Full chapter
Circular Motion
All 3 topics and the revision summary on one page.
01
Describe a turn
An angle tells you how far around a circle a body has moved. Its radius then connects that turn to the distance travelled and the speed along the path.
Radians measure arc length relative to radius
An arc of length sarc on a circle of radius r subtends an angle of magnitude θ, measured in radians:
sarc = rθ
One complete turn: 2π rad = 360°
An angle of one radian cuts off an arc equal in length to the radius. The arc follows the circle; the straight chord between its endpoints is shorter. Use the radius from the rotation axis, not the diameter.
Radians are a ratio of two lengths, so the ratio is dimensionless. Keep the label rad to show that the number represents an angle. Convert degrees using θrad = θdegrees × π/180 before using sarc = rθ or the circular-motion equations.
Angular velocity, period and frequency
Angular velocity is the rate of change of angular displacement. In a stated plane, choose a viewing direction and positive rotation sense to give angular displacement a sign. For example, take anticlockwise viewed from above as positive; clockwise rotation then has negative angular velocity. An arc length remains non-negative.
For uniform rotation, this equals the constant angular velocity ω. Its units are rad/s. Angular speed is its magnitude. In the speed equations below, v and ω denote non-negative speed and angular-speed magnitudes; the chosen rotation sense is stated separately.
The period T is the time for one complete revolution, in seconds. The frequency f is the number of revolutions per second, in hertz (Hz = s-1).
ω = 2π/T = 2πf
Angular frequency is also written ω in rad/s: one cycle corresponds to 2π radians. A frequency in Hz is therefore not numerically equal to the angular rate in rad/s. Divide revolutions per minute by 60 to obtain revolutions per second before multiplying by 2π.
For a fixed radius, speed along the circle is arc distance divided by time. Substituting sarc = rθ gives:
Instantaneous velocity is tangent to the circle. The curved arrow used to mark rotation sense is not a velocity vector pointing around the whole path.
Estimate the size first
A radius of about 0.4 m and a period of about 2 s give an angular rate of about 3 rad/s and speed of about 1 m/s. The inward acceleration is of order a few m/s2, so a roughly 0.2 kg marker would need an inward resultant of order 1 N.
These assumed rough scales are a check on magnitude. The acceleration and force pages explain why those quantities point inward and how to calculate them.
Worked uniform turn
Two markers on one rigid platform
A marker at radius 0.400 m completes one anticlockwise turn every 2.00 s, viewed from above. The rate is uniform.
ω = 2π/2.00 = π rad/s
v = 0.400π = 1.26 m/s
In 0.750 s, starting from the radius pointing right:
sarc = rΔθ = 0.300π = 0.942 m
A turn measures an arc, not its chord
View the platform from above, with anticlockwise positive. The outer marker starts at the hollow point on the right; both markers advance through the same angle in 0.750 s.
Both markers have period 2.00 s and angular speed π rad/s. The outer speed is 1.26 m/s; the inner speed is 0.628 m/s. The brown arc measures distance travelled; the blue straight arrow shows instantaneous velocity.
A second marker at radius 0.200 m on the same rigid platform also takes 2.00 s per turn. Its angular speed is still π rad/s, but its tangential speed is 0.200π = 0.628 m/s, half the outer marker's. Equal period does not mean equal distance travelled.
Optional check Two markers at radii 0.200 m and 0.400 m share a rigid platform that turns uniformly once every 2.00 s. How do their angular speeds and tangential speeds compare?
Read the coordinates around a circle
Put the centre at the origin, with x to the right and y upwards on the top-view drawing. Measure θ anticlockwise from +x. Resolving the radius gives x = r cos θ and y = r sin θ. The radius r stays positive while the coordinates can be positive, zero or negative.
Position components depend on the angle
Measure x to the right and y upward from the circle's centre. The angle θ starts at the positive horizontal radius and increases anticlockwise. Then x/r = cos θ and y/r = sin θ.
x/r = cos θ
y/r = sin θ
The vertical ratios are dimensionless. These curves plot position against angle; the horizontal axis is not time.
At 0, π/2, π, 3π/2 and 2π, the marker passes right, top, left, bottom and right again. Accordingly, x/r follows 1, 0, -1, 0, 1 while y/r follows 0, 1, 0, -1, 0. These sine and cosine curves describe the coordinates of the circular position; the two-dimensional path is still a circle.
