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2.9 - Circular Motion

Acceleration from Changing Direction

An object accelerates whenever its velocity changes, and since velocity is a vector with both magnitude (speed) and direction, that acceleration can come from speed changing (speeding up or slowing down), direction changing (turning, even at constant speed), or both changing at once (typical real-world motion).
Here’s the key insight: an object moving in a circle at constant speed is still accelerating, because its direction is continuously changing. This is called centripetal acceleration.

Centripetal Acceleration

Centripetal acceleration is the piece of an object’s acceleration pointed toward the center of its circular path. The word comes from Latin, centrum meaning center and petere meaning to seek, so centripetal essentially means “center-seeking.”
ac=v2ra_c = \frac{v^2}{r}
where aca_c is centripetal acceleration in m/s², vv is the linear speed along the circular path in m/s, and rr is the radius of the circle in meters. It always points toward the center of the circle, perpendicular to the velocity.
2.9 - Circular Motion
Why speed squared? Higher speeds need sharper changes in direction to stay on the circle, creating more acceleration, and the relationship is quadratic: double the speed and you get quadruple the acceleration, triple the speed and you get nine times the acceleration.
Why radius in the denominator? Tighter circles (smaller radius) need more dramatic direction changes: smaller radius means sharper turn means greater acceleration, and larger radius means gentler turn means less acceleration. A highway off-ramp at 60 mph uses a large radius (a gentle, comfortable curve), while a parking lot turn at 10 mph uses a small radius (a sharp but still manageable turn), and taking highway speeds through that same parking lot radius would produce dangerously high acceleration.
Any curved path, even non-circular ones, can be analyzed as circular motion, or approximated locally as a series of circular arcs, each with its own radius. A sharp turn is a small radius circle (high centripetal acceleration), a gentle curve is a large radius circle (low centripetal acceleration), and a lane change is an extremely large radius (barely any centripetal acceleration at all).

Tangential Acceleration and Net Acceleration

Centripetal acceleration only handles the change in direction. If the object’s speed is also changing, it has a second piece of acceleration, the tangential acceleration (ata_t). It’s the rate at which the object’s speed changes, and it points along the tangent to the circle: forward (the same way as the velocity) when speeding up, backward when slowing down.
The object’s net acceleration is the vector sum of the two pieces. Since centripetal (toward the center) and tangential (along the path) are always perpendicular, the Pythagorean theorem gives its size:
a=ac2+at2a = \sqrt{a_c^2 + a_t^2}
In uniform circular motion (constant speed), at=0a_t = 0 and the net acceleration points straight at the center. If the object is speeding up or slowing down, the net acceleration tilts away from the center, forward or backward.
Example: a car on a circular track of radius 50.0 m is moving at 10.0 m/s and speeding up at 1.50 m/s². Its centripetal acceleration is ac=v2r=(10.0)250.0=2.00 m/s2a_c = \frac{v^2}{r} = \frac{(10.0)^2}{50.0} = 2.00 \text{ m/s}^2 toward the center, and its tangential acceleration is 1.50 m/s² forward. The net acceleration is:
a=(2.00)2+(1.50)2=2.50 m/s2a = \sqrt{(2.00)^2 + (1.50)^2} = 2.50 \text{ m/s}^2
It points inward and tilted forward, tan⁡−1(1.502.00)=36.9∘\tan^{-1}\left(\frac{1.50}{2.00}\right) = 36.9^\circ away from the direction toward the center.
2.9 - Circular Motion

Centripetal Force

Since acceleration requires force, circular motion requires a centripetal force.
Fc=mac=mv2rF_c = ma_c = m\frac{v^2}{r}
where FcF_c is centripetal force in Newtons, mm is mass in kg, vv is speed in m/s, and rr is radius in meters.
Here’s an important idea: centripetal force is not a new type of force to add to your free body diagram. Instead, it’s the net force pointing toward the center, and it can come from a single force (like tension in a string), multiple forces combined (like normal force minus weight), or components of forces (like friction during a turn). “Centripetal force” describes the role a force is playing (causing circular motion), not a brand new physical force.

Common Sources of Centripetal Force

Friction: in a car turning on a flat road, friction between the tires and the road provides the inward force needed to change direction. Take a turn at 15 m/s and friction points toward the center of the turn, but with no friction, the car just continues straight (Newton’s First Law).
Tension: for an object attached to a string swung in a horizontal or vertical circle, the string pulls inward, providing centripetal force. Spin a ball on a rope overhead and tension points toward the center along the rope, but break the rope and the ball flies off tangentially.
Normal force: in a roller coaster loop, a car going over a hill, or an object on a spinning platform, the contact force from a surface provides part or all of the centripetal force. At the top of a roller coaster loop, both normal force and weight point downward, toward the center, and together they provide the centripetal force.
Gravity: for a satellite orbiting Earth, or planets orbiting the Sun, gravitational attraction provides the centripetal force for orbital motion. For the Moon orbiting Earth, gravity pulls it toward Earth (centripetal), while its tangential velocity keeps it from actually falling in.

