Uniform Circular Motion: Centripetal Acceleration and Real Forces

1) Angular ideas you can feel: revolution, frequency, period

In uniform circular motion an object moves around a circle at constant speed, but its motion is still changing because its direction changes continuously. To describe “how fast it goes around,” we use a few everyday angular ideas:

• One revolution: one full trip around the circle (360°).
• Period T: time for one revolution (seconds per revolution).
• Frequency f: revolutions per second (Hz). It is the inverse of period: f = 1/T.

The distance traveled in one revolution is the circumference 2πr. If the object takes time T to go around once, its speed is

v = (distance per revolution)/(time per revolution) = (2πr)/T

Equivalently, using f = 1/T:

v = 2πr f

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Intuition check: for the same period T, a larger radius means a larger circumference, so the object must move faster to complete a lap in the same time.

2) Why constant speed still means acceleration: centripetal acceleration

Direction: inward, toward the center

Acceleration is about how the velocity vector changes. In uniform circular motion, the speed (the length of the velocity vector) stays the same, but the velocity vector rotates as the object moves around the circle.

Pick two nearby points on the circle. At each point, the velocity vector is tangent to the circle. Because the tangents point in different directions, the velocity vectors differ. The change in velocity Δv points roughly toward the center for small time steps, and in the limit of very small time steps it points exactly inward. Therefore the acceleration points inward.

This inward acceleration is called centripetal acceleration (“center-seeking”).

Magnitude: a_c = v²/r

We can get the size of the acceleration using a vector-geometry argument. Consider two velocity vectors of equal magnitude v separated by a small angle Δθ. The tip-to-tip difference vector has magnitude approximately

|Δv| ≈ v Δθ

for small angles (because the arc length on a circle of radius v is vΔθ). Over a small time Δt, the acceleration magnitude is

a = |Δv|/Δt ≈ (v Δθ)/Δt

Now connect Δθ to how far the object moved along the circle. In time Δt, it travels arc length Δs = vΔt. On a circle, arc length and angle satisfy Δs = rΔθ, so Δθ = Δs/r = (vΔt)/r. Substitute into the acceleration expression:

a ≈ (v (vΔt/r))/Δt = v²/r

In the limit as Δt → 0, this becomes exact:

a_c = v²/r

Direction: toward the center. Magnitude: v²/r.

Limiting cases to build intuition

• Bigger speed: doubling v makes a_c four times larger. Fast turns require much more inward acceleration.
• Bigger radius: for the same speed, a larger r reduces a_c. Wide turns feel gentler.
• Straight-line limit: as r → ∞, a_c → 0. A huge-radius “circle” looks like straight motion with almost no direction change.

3) “Centripetal force” is not a new force

Newton’s second law says the net force equals mass times acceleration. For uniform circular motion, the acceleration is inward with magnitude v²/r, so the net inward force must be

F_net,inward = m a_c = m v²/r

The phrase centripetal force is just a name for whatever combination of real forces adds up to this inward net force. Depending on the situation, the inward net force might be provided by:

• Tension (mass on a string)
• Static friction (car turning on a flat road)
• Gravity (satellite orbiting, or a mass moving in a vertical circle at some points)
• Normal force (banked curve, roller coaster track)
• Or a combination (banked curve with friction)

Key habit: never draw “centripetal force” as an extra arrow on a free-body diagram. Draw only real forces. Then choose a radial (inward/outward) axis and set the net radial component equal to m v²/r inward.

4) Classic scenarios (with step-by-step setups)

A) Car rounding a flat curve: friction limit

A car of mass m goes around a flat curve of radius r at speed v. On a flat road, the only horizontal force available to turn the car is static friction. Weight and normal cancel vertically (no vertical acceleration).

Step-by-step:

• Choose axes: radial inward (toward the center of the curve) and vertical.
• List real forces: weight mg down, normal N up, static friction f_s horizontally toward the center (if the car is turning without slipping).
• Vertical equation: N - mg = 0 so N = mg.
• Radial equation: net inward force must be m v²/r. Only friction contributes radially, so f_s = m v²/r.
• Friction limit: f_s ≤ μ_s N = μ_s mg.

Combine the last two lines:

m v²/r ≤ μ_s m g

Mass cancels, giving the maximum safe speed:

v_max = √(μ_s g r)

Misconception to avoid: if you go too fast, the car does not “fly outward because of an outward force.” Instead, friction cannot supply enough inward force, so the car cannot curve tightly enough and it slides roughly tangentially (straight-ish) relative to the curve.

