Skip to main content
New YorkPhysicsSyllabus dot point

Why does an object moving in a circle at constant speed still accelerate, and what supplies the force?

Describe uniform circular motion, calculate centripetal acceleration with ac=v2/ra_c = v^2/r and centripetal force with Fc=mv2/rF_c = mv^2/r, and identify the real force that supplies the centripetal force in a given situation.

A Regents Physics answer on uniform circular motion: why circular motion is accelerated even at constant speed, how to calculate centripetal acceleration and force with the Reference-Table equations, and what real forces supply the centripetal force, with worked examples.

Generated by Claude Opus 4.811 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this topic is asking
  2. Why circular motion is accelerated
  3. Centripetal acceleration and force
  4. What supplies the centripetal force
  5. Reference Tables note
  6. Try this

What this topic is asking

Uniform circular motion is motion in a circle at constant speed, and the Regents tests one central, counterintuitive idea: such motion is accelerated, because the direction of the velocity is always changing. The Physical Setting/Physics course asks you to calculate the centripetal acceleration and centripetal force directed toward the center, and to identify which real force (tension, gravity, friction, the normal force) supplies that centripetal force in a given situation.

Why circular motion is accelerated

This is the conceptual heart of the topic. Speed (a scalar) does not change, so kinetic energy does not change, but velocity (a vector) does, so there is acceleration. Because the acceleration is always perpendicular to the motion, it changes only the direction of the velocity, not its size, which is exactly why the speed stays constant.

Centripetal acceleration and force

The centripetal force is just Newton's second law applied to circular motion: Fc=mac=mv2rF_c = ma_c = m\dfrac{v^2}{r}. Note the strong dependence on speed (squared) and the inverse dependence on radius: a tighter turn (smaller rr) or a faster speed needs a much larger inward force. This is why fast, tight turns are hard to make.

What supplies the centripetal force

A frequent Regents error is to invent an outward "centrifugal force". There is no such force on the object; the outward feeling (being pressed against a car door on a turn) is the body's inertia resisting the inward acceleration. The only force on the object is inward.

Reference Tables note

The Reference Tables print ac=v2ra_c = \dfrac{v^2}{r}, Fc=mv2rF_c = \dfrac{mv^2}{r} and v=2Ï€rTv = \dfrac{2\pi r}{T} in the Mechanics section. You supply the recognition that circular motion is accelerated, the identification of the real force providing the centripetal force, and the understanding that no work is done by the centripetal force (it is perpendicular to the velocity), so the speed and kinetic energy are constant, as discussed in work and power.

Try this

Q1. State why an object moving at constant speed in a circle is accelerating. [2 points]

  • Cue. Its velocity direction is constantly changing; a changing velocity (a vector) means acceleration, directed toward the center.

Q2. A 2.02.0 kg object moves in a circle of radius 4.04.0 m at 6.06.0 m/s. Calculate the centripetal force. [2 points]

  • Cue. Fc=mv2r=(2.0)(6.0)24.0=(2.0)(36)4.0=18F_c = \dfrac{mv^2}{r} = \dfrac{(2.0)(6.0)^2}{4.0} = \dfrac{(2.0)(36)}{4.0} = 18 N toward the center.

Exam-style practice questions

Practice questions written in the style of NYSED exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Regents (style)2 marksPart B-2 (constructed response). A 0.500.50 kg ball moves in a horizontal circle of radius 2.02.0 m at a constant speed of 4.04.0 m/s. Calculate the magnitude of the centripetal force on the ball. Show the equation, substitution and answer.
Show worked answer →

A 2-point constructed-response calculation using the Reference-Table equation Fc=mv2rF_c = \dfrac{mv^2}{r}.

Equation: Fc=mv2rF_c = \dfrac{mv^2}{r}.
Substitution: Fc=(0.50)(4.0)22.0=(0.50)(16)2.0F_c = \dfrac{(0.50)(4.0)^2}{2.0} = \dfrac{(0.50)(16)}{2.0}.
Answer: Fc=4.0F_c = 4.0 N, directed toward the center of the circle.

Markers reward the equation from the tables, correct substitution with units, and an answer that states the force points toward the center. A common error is forgetting to square the speed.

Regents (style)1 marksPart A (multiple choice). A car rounds a flat curve at constant speed. In which direction is the net force on the car? (1) forward, in the direction of motion (2) backward (3) toward the center of the curve (4) away from the center. Justify your choice.
Show worked answer →

A 1-point Part A item on the direction of the centripetal force. The answer is (3).

In uniform circular motion the acceleration, and therefore the net force, points toward the center of the circle (centripetal). For a car on a flat curve this inward force is supplied by friction between the tyres and the road. There is no outward force; the feeling of being thrown outward is the car's inertia, not a real force.

Related dot points

Sources & how we know this