Small angles must be in radians
Near zero, the following useful approximations hold:
cos θ ≈ 1
for small |θ|, with θ in radians
Here ≈ means approximately equal. At θ = 0.10 rad, sin θ = 0.0998334, tan θ = 0.1003347 and cos θ = 0.9950042. The approximations become closer as |θ| gets smaller; whether they are accurate enough depends on the required precision.
For r = 0.400 m and θ = 0.10 rad, the arc is rθ = 0.0400 m. The chord is 2r sin(θ/2) = 0.0399833 m, nearly the same length. This explains why a short chord can approximate a small arc. The earlier 135° turn is not a small angle; use the full trigonometric functions and arc relation there.
Measure repeated turns
On a secured rotating platform at a controlled steady rate, choose a visible reference line. Start and stop as the marker crosses that line in the same sense, counting a stated number of complete revolutions. Measure the radius from the axis to the marker's centre with a rule of suitable range and resolution.
Illustrative stopwatch readings for 20 complete turns are 39.8 s, 40.2 s and 40.0 s. Their mean is 40.0 s, giving T = 40.0/20 = 2.00 s. With radius 0.400 m, these values reproduce ω = π rad/s and v = 1.26 m/s. They are supplied example readings, not a claim that every platform turns at this rate.
Timing many turns reduces the fractional effect of start/stop timing. Repeats reveal scatter, but do not correct a wrongly located axis or a repeated miscount of whole turns. Check whether the rotation rate changes across the interval rather than assuming uniform motion from one average.
For an independent speed comparison, calibrated video can provide displacement over short known frame intervals. View perpendicular to the rotation plane, calibrate distance in that plane, identify the centre and verify the frame timing. A tilted camera distorts the geometry. A finite chord divided by time approximates arc speed; smaller angular intervals improve that approximation but must still be large enough to resolve position and time reliably.
Compare this independently estimated speed with rω obtained from the radius and a separately timed full revolution at the same steady rate. Calculating v from rω and substituting it back is not an independent experimental test of the relationship.
02
Why acceleration points inward
A body can accelerate without speeding up. In uniform circular motion, its velocity changes direction continuously, so its acceleration is nonzero and points towards the centre.
Speed is the magnitude of velocity. The instantaneous velocity is tangent to the path. To find a change of velocity, subtract the vectors rather than their magnitudes; the vector method lets you translate an arrow without rotating or changing its length.
Δv = vlater - vearlier
The symbols v here represent vectors. Place the two velocities at a common origin: the arrow from the earlier tip to the later tip is their difference.
Worked velocity change
Equal speeds, different velocity vectors
A body travels anticlockwise at 2.00 m/s on a circle of radius 1.00 m. Consider positions 10° below and 10° above the right-pointing radius, with x right and y up.
Equal speeds can have different velocity vectors
Use radius 1.00 m, speed 2.00 m/s and anticlockwise motion. The two positions are at -10° and +10° from the rightward radius. Right and up define the positive component directions.
Velocities are tangent at their own positions
Both speeds are 2.00 m/s, but the directions differ. At M, the instantaneous acceleration is 4.00 m/s2 left, perpendicular to the upward velocity there.
Translate the vectors without rotating them
The vertical velocity components cancel in the subtraction. Δv = (-0.694593, 0) m/s and Δt = 0.174533 s give average acceleration 3.97972 m/s2 left. A smaller angular interval brings this average closer to the instantaneous value, 4.00 m/s2.
Keeping extra digits during the calculation, the velocity components in m/s are:
vlater = (-0.347296, +1.969616)
Δv = (-0.694593, 0) m/s
The angular separation is 20° = π/9 rad. With ω = v/r = 2.00 rad/s, its duration is Δt = (π/9)/2 = 0.174533 s. Therefore:
= (-3.97972, 0) m/s2
This finite-interval average points left, towards the centre from the arc's midpoint. The instantaneous acceleration at that midpoint is (-4.00, 0) m/s2. They are close but not identical. Reducing the angular interval makes the average approach the instantaneous acceleration.
Use the magnitude and give its direction separately
For a body travelling uniformly around a circle, the instantaneous centripetal, or inward, acceleration has magnitude:
The second form follows by substituting v = rω into the first. Use radians per second for ω. These expressions give a magnitude, not a fixed x or y component. The inward direction itself changes as the body travels around the circle.
