \(36\,\text{km/h}\) on a circle of radius \(20\,\text{m}\). What is its centripetal acceleration, in m/s²?
A car goes round the same bend at twice its earlier speed. What happens to its centripetal acceleration?
Which statements about centripetal acceleration are true?
Three standard examples
Three examples of uniform circular motion appear again and again in physics. They span a range of sizes from a playground to an atom. In each, a body goes round a centre at a nearly steady speed, so each has a centripetal acceleration that points at its centre.
The first is a stone whirled in a horizontal circle on a string, as in the example above. The second is the Moon going round the Earth. Its orbit is very nearly a circle of radius \(3.84 \times 10^8\,\text{m}\), covered in \(27.3\) days. The third is the electron in the simplest model of the hydrogen atom, the Bohr model. There it moves at \(2.19 \times 10^6\,\text{m/s}\) on a circle of radius \(5.29 \times 10^{-11}\,\text{m}\).
Find the centripetal acceleration of the Moon, using the numbers above, and of the electron in the Bohr model.
Moon: \(T = 27.3 \times 86400 \approx 2.36 \times 10^6\,\text{s}\). Then \(a_c = \omega^2 r = 4\pi^2 r/T^2 = 4\pi^2 \times 3.84 \times 10^8/(2.36 \times 10^6)^2 \approx 2.7 \times 10^{-3}\,\text{m/s}^2\).
Electron: \(a_c = v^2/r = (2.19 \times 10^6)^2/(5.29 \times 10^{-11}) \approx 9.1 \times 10^{22}\,\text{m/s}^2\).
The Moon's acceleration is about \(3600\) times smaller than \(g\). The electron's is larger than \(g\) by a factor of about \(10^{22}\). The same formula covers both.
Kinematics can find these accelerations, but it cannot say what produces them. The string pulls on the stone. The Earth's gravity pulls on the Moon. The electric attraction of the nucleus pulls on the electron. The next chapter, Dynamics of Circular Motion, sets each of these forces equal to mass times the centripetal acceleration. The tiny number for the Moon is the one Newton used to test his law of gravitation.
A person stands on the equator. The Earth's spin carries the person round a circle of radius about \(6400\,\text{km}\) once a day. Estimate the centripetal acceleration, in m/s².
In 1659 the Dutch physicist Christiaan Huygens worked out the pull that a body going round a circle needs. It points to the centre and grows as \(v^2/r\). He was studying pendulum clocks, and he published the result in 1673 in his book on clocks. Newton reached the same result on his own in the 1660s, in his notebooks. Newton then did something Huygens did not. He combined \(v^2/r\) for the planets with Kepler's rule for their periods. The result was that the pull of the Sun must fall as the square of the distance.
A washing machine drum of radius about \(0.25\,\text{m}\) spins at up to \(1200\,\text{rpm}\), which is \(40\pi \approx 126\,\text{rad/s}\). The centripetal acceleration of the clothes against the drum wall is then \(\omega^2 r \approx 3900\,\text{m/s}^2\), nearly \(400\) times \(g\). The drum pushes the clothes inward to give them this acceleration. Nothing holds back the water at the holes in the wall. So the water leaves along the tangent, and the clothes come out nearly dry.
- Saying that a body in uniform circular motion has constant velocity, or no acceleration.Its speed is constant, but the direction of the velocity turns, so the velocity changes and \(a_c = v^2/r\) is present.
- Drawing the centripetal acceleration outwards, the way passengers feel pushed.The acceleration points towards the centre. The outward push that passengers feel is the subject of the next chapter.
- Using \(a_c = v^2/r\) with the speed in km/h, or \(\omega^2 r\) with \(\omega\) in rpm.Convert first: km/h to m/s, rpm to rad/s.
- Thinking that the size of the acceleration changes as the body goes round.In uniform circular motion the size \(v^2/r\) is fixed; only the direction turns, always towards the centre.