Free Body Diagrams for Circular Motion

The approach is similar to other Newton’s Law problems, but with a key difference: instead of horizontal/vertical or parallel/perpendicular to a surface, use circular and non-circular directions. The circular direction is toward or away from the center of the path (radial), and the non-circular direction is tangent to the path (tangential). The convention is that pointing into the center (the centripetal direction) counts as positive.
Example: car taking a turn on a flat road. A car turns left on a horizontal, flat road at constant speed. The forces are weight (Fg=mgF_g = mg, down), normal force (NN, up), and friction (ff, toward the center of the turn, providing the centripetal force).
2.9 - Circular Motion
In the non-circular (vertical) direction there’s no acceleration since the car doesn’t leave the road:
∑Fvertical=0  ⟹  N−mg=0  ⟹  N=mg\sum F_{\text{vertical}} = 0 \implies N - mg = 0 \implies N = mg
In the circular (horizontal, toward center) direction:
∑Fradial=Fc  ⟹  f=mv2r\sum F_{\text{radial}} = F_c \implies f = m\frac{v^2}{r}
Friction provides the centripetal force. Substituting the friction formula, μN=mv2r\mu N = m\frac{v^2}{r}, and then N=mgN = mg:
μmg=mv2r  ⟹  μg=v2r  ⟹  v=μgr\mu mg = m\frac{v^2}{r} \implies \mu g = \frac{v^2}{r} \implies v = \sqrt{\mu g r}
The maximum safe turning speed depends on the friction coefficient, gravity, and the turn’s radius, but not on the car’s mass. Heavy trucks and light cars can take the exact same turn at the exact same maximum speed as long as their tires have the same friction coefficient.
Example: roller coaster at the top of a loop. At the highest point of a vertical loop, the forces are weight (Fg=mgF_g = mg, downward, toward the center) and normal force (NN, downward, toward the center, since the seat pushes down on the rider). At the top of the loop, both forces point downward, toward the center of the circular path.
2.9 - Circular Motion
In the circular direction, since both forces point toward the center:
∑Ftoward center=Fc  ⟹  N+mg=mv2r\sum F_{\text{toward center}} = F_c \implies N + mg = m\frac{v^2}{r}
Solving for normal force:
N=mv2r−mg=m(v2r−g)N = m\frac{v^2}{r} - mg = m\left(\frac{v^2}{r} - g\right)
At high speed, v2/r>gv^2/r > g, so N>0N > 0, and the rider feels pressed into the seat. At the minimum speed, v2/r=gv^2/r = g, so N=0N = 0, and the rider feels weightless, just barely staying in the seat. If it’s too slow, v2/r<gv^2/r < g, giving N<0N < 0, which is physically impossible, the rider would fall (which is why restraints are needed).
The minimum speed to maintain contact comes from setting vmin⁡2r=g\frac{v_{\min}^2}{r} = g:
vmin⁡=grv_{\min} = \sqrt{gr}

Banked Curves

Highway curves and racetracks are often tilted (banked) toward the center of the turn. On a banked road, the normal force is perpendicular to the tilted surface, so it points partly up and partly toward the center. That inward part of the normal force can supply the centripetal force, even with no friction at all.
2.9 - Circular Motion
No friction needed (the design speed): take a car moving at speed vv around a curve of radius rr banked at angle θ\theta, on an icy (frictionless) road. The only forces are NN (perpendicular to the road) and mgmg (down). Split NN into a vertical part and a horizontal part that points toward the center:
Vertical: Ncos⁡θ−mg=0  ⟹  Ncos⁡θ=mg\text{Vertical: } N\cos\theta - mg = 0 \implies N\cos\theta = mgToward center: Nsin⁡θ=mv2r\text{Toward center: } N\sin\theta = m\frac{v^2}{r}
Dividing the second equation by the first cancels NN and mm:
tan⁡θ=v2gr\tan\theta = \frac{v^2}{gr}
So for a given radius and bank angle, there’s exactly one speed at which a car can take the curve with no friction, and it doesn’t depend on the car’s mass.
Example: a highway curve has a radius of 125 m. What bank angle lets a car take the curve at 20.0 m/s with no help from friction? Use g=10 m/s2g = 10 \text{ m/s}^2.
tan⁡θ=v2gr=(20.0)2(10)(125)=0.320  ⟹  θ=17.7∘\tan\theta = \frac{v^2}{gr} = \frac{(20.0)^2}{(10)(125)} = 0.320 \implies \theta = 17.7^\circ
With friction (qualitative only on the AP exam): if a car goes faster than the design speed, the inward part of the normal force isn’t enough by itself, and the car tends to slide up and out, so static friction points down the slope to help pull it inward. If the car goes slower than the design speed, it tends to slide down the bank, so static friction points up the slope. Either way, components of both the normal force and static friction add up to the net force toward the center.