In the 1.00 m circle above, a = 2.002/1.00 = 4.00 m/s2. For the 0.400 m rotating marker with ω = π rad/s, a = 0.400π2 = 3.95 m/s2, inward.
A resultant perpendicular to velocity changes its direction without changing its speed, producing the curved path in uniform circular motion. The perpendicular force does no instantaneous work, consistent with constant kinetic energy.
If a body speeds up or slows down while following a curved path, it also has a tangential acceleration component. Its total acceleration and resultant force need not then be perpendicular to velocity. Do not apply the purely inward description to every changing-speed turn.
Optional check A body travels at constant speed around a circle. Which statement correctly describes its instantaneous velocity and acceleration?
Next, identify which actual forces supply that inward resultant. A required acceleration does not introduce a new interaction by itself.
03
Choose the radial force
Centripetal force means the inward resultant required for circular motion. Identify the real interactions first, then resolve them towards the centre.
A body moving uniformly around a circle has inward acceleration a = v2/r = rω2. Newton's second law in the inward direction gives:
= mv2/r = mrω2
In a free-body diagram, include the actual forces on the selected body: for example, tension, weight, friction or a normal contact force. Do not add a separate centripetal-force arrow to these interactions. It names the resultant's role.
Choose inward as positive for the radial equation and deal separately with other directions. In the laboratory frame used here, no outward force is needed to balance the inward resultant: the body is accelerating.
Worked horizontal tether
Tension supplies the horizontal resultant
A 0.200 kg body moves at 2.00 m/s in a circle of radius 0.500 m on a smooth horizontal table. A light taut horizontal string pulls it towards the centre. Neglect resistance and take local g = 9.81 N/kg.
The string supplies the inward horizontal force
A 0.200 kg body travels at 2.00 m/s on a smooth horizontal table, with a taut string of radius 0.500 m. Resistance is negligible; g = 9.81 N/kg.
Top view: tension before release; tangent path after removal
The purple arrow applies before removal. Removing the string removes this tension; then the dashed tangent path applies. No extra outward or inward force is added.
Side view: only the vertical balance is shown
Vertical acceleration is zero. While the string remains taut, the horizontal resultant is still 1.60 N inward, giving horizontal acceleration 8.00 m/s2.
There is no vertical acceleration, so support N balances weight mg. The only horizontal force is the string tension S:
S = mv2/r
= 0.200(2.00)2/0.500 = 1.60 N
The tension itself supplies the 1.60 N inward resultant. An additional 1.60 N centripetal force would count the same effect twice.
If the string is removed and no horizontal force remains, the body initially continues along the tangent in the table's plane. It does not depart radially outward or keep following the circle without a horizontal force.
State what remains fixed
At the same mass and radius, doubling speed from 2.00 to 4.00 m/s makes the required force 6.40 N, four times as large. By contrast:
- At fixed mass and speed, doubling radius halves mv2/r.
- At fixed mass and angular speed, doubling radius doubles mrω2.
Both conclusions are correct under their stated conditions. Speed and angular speed are connected by v = rω; they cannot both stay fixed when radius changes.
Optional check A 0.200 kg body moves on a smooth horizontal table, attached to a light horizontal string of radius 0.500 m. The string can provide at most 3.60 N tension. What is the maximum speed in this model?
Worked conical pendulum
Only one component of tension is inward
A 0.200 kg bob on a 1.00 m light inextensible string travels uniformly around a horizontal circle of radius 0.600 m. The fixed pivot is 0.800 m above the orbit plane. The string is taut and air resistance is negligible. For this example, use g = 10.0 N/kg.
Let θ be the string's angle to the vertical. The right triangle gives sin θ = 0.600/1.00 and cos θ = 0.800/1.00, so θ = 36.8699°. Name tension S to distinguish it from period T.
The circle's radius is not the string length
A 0.200 kg bob moves uniformly on a taut 1.00 m string. This example uses g = 10.0 N/kg. Use S for tension, reserving T for the period.
Side view: a 0.600 / 0.800 / 1.00 m triangle
Two real forces; tension is resolved into components
Purple arrows are the two real forces on the bob. Brown dashed arrows are components of that same tension, not additional forces.
