An object moving in a circle at constant speed is still accelerating. Change the speed, radius and mass and watch what the arrows do.
Written for AQA A-level Physics (7408), spec 3.6.1.1.
The motion
ω = vr
T = 2πrv
f = 1T
a = v2r
F = mv2r
How the velocity changes
|Δv| = 2v sin(Δθ/2)
Δt = Δθω
Average: |Δv|Δt
Exact: v2r
Shrink Δθ and the average meets v2/r. Δv points to the centre of the circle, measured halfway between the two positions.
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Animation speed is scaled so you can watch it, and the slow-motion button slows it five times more. The velocity arrows depend only on the speed, and are the same length in both panes. The acceleration and force arrows are compressed to fit. Use the numbers for exact values.
Radians and angular speed
Angles in physics are measured in radians. One radian is the angle at the centre when the arc is as long as the radius. Drag the angle, then spin the disc.
Radians: wrap the radius round the circle
θ (radians)
θ (degrees)
θ ÷ π
s = rθ
θ = sr
The angle in radians is the arc length counted in radii. To convert from degrees:
θ(rad) = θ(°) × π180
Angular speed: same ω, different v
ω = ΔθΔt
f = ω2π
T = 2πω
vA = ωrA
vB = ωrB
Point B is twice as far out as point A, so it covers twice the distance in the same time: v = ωr.
Key equations for circular motion
Angle in radians
Useful relationship
θ = sr
s arc length (m), r radius (m). 2π rad = 360°
Angular speed
Definition + AQA formula sheet
ω = ΔθΔt
ω = vr = 2πf
ω in rad s−1, f in Hz. The second line is the AQA formula-sheet relationship; the first is the definition of angular speed.
Period and linear speed
Useful rearrangements
f = 1T
v = ωr
T in s, v in m s−1
Centripetal acceleration
AQA formula sheet
a = v2r = ω2r
a in m s−2, towards the centre
Centripetal force
AQA formula sheet
F = mv2r = mω2r
F in N, the resultant towards the centre
Quick catch
Five multiple-choice questions in the style of the AQA paper. Pick A, B, C or D and the reasons appear straight away.
0 of 5 right
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Where it comes from
The equation for centripetal acceleration does not appear out of nowhere. Here are two ways to get it. Pick the one that suits you.
1
Two positions, a short time apart
In a short time Δt the object moves from A to B. The radius turns through Δθ. The displacement from A to B is the straight line, of length |Δr|, and the distance along the path is vΔt. For a small angle the two are almost equal.
|Δr| ≈ arc AB = vΔt
2
Draw the velocities
Velocity is always at right angles to the radius, so v1 is perpendicular to OA and v2 to OB (see the right-angle marks). Now slide both arrows, keeping their lengths and directions, until their tails meet. Both have length v, and the angle between them is Δθ.
3
Two similar triangles
Triangle OAB (sides r, r and |Δr|) and the velocity triangle (sides v, v and |Δv|) are both isosceles with the same angle Δθ between the equal sides. They are similar, though not the same size, so their sides are in the same ratio.
|Δv|v = |Δr|r
4
Put the displacement in
Replace |Δr| by vΔt, then rearrange for the acceleration, a = |Δv|/Δt.
|Δv|v = vΔtr
|Δv|Δt = v2r
5
Which way does it point?
In the velocity triangle the angle at the tip of v1 is 90° − Δθ/2 (marked on the diagram). As Δθ shrinks it tends to 90°, so Δv, and therefore the acceleration, points at the centre of the circle, along the same line as the a arrow. With v = ωr:
a = |Δv|Δt = v2r = ω2r
6
Add Newton’s second law
F = ma, and the mass is positive, so F points the same way as a: towards the centre. That is why the centripetal force is not a new force. It is the resultant of the real forces (tension, friction, gravity …) in that direction.
F = ma = mv2r = mω2r
Positions and velocities
1
Write the position as a vector
Put the centre at the origin. After time t the object has turned through the angle ωt.
r = (r cos ωt, r sin ωt)
2
Differentiate once: velocity
Differentiating each component with respect to time gives the velocity.
v = drdt = (−ωr sin ωt, ωr cos ωt)
3
Check it
The size is ωr√(sin²ωt + cos²ωt) = ωr, which is the speed v. And r · v = 0, so the velocity is at right angles to the radius, along the tangent.
4
Differentiate again: acceleration
Differentiate the velocity components once more.
a = dvdt = (−ω2r cos ωt, −ω2r sin ωt) = −ω2r
5
Read the result
The minus sign says a points the opposite way to r, that is, towards the centre. Its size is ω2r, and since v = ωr that is v2/r.
a = v2r = ω2r
6
Add Newton’s second law
F = ma, so F points the same way as a, towards the centre. It is the resultant of the real forces, not an extra one.
F = ma = mv2r = mω2r
Position, velocity and acceleration
Test yourself
Ten questions in four levels, easiest first. Type a number, then press Check. Each question tells you how to present the final answer. The checker separates the numerical value from the required form, so correct physics is not confused with incorrect rounding. Use the equations above if you need them.
0 of 10 correct
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