Circular motion.

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Circular motion always involves acceleration

    Let's treat the Earth's orbit around the sun as a circle -- which is a reasonably good approximation.

    The Earth's speed on this orbit is constant. Nevertheless, the Earth is accelerating. The gravitational force exerted by the sun on the Earth attracts the latter towards the sun: the Earth is accelerating towards the sun. We shall show that the acceleration is towards the centre of the circle: it is centripetal acceleration. The force that causes it (gravity, in this case) is called centripetal force.

    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.

    diagram illustrating centripetal acceleration

    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.

diagram illustrating centripetal acceleration

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.

      equation
    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,

      |v|  ≅  vΔθ.
    In the differential limit, this gives us directly the magnitude of a:
      equation
    So:
      equation
    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:
      equation
    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.

Airborne automobiles

What is the acceleration of your room?

    This is not an idle puzzle. Aristarchus of Samos and, much later, Copernicus argued the earth orbits the sun. On the other side Aristotle, Ptolemy and many others believed that the sun must go around the earth, because they did not feel the motion of the earth. Who is correct? We do not feel constant velocity (try closing your eyes on a smooth train or plane ride), but we do feel accelerations, both by the forces on us that cause accelerations and by the forces acting in the middle ear when the head accelerates. So, what is your acceleration due to the rotation and orbital motion of the Earth?

    You will need to know some values: the period of the Earth's rotation is (slightly less than) 24 hours. The actual value depends on your latitude, but your distance from the Earth's axis is not greater than six thousand kilometres. So

      What is the acceleration due to the Earth's rotation about its axis?
    To get an idea whether you would feel this, you could express your answer as a fraction of g.

    You should do a similar calculation for the acceleration due to the Earth's orbit, which has a radius of 1.5 x 1011 m.

Components of uniform circular motion

    Let's consider a point at vector r undergoing uniform circular motion. Circular, so its magnitude r is constant, and uniform, so its angular displacement θ = ωt. From trignometry, its x and y components are
      x  =  r cos θ,    y  =  r sin θ   so:
      x  =  r cos ωt,    y  =  r sin ωt  
    and it's left as an exercise to check that x2 y2 = 0.

What if the circular motion is not uniform?

This page supports the multimedia tutorial Circular Motion

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