Conical Pendulum

A conical pendulum is a mass on a string swinging in a horizontal circle, so the string traces out a cone. The forces on the mass are tension FTF_T along the string and weight mgmg straight down. Gravity has no horizontal part, so the horizontal component of the tension is the only inward force, and it provides the centripetal force, while the vertical component of the tension holds the mass up.
2.9 - Circular Motion
With the string at angle θ\theta from vertical and a circle of radius rr:
Vertical: FTcos⁡θ=mg\text{Vertical: } F_T\cos\theta = mgToward center: FTsin⁡θ=mv2r\text{Toward center: } F_T\sin\theta = m\frac{v^2}{r}
Dividing gives tan⁡θ=v2gr\tan\theta = \frac{v^2}{gr}, the same relationship as the frictionless banked curve. In both cases, one force tilted at θ\theta from vertical balances gravity with its vertical part and supplies the centripetal force with its horizontal part.
Example: a 0.250 kg ball on a 1.20 m string swings in a horizontal circle with the string at 30.0° from vertical. Use g=10 m/s2g = 10 \text{ m/s}^2.
Tension: FT=mgcos⁡θ=(0.250)(10)cos⁡30.0∘=2.89F_T = \frac{mg}{\cos\theta} = \frac{(0.250)(10)}{\cos 30.0^\circ} = 2.89 N, more than the ball’s 2.50 N weight, because only part of the tension holds the ball up.
Radius: r=Lsin⁡θ=(1.20)sin⁡30.0∘=0.600r = L\sin\theta = (1.20)\sin 30.0^\circ = 0.600 m
Speed: v=grtan⁡θ=(10)(0.600)tan⁡30.0∘=1.86v = \sqrt{gr\tan\theta} = \sqrt{(10)(0.600)\tan 30.0^\circ} = 1.86 m/s
Period: T=2πrv=2π(0.600)1.86=2.03T = \frac{2\pi r}{v} = \frac{2\pi(0.600)}{1.86} = 2.03 s

Uniform Circular Motion: Period and Frequency

Uniform circular motion happens when an object travels in a circle at constant speed. Even though speed stays constant, velocity keeps changing direction, so acceleration still exists.
Period (TT) is the time to complete one full revolution, measured in seconds. For example, a merry-go-round completing one rotation in 8 seconds has T=8T = 8 s.
Frequency (ff) is the number of complete cycles per unit time, measured in cycles per second, or Hertz (Hz). For example, a ceiling fan completing 3 rotations per second has f=3f = 3 Hz.
Period and frequency are reciprocals of each other:
f=1TT=1ff = \frac{1}{T} \qquad T = \frac{1}{f}
A high frequency means a short period (many cycles per second), while a low frequency means a long period (few cycles per second).
The distance traveled in one revolution is the circle’s circumference, 2πr2\pi r. Since speed is distance over time, for one complete cycle:
v=2πrTv = \frac{2\pi r}{T}
Or in terms of frequency, since T=1/fT = 1/f:
v=2πrfv = 2\pi r f
These connect linear speed to the rotational quantities of period and frequency.