The inward resultant is 1.50 N, giving acceleration 7.50 m/s2. Using r = 0.600 m gives speed 2.12 m/s; neither the full 2.50 N tension nor the 1.00 m string length is substituted as the radial quantity.
The bob stays at the same height, so the vertical component of tension balances its weight:
0.800S = 0.200 × 10.0 = 2.00 N
S = 2.50 N
The horizontal component points towards the centre of the horizontal orbit:
= 2.50 × 0.600 = 1.50 N
mv2/r = 1.50
Use the orbit radius 0.600 m, not the 1.00 m string length:
v = √(1.50 × 0.600/0.200)
= √4.50 = 2.12 m/s
ω = v/r = 3.54 rad/s
T = 2π/ω = 1.78 s
The full tension is 2.50 N, but the inward resultant is only 1.50 N. Its 2.00 N upward component balances the 2.00 N downward weight. Keep extra digits until the final result when moving from speed to angular speed and period.
Optional check A 0.200 kg conical-pendulum bob moves in a horizontal circle. Its tension is 2.50 N at 36.87 degrees to the vertical; take g = 10.0 N/kg. What supplies the inward resultant?
Choose a graph that tests the relationship
For the smooth-table model with fixed m = 0.200 kg and r = 0.500 m, the inward force is F = (m/r)v2. The following values are calculated model data, not force-sensor observations.
| v / (m/s) | v2 / (m2/s2) | F / N |
|---|---|---|
| 1.00 | 1.00 | 0.400 |
| 1.50 | 2.25 | 0.900 |
| 2.00 | 4.00 | 1.60 |
| 2.50 | 6.25 | 2.50 |
F plotted vertically against v horizontally is curved. Plotting F against v2 instead gives a straight line through the origin, with gradient:
Gradient units: N/(m2/s2) = kg/m
The transformed horizontal quantity is squared speed, not speed with a relabelled axis. A straight line with this gradient is consistent with the supplied model.
Use logarithms to find an exponent
Take reference values v0 = 1.00 m/s and F0 = 0.400 N from the first row. The ratios v/v0 and F/F0 have no units, so their logarithms are meaningful.
ln(F/F0) = 2 ln(v/v0)
A plot of ln(F/F0) vertically against ln(v/v0) horizontally therefore has gradient 2. At v = 2.00 m/s, the force ratio is 4, giving ln 4 / ln 2 = 2. Both ratios equal 1 at the reference row, so that transformed point is (0, 0).
More generally, a supplied power law F/F0 = (v/v0)n becomes a straight line of gradient n on these logarithmic axes. Using log10 consistently on both axes gives the same exponent. Logarithms here require positive inputs; do not take a logarithm of a zero or negative force ratio, or of an unnormalised quantity carrying units.
Worked logarithms of combined changes
Change mass, speed and radius together
Starting from m0 = 0.200 kg, v0 = 1.00 m/s and r0 = 0.500 m, the inward force is F0 = 0.400 N. In a second circular-motion model, triple the mass, double the speed and increase the radius by a factor of 1.5. Each force uses F = mv2/r.
= 3 × 22/1.5 = 8
F = 8 × 0.400 = 3.20 N
To separate these contributions using logarithms, a product becomes a sum and a quotient becomes a difference. For positive dimensionless a and b:
ln(a/b) = ln a - ln b
ln(an) = n ln a
Apply all three rules to the force ratio:
+ 2 ln(v/v0) - ln(r/r0)
= ln 3 + 2 ln 2 - ln 1.5
= 2.07944
Exponentiating recovers F/F0 = exp(2.07944) ≈ 8. Keep the unrounded logarithm for a precise calculation. Adding the logarithms accounts for multiplying the positive ratios; ln(a + b) is not ln a + ln b.
Read an orders-of-magnitude scale
lg means log10, whereas ln uses base e. The product, quotient and power rules work with either base when used consistently. On a graph of lg(F/F0), consider these positive force ratios:
| F/F0 | lg(F/F0) |
|---|---|
| 0.001 | -3 |
| 0.01 | -2 |
| 0.1 | -1 |
| 1 | 0 |
Each step of 1 on this logarithmic axis means multiplying the force ratio by 10. Moving from -3 to 0 spans three orders of magnitude: the ratio increases from 0.001 to 1, a factor of 103 = 1000, not an increase of 3 N. Such a scale displays multiplicative comparisons across a wide range; it cannot include zero or negative ratios.