B) Banked curve (no friction): the road supplies the inward force via the normal

Now the road is tilted (banked) by angle θ. Assume no friction. The normal force is perpendicular to the road surface, so it has both vertical and horizontal components. The horizontal component points inward and can provide the centripetal requirement.

Step-by-step:

• Axes: vertical and radial inward.
• Forces: weight mg down, normal N perpendicular to the surface.
• Resolve N: vertical component N cosθ upward, inward radial component N sinθ.
• Vertical balance: N cosθ - mg = 0 so N = mg / cosθ.
• Radial requirement: N sinθ = m v²/r.

Substitute N:

(mg/cosθ) sinθ = m v²/r

g tanθ = v²/r

v = √(r g tanθ)

This is the design speed for a frictionless banked turn: at this speed, the banking alone provides exactly the needed inward net force.

C) Banked curve with friction: a range of possible speeds

With friction available, the car can take the same banked curve at speeds different from the design speed. Friction can point either up the slope or down the slope depending on whether the car would otherwise tend to slide.

How to decide friction direction (practical rule):

• If v is too low compared to the design speed, the car tends to slide down the bank (toward the inside). Static friction acts up the bank.
• If v is too high, the car tends to slide up the bank (toward the outside). Static friction acts down the bank.

Step-by-step setup (symbolic):

• Forces: mg down, N perpendicular to surface, f_s along surface (up or down the slope).
• Resolve into vertical and radial inward components. Take inward as positive radial.
• Use: f_s ≤ μ_s N and at the limit f_s = μ_s N to find extreme speeds.

Using components (with friction magnitude at its maximum f = μ_s N):

• Case 1 (high speed): friction acts down the slope (adds inward component). Equations: N cosθ - f sinθ = mg and N sinθ + f cosθ = m v²/r.
• Case 2 (low speed): friction acts up the slope (reduces inward component). Equations: N cosθ + f sinθ = mg and N sinθ - f cosθ = m v²/r.

Solving each case with f = μ_s N gives the maximum and minimum speeds (often presented as):

v_max = √( r g (sinθ + μ_s cosθ) / (cosθ - μ_s sinθ) )

v_min = √( r g (sinθ - μ_s cosθ) / (cosθ + μ_s sinθ) )

Practical note: if a denominator becomes zero or negative, that indicates the chosen parameters (θ, μ_s) make that extreme unattainable (the required friction would exceed what’s available, or the direction assumptions break down).

D) Mass on a string in a horizontal circle: tension provides the inward net force

A mass m moves in a horizontal circle of radius r at speed v, attached to a string. If the circle is horizontal and the string is horizontal (idealized), the inward force is the string tension T.

Step-by-step:

• Radial axis: inward along the string.
• Forces: tension T inward (and other forces depending on the exact setup; in this idealized horizontal-string model, we focus on radial).
• Radial equation: T = m v²/r.

What this tells you: for a given radius, tension grows like v². Doubling speed requires four times the tension, which is why strings (or seat belts, tires, track walls) must handle rapidly increasing loads at higher speeds.

If instead the mass forms a conical pendulum (string at angle, mass still moving in a horizontal circle), then tension has both vertical and horizontal components: the vertical component balances weight, and the horizontal component supplies m v²/r. The same strategy applies: resolve tension into components and apply the radial requirement.

Misconceptions to address explicitly

1) “If speed is constant, acceleration is zero”

False in circular motion. Acceleration depends on the change of the velocity vector, not just its magnitude. Constant speed with changing direction still means nonzero acceleration.

2) Confusing velocity direction with force direction

In uniform circular motion:

• Velocity is tangent to the circle.
• Net force (and acceleration) points inward.

An inward force does not mean the object moves inward; it means the object’s direction is continuously being turned.

3) “Centrifugal force”

In an inertial frame (a non-accelerating frame like the ground, approximately), there is no real outward force on the turning object. The object requires an inward net force to curve; if that inward force disappears, the object moves off tangentially.

In a non-inertial rotating frame (for example, analyzing motion while you rotate with the car), you may introduce a fictitious outward force called the centrifugal force to make Newton’s second law look like it works in that accelerating frame. It is not an interaction force from another object; it is a bookkeeping tool tied to the choice of frame.

Practical guidance: for beginner problem-solving, stay in an inertial frame whenever possible: draw only real forces, then enforce F_net,inward = m v²/r.