Orbital Motion

Circular motion principles apply beautifully to orbital mechanics, where gravity supplies the centripetal force.
For a satellite orbiting a planet at constant speed, gravitational force provides the centripetal force:
Fgravity=Fcentripetal  ⟹  GMmr2=mv2rF_{\text{gravity}} = F_{\text{centripetal}} \implies G\frac{Mm}{r^2} = m\frac{v^2}{r}
where MM is the planet’s mass, mm is the satellite’s mass, rr is the orbital radius (center of planet to satellite), and GG is the universal gravitational constant. Canceling one mm and one rr:
GMr=v2  ⟹  v=GMrG\frac{M}{r} = v^2 \implies v = \sqrt{\frac{GM}{r}}
Orbital speed depends only on the planet’s mass and the orbital radius, not on the satellite’s own mass.
Finding orbital period: substitute v=2πr/Tv = 2\pi r/T into the orbital speed equation:
2πrT=GMr\frac{2\pi r}{T} = \sqrt{\frac{GM}{r}}
Squaring both sides:
4π2r2T2=GMr\frac{4\pi^2 r^2}{T^2} = \frac{GM}{r}
Solving for T2T^2:
T2=4π2r3GM  ⟹  T=2πr3GMT^2 = \frac{4\pi^2 r^3}{GM} \implies T = 2\pi\sqrt{\frac{r^3}{GM}}
This is Kepler’s Third Law of Planetary Motion: the square of the orbital period is proportional to the cube of the orbital radius, T2∝r3T^2 \propto r^3. Physically, satellites farther from a planet move slower and take longer to orbit, and both effects add up to a longer period.

Worked Example: Orbital Period of a Satellite

A satellite orbits Earth in a circle 4.00×1054.00 \times 10^5 m (400 km) above the surface, about the height of the International Space Station. Earth’s mass is 5.97×10245.97 \times 10^{24} kg and its radius is 6.37×1066.37 \times 10^6 m. Find the orbital period.
Orbital radius (center to center): r=6.37×106+4.00×105=6.77×106r = 6.37 \times 10^6 + 4.00 \times 10^5 = 6.77 \times 10^6 m
Kepler’s third law:
T=2πr3GM=2π(6.77×106)3(6.67×10−11)(5.97×1024)≈5.55×103 sT = 2\pi\sqrt{\frac{r^3}{GM}} = 2\pi\sqrt{\frac{(6.77 \times 10^6)^3}{(6.67 \times 10^{-11})(5.97 \times 10^{24})}} \approx 5.55 \times 10^3 \text{ s}
That’s about 92.4 minutes per orbit. The satellite’s own mass never showed up, so anything orbiting at this radius takes the same time. Its speed is v=2πrT≈7.67×103v = \frac{2\pi r}{T} \approx 7.67 \times 10^3 m/s, and its centripetal acceleration v2r≈8.69 m/s2\frac{v^2}{r} \approx 8.69 \text{ m/s}^2 matches the gravitational field strength at that height (see 2.6), as it should, since gravity is the only force on it.

Worked Example: Object on Rotating Platform

A 2.5 kg block sits on a horizontal platform rotating about a vertical axis, 1.2 m from the axis. The coefficient of static friction between block and platform is 0.40. Use g=10 m/s2g = 10 \, \text{m/s}^2. Find the maximum speed before the block slides, and determine whether the block slides if the platform’s period is 4.0 seconds.
Draw: the forces on the block are weight (Fg=mgF_g = mg, down), normal force (NN, up), and friction (fsf_s, toward the center, providing the centripetal force).
2.9 - Circular Motion
Sum: in the non-circular (vertical) direction, ∑Fy=0  ⟹  N−mg=0  ⟹  N=mg\sum F_y = 0 \implies N - mg = 0 \implies N = mg.
In the circular (horizontal, toward center) direction, ∑Fradial=Fc  ⟹  fs=mv2r\sum F_{\text{radial}} = F_c \implies f_s = m\frac{v^2}{r}.
At the maximum speed, static friction hits its cap, fs,max⁡=μsN=μsmgf_{s,\max} = \mu_s N = \mu_s mg.
Solve for maximum speed: setting maximum friction equal to the required centripetal force,
μsmg=mvmax⁡2r  ⟹  μsg=vmax⁡2r  ⟹  vmax⁡=μsgr\mu_s mg = m\frac{v_{\max}^2}{r} \implies \mu_s g = \frac{v_{\max}^2}{r} \implies v_{\max} = \sqrt{\mu_s g r}
Plugging in numbers:
vmax⁡=(0.40)(10)(1.2)=4.8≈2.19 m/sv_{\max} = \sqrt{(0.40)(10)(1.2)} = \sqrt{4.8} \approx 2.19 \text{ m/s}
The block can spin at speeds up to 2.19 m/s without sliding. Beyond that, friction can’t supply enough centripetal force, and the block slides outward.
Checking T=4.0T = 4.0 s: find the actual speed at this period.
v=2πrT=2π(1.2)4.0=7.544.0≈1.88 m/sv = \frac{2\pi r}{T} = \frac{2\pi (1.2)}{4.0} = \frac{7.54}{4.0} \approx 1.88 \text{ m/s}
Comparing to the maximum: actual speed is 1.88 m/s, maximum speed is 2.19 m/s. Since the actual speed is less than the maximum, the block does not slide, it stays below the threshold where friction maxes out. The required friction is frequired=mv2r=(2.5)(1.88)21.2≈7.4f_{\text{required}} = m\frac{v^2}{r} = (2.5)\frac{(1.88)^2}{1.2} \approx 7.4 N, which is less than the maximum static friction of fs,max⁡=μsmg=(0.40)(2.5)(10)=10f_{s,\max} = \mu_s mg = (0.40)(2.5)(10) = 10 N, confirming that static friction can handle it.