What a real force comparison needs
An investigation needs independently measured speed and radial force, with mass and radius controlled. Use an appropriately secured apparatus at a controlled rate, a calibrated force instrument with a suitable range, and a clear account of the forces it measures.
A measured total tension is not automatically the radial resultant, as the conical example shows. A hanging mass's weight is not automatically the force transmitted to the moving body either: establish the apparatus force balance and account for intervening friction before equating them.
A fitted intercept can prompt a check for force-zero or background-force errors; scatter alone does not identify its cause. A changing radius or rotation rate can also spoil the intended comparison. Repeats help assess consistency but cannot repair the wrong force model.
Revision summary
Identify the centre and radius, distinguish speed from velocity, and name the actual forces before writing an inward-force equation.
Angle, period and frequency
2π rad = 360°
f = 1/T
ω = 2πf = 2π/T
v = rω
T is time per complete revolution; f is revolutions per second. In these magnitude equations, ω is angular speed. For signed angular velocity, state the viewing direction and positive rotation sense, then use the rate of signed angular displacement. Convert degrees to radians and revolutions/min to revolutions/s as needed.
Two points on one rigid rotating platform have equal angular speed and period. Their tangential speeds are proportional to their distances from the axis.
With angle measured anticlockwise from +x, x/r = cos θ and y/r = sin θ. Coordinates can be signed even though radius is positive. For small |θ| in radians, sin θ ≈ tan θ ≈ θ and cos θ ≈ 1. A small chord approximates its arc; these approximations are unsuitable for a 135° turn.
Time many complete turns between same-sense crossings to obtain T. Check the radius from the axis, counting, timing, rate stability and repeatability. A calibrated video chord gives an interval estimate, not an exact finite-arc distance; calculate an independent speed if testing v = rω.
Changing velocity at constant speed
Instantaneous velocity is tangent; acceleration is inward and perpendicular to velocity for uniform circular motion. Their directions change around the circle. Constant speed does not mean zero acceleration.
Form Δv by vector subtraction, then divide by Δt for average acceleration. A finite-interval average approaches the instantaneous value as the interval shrinks; do not label the finite construction as exact instantaneous acceleration.
A purely perpendicular resultant changes direction without changing speed and does no instantaneous work. If speed changes on a curved path, a tangential component is also present, so the total resultant need not be perpendicular.
Use the inward resultant, not an extra force
Draw actual interactions on the selected body, resolve them inward and account for other directions. Tension, gravity, friction or contact forces can contribute. There is no additional centripetal interaction to add or outward balancing force to invent in the stated laboratory frame.
For a horizontal string on a smooth table, tension supplies the radial resultant and support balances weight. For a conical pendulum with angle θ from vertical, S cos θ = mg and S sin θ = mv2/r. Use the horizontal orbit radius, not the slanted string length.
At fixed m and r, F is proportional to v2. At fixed m and v it is proportional to 1/r; at fixed m and ω it is proportional to r. State what is held fixed. If the horizontal tether is removed and no horizontal force remains, the body initially follows the tangent.
A plot of F against v2 has gradient m/r in kg/m. A plot of ln(F/F0) against ln(v/v0) has dimensionless gradient 2 for the same fixed-mass, fixed-radius law. Use positive dimensionless ratios and the same logarithm base on both axes. Model-data agreement alone is not experimental verification.
For positive ratios, logarithms turn multiplication into addition, division into subtraction and powers into factors. lg means log10: a difference of 3 in lg values represents a factor of 1000. Do not apply the product rule to a sum.
| Quantity | Symbol | Units |
|---|---|---|
| Angle; angular displacement | θ | rad, or ° before conversion |
| Angular speed, angular velocity; angular frequency | ω | rad/s |
| Period; frequency | T; f | s; Hz = s-1 |
| Radius; arc length | r; sarc | m |
| Speed; acceleration | v; a | m/s; m/s2 |
| Mass | m | kg |
| Resultant force; tension | F; S here | N |
Read the named quantity and units: θ and T can denote other quantities elsewhere. Here T is period, while S labels tension. Angular velocity includes a stated rotation sense; angular speed is its magnitude.
Back to describing a turn