Circular motion.

This page supports the multimedia tutorial Circular Motion

Acceleration is, by definition, the time rate of change of the velocity. While speed is a scalar, velocity is a vector: velocity has magnitude and direction. (See the module on constant acceleration, and the page on vectors.) In the animation below, the velocity vector is shown as an arrow. It is changing direction, but not changing magnitude.

Let's look at the angular position, θ, measured from the positive x axis. The distance travelled from the x axis is the arc with length s = rθ. The speed v, written without a vector bar, is the magnitude of the velocity. By definition, the speed is ds/dt. (See An introduction to calculus.) The radius, r, is constant, so v is r times dθ/dt. v is the rate of change of linear position, and dθ/dt is the rate of change of angular position. We call it the angular velocity and use the symbol ω. So the magnitude of the velocity is rω, and its direction is tangential. If it completes one circle in a period T, then the angle increases by 2π radians, so ω is 2π/T and v  =  rω  =  2πr/T. This is no surprise: it is one circumference per period.

Calculating acceleration Let's look at the particle at two times. The first is at time t, when its displacement r(t) from the centre makes an angle θ with the x axis, when it has velocity v(t). Later, at time t+Δ, its new displacement r(r+Δt) from the centre makes an angle θ+Δθ with the x axis, and it has velocity v(t+Δt). To work out the acceleration, we need the change in velocity, so we redraw the two velocity vectors and use the triangle method to find the difference, Δv. Δv is the vector you must add to v(t) to get v(t+Δt). This is shown in the small diagram at top right. First let's look at the direction of Δv. The speed -- the magnitude of the velocity -- is constant. So the two long sides of this triangle are equal, as are two of its angles. If Δt is very small, then the angle Δθ is also very small. In the limit of very small time and angle, the change in velocity Δv is at right angles to the velocity. The two velocity vectors are tangents to the circle, ie at right angles to their radii, therefore Δv is towards the centre of the circle. The acceleration is the time rate of change of velocity: it is the ratio of the change in velocity to the change in time, in the limit of very small change in time. So a is parallel to Δv, which means that a, like Δv, is also in the inward radial direction. This is an important result: in uniform circular motion, the acceleration is towards the centre: it has only centripital acceleration. Now we must find |a|, the magnitude of a. For that we the the magnitude of the change in velocity, |Δv|. Provided that the angle Δθ in the triangle is very small, |Δv|  ≅  vΔθ. In the differential limit, this gives us directly the magnitude of a: So: We may now use v = rω, which allows us to substitute either for ω or for v to obtain two useful expressions for the magnitude a of the acceleration: Remember that, by definition, a is parallel to Δv and so, as the previous diagram shows, its direction is centripetal, or towards the centre. In other words, its direction is opposite that of the radius vector r. So, finally, we may write an expression for the vector quantity a a  =  − ω2r.

The animation below is introduced in the Circular Motion module, and is reproduced here for those who might like to view it in slow motion.

The car drives at constant speed over a hill at whose summit the radius of curvature, in the vertical plane, is 30 m, as shown. Let's find the critical speed v_crit at which the car loses contact with the road.

Following the road at constant speed, the car is in uniform circular motion. At the top of the hill, its acceleration is therefore downwards and its magnitude is v²/r. What are the downwards forces on it? Its weight is one such force. In general, there is also a downwards component of the force that the air exerts on the car*. For our case, this second force is negligible in comparison with the weight. (See Air resistance to obtain an upper limit.)

So, at the top of the hill, the weight is the only downwards force, and so the maximum downwards acceleration is g, which, as we saw in the Projectile module, is 9.8 m.s⁻². Now the downwards centripital acceleration when the car is at the top of the hill is v²/r, and this acceleration cannot be greater than g. So the critical acceleration and velocity are

a_crit  =  v_crit²/r  =  g.
Rearranging this equation gives the critical speed:
v_crit  =  √(rg)  =  17 m.s⁻¹   =  62 k.p.h.
Note that, while in contact with the road, the car is rotating in the vertical plane with angular velocity ω = v/r. At the point near where it loses contact with the road, the forces exerted on the wheels are small, and so too are the torques (which we shall meet when we come to Rotation). Consequently, the car, whilst airborne, continues to rotate at this rate. This rotation adds extra danger to the manoeuvre, which is just one of the reasons for the advice, rarely given in physics, that one should trust the calculation here, rather than perform the experiment.

* Racing cars often have a large foil whose effect, at high speed, is to provide a downwards force and thus allow greater frictional forces for cornering, braking and even acceleration. Some street cars have a miniature version of this foil. These have several possible functions. First, they identify the style of the driver, which may provide a warning to other drivers. Second, they increase drag and petrol consumption, which benefits oil companies (in the short term, at least). Do they have a significant effect in producing a downwards force? Taking the weight of the car as 20 kN, a 10% increase in downwards force would require the foil to exert 2 kN downwards. I do not recommend the following experiment, but if it can exert 2 kN, then it ought to support, elastically, the weight of a few people standing on it.

In the derivation above, we assumed that the angular velocity was constant. Consequently, there was no tangential component of acceleration: the acceleration was purely centripital. In nonuniform circular motion, which we shall meet again when studying rotation, the motion is still circular (r is still constant), but v and ω may be functions of time.

First, let's note that, in circular motion of any sort, the velocity is always tangential: in other words, v and r are at right angles. If v had a component in the r direction, then r would change in time, so it would not be circular motion. We can now derive some important results here using the scalar product.

We have just argued that v and r are at right angles to each other, so

v. r  =  0
If we take the derivative of this equation, using a  =  dv/dt and v  =  dr/dt, we have
a. r + v. v  =  0,       so

a. r  =  - v. v

Let a_r be the radial component of a. Again, using results from the vectors page, the above equation may be written
a_r r  =  - v²      so

a_r  =  - v²/r

This is the same result that we obtained above for uniform circular motion, using geometry, but here we show it to be more general. In nonuniform circular motion, the motion is still circular (r is still constant, v_r is still zero and a_r is still - v²/r), but ω and v_t may be functions of time.

This page supports the multimedia tutorial Circular Motion