Problem-Solving Strategy

Draw: free body diagram with all forces, identify the center of the circular path, and figure out which forces point toward or away from center.
Sum: split forces into circular (radial) and non-circular (tangential) directions. Circular: ∑Fradial=mv2r\sum F_{\text{radial}} = m\frac{v^2}{r} (toward center is positive). Non-circular: often ∑Ftangential=0\sum F_{\text{tangential}} = 0 for uniform circular motion.
Solve: use algebra to find the unknowns, possibly relating speed to period through v=2πr/Tv = 2\pi r/T, and check that the answer makes sense.

Common Problem Scenarios

Maximum safe speed: given a friction coefficient and radius, find the maximum speed before sliding or skidding, with vmax⁡=μgrv_{\max} = \sqrt{\mu g r}.
Minimum speed in vertical circles: given a loop’s radius, find the minimum speed to maintain contact at the top, with vmin⁡=grv_{\min} = \sqrt{gr}.
Tension in a string: given mass, speed, and radius, find the tension, which provides or contributes to the centripetal force.
Orbital period: given the mass of a planet and orbital radius, find the satellite’s period, using T=2πr3GMT = 2\pi\sqrt{\frac{r^3}{GM}}.

Real-World Applications

Banked highway curves: roads tilt inward on curves so the normal force, not just friction, contributes to centripetal force, allowing safer high-speed turns even on icy roads with low friction.
Washing machine spin cycles: rapid rotation creates high centripetal acceleration. Water droplets can’t maintain that circular motion since nothing pulls them inward, so they fly outward through holes in the drum, leaving clothes drier.
Gravitational assist maneuvers: spacecraft use a planet’s gravity (which provides centripetal force during a close approach) to change direction and speed, reaching distant destinations without burning excessive fuel.
Centrifuges: lab and industrial centrifuges spin samples at very high speeds. Denser materials need more centripetal force to keep up, so they move outward, separating from less dense materials.

Key Takeaways

  • Centripetal acceleration: ac=v2ra_c = \frac{v^2}{r}, always pointing toward the center of a circular path
  • Acceleration exists even at constant speed, because direction is always changing
  • Centripetal force: Fc=mv2rF_c = m\frac{v^2}{r}, is the net force toward the center, not a new type of force
  • Common sources include friction, tension, normal force, and gravity, depending on the scenario
  • Positive direction is toward the center (centripetal direction)
  • Split forces into circular (radial, toward/away from center) and non-circular (tangential)
  • Period: time for one full revolution, T=2πrvT = \frac{2\pi r}{v}
  • Frequency: cycles per unit time, f=1Tf = \frac{1}{T}
  • Speed from period: v=2πrT=2πrfv = \frac{2\pi r}{T} = 2\pi rf
  • Maximum turning speed with friction: vmax⁡=μgrv_{\max} = \sqrt{\mu g r}, independent of mass
  • Minimum loop speed at the top of a vertical circle: vmin⁡=grv_{\min} = \sqrt{gr} to maintain contact
  • Orbital motion: gravity provides the centripetal force, with v=GMrv = \sqrt{\frac{GM}{r}} and T=2πr3GMT = 2\pi\sqrt{\frac{r^3}{GM}}
  • Mass cancels out in many circular motion problems, similar to inclines and gravitational orbits
  • Centripetal means center-seeking, the force always points toward the circle’s center
  • Use the Draw-Sum-Solve framework with circular/non-circular force analysis
  • Tangential acceleration changes the speed and points along the path; net acceleration a=ac2+at2a = \sqrt{a_c^2 + a_t^2}
  • Frictionless banked curve: the inward part of the normal force is the centripetal force, and tan⁡θ=v2gr\tan\theta = \frac{v^2}{gr}
  • On a banked curve with friction, static friction points down the slope above the design speed and up the slope below it
  • Conical pendulum: the horizontal part of the tension is the centripetal force, and tan⁡θ=v2gr\tan\theta = \frac{v^2}{gr}
  • Kepler’s third law: T2=4π2GMr3T^2 = \frac{4\pi^2}{GM}r^3, which doesn’t depend on the satellite’s mass

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