Unit 2 · Topic 2.9 Beta

Circular Motion

Circular motion at constant speed is accelerated motion: a_c = v²/r toward the center, with v = 2πr/T.

Practice 1: Creating RepresentationsPractice 2: Mathematical RoutinesPractice 3: Scientific Questioning and Argumentation

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Question set for this topic

Part 1 · Hook

Why this matters

A car takes a curve, a planet circles the Sun, a bucket of water whirls overhead without spilling. Each object moves at a steady speed yet is accelerating, because its direction keeps changing. Something has to pull it toward the center. Finding that something, among the real forces, is the whole skill.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. An object's speed stays the same but its direction changes. It is

  1. accelerating
  2. in equilibrium
  3. at constant velocity
  4. unaffected by any net force
Show the answer

Any change in velocity, size or direction, is an acceleration.

  • Correct: accelerating:
  • in equilibrium:
  • at constant velocity:
  • unaffected by any net force:

2. An object with nonzero net force on it

  1. accelerates in the direction of the net force
  2. moves in the direction of the net force
  3. keeps its velocity
  4. slows to a stop
Show the answer

Newton's second law: a = ΣF/m, along ΣF.

  • Correct: accelerates in the direction of the net force:
  • moves in the direction of the net force:
  • keeps its velocity:
  • slows to a stop:

3. The gravitational force between two masses is

  1. Gm₁m₂/r²
  2. Gm₁m₂/r
  3. mg at any distance
  4. zero in orbit
Show the answer

Newton's law of universal gravitation.

  • Correct: Gm₁m₂/r²:
  • Gm₁m₂/r:
  • mg at any distance:
  • zero in orbit:

Part 4 · See it

See it first

A car on a flat curve seen from above, driving along the road, with velocity tangent to the road and acceleration toward the center of the curve. Beside it, the free-body diagram seen from behind: F_N, road on car, up and F_g, Earth on car, down, equal; F_f, road on tires, horizontal toward the center and shorter than the weight. Static friction supplies the centripetal net force; no separate centripetal arrow is drawn.
Velocity is tangent, acceleration points to the center, and static friction is the real force that turns the car. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. An object moves in a circleits velocity direction changes continuously, so it accelerates even at constant speed
  2. The speed is v and the radius rthe centripetal acceleration points to the center with size a_c = v²/r
  3. Newton's second law applies along the radiusthe real forces must add to a net force toward the center of size mv²/r
  4. That net force disappears (the string breaks)the object moves off in a straight line along the tangent

Part 6 · Key ideas

Key ideas

  • Uniform circular motion: constant speed v = 2πr/T on a circle; the acceleration ac = v²/r points to the center. The period T is the time for one trip; the frequency f = 1/T, in Hz.
  • The centripetal net force, ΣF toward the center = mv²/r, is supplied by real forces: tension, friction, normal force, gravity, or their parts. Never draw it as an extra arrow.
  • Speeding up or slowing on a circle adds a tangential acceleration; the total is the vector sum.
  • Circular orbits: GMm/r² = mv²/r gives v = √(GM/r) and Kepler's third law, T² = (4π²/GM)r³.

Part 7 · Misconception

A common mistake

The wrong idea: An object moving in a circle has a centripetal force pulling it in and a centrifugal force pushing it out, and both go on the free-body diagram.

What actually happens: Only real forces go on the diagram. "Centripetal" names the job of the net force toward the center; there is no outward force. A released object leaves along the tangent, not outward.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Diagram

A puck on a string

A 0.40 kg puck slides in a horizontal circle of radius 0.50 m on a frictionless table, held by a light string tied to a peg at the center. It makes 2.0 revolutions each second at a constant speed.

pegvtop view of africtionless table

1. What is the speed of the puck, in m/s?

Type a number in m/s.

Show the answer

The period is T = 1/f = 1 ÷ 2.0 s⁻¹ = 0.50 s. In one period the puck goes once around: v = 2πr ÷ T = 2π(0.50 m) ÷ 0.50 s = 6.3 m/s.

  • Answer: 6.3 m/s

2. What is the tension in the string, in newtons?

Type a number in N.

Show the answer

The string's pull is the only horizontal force, so it supplies the centripetal net force: F_T = mv²/r = (0.40 kg)(6.28 m/s)² ÷ 0.50 m = 32 N (31.6 N).

  • Answer: 32 N

3. The string breaks at the moment shown in the diagram. Which path does the puck follow next, seen from above?

  1. A straight line along the velocity arrow
  2. A straight line outward, along the line from the peg
  3. It keeps moving around the same circle
  4. A curve that spirals slowly outward
Show the answer

With the string gone, no horizontal force acts on the puck. By the first law it keeps its velocity at that instant, which is tangent to the circle, and moves in a straight line.

  • Correct: A straight line along the velocity arrow: Right: zero net force, constant velocity, along the tangent.
  • A straight line outward, along the line from the peg: Nothing pushes it outward; there is no outward force. It only looks like it "flies out" because the circle curves away from its straight path.
  • It keeps moving around the same circle: Going around a circle needs a net force toward the center; without the string there is none.
  • A curve that spirals slowly outward: Any curved path needs a sideways force; with none, the path is straight.

4. Which free-body diagram of the puck, drawn from the side, is correct?

  1. F_T toward the peg, F_N up and F_g down
  2. F_T toward the peg, a centripetal force toward the peg, F_N up and F_g down
  3. F_T toward the peg, an outward force, F_N up and F_g down
  4. F_T toward the peg, a forward force along v, F_N up and F_g down
Show the answer

The puck touches the string and the table, and Earth pulls on it: three forces. The string's pull is the centripetal net force; "centripetal" names a role, not an extra force.

  • Correct: F_T toward the peg, F_N up and F_g down: Right: only real forces with agents.
  • F_T toward the peg, a centripetal force toward the peg, F_N up and F_g down: This counts the string's pull twice. The centripetal net force IS the tension here; it is not a separate arrow.
  • F_T toward the peg, an outward force, F_N up and F_g down: There is no outward force with an agent; the puck tends to go straight, which is not a force.
  • F_T toward the peg, a forward force along v, F_N up and F_g down: Nothing pushes the puck forward; at constant speed no force acts along the velocity.

Diagram

A car over a hill

A 1200 kg car drives over the crest of a rounded hill at a constant speed of 12 m/s. Near the crest the road follows a vertical circle of radius 45 m. Use g = 9.8 m/s².

vcrest: part of a vertical circle

5. What is the normal force of the road on the car at the crest, in newtons?

Type a number in N.

Show the answer

At the crest the center of the circle is below the car, so the net force must point down: mg − F_N = mv²/r. F_N = m(g − v²/r) = 1200 kg × (9.8 − 144/45) m/s² = 1200 kg × 6.6 m/s² = 7900 N (7920 N).

  • Answer: 7.9 × 103 N

6. A car rounds a flat, unbanked curve at a constant speed. Which force supplies the centripetal net force?

  1. Static friction from the road on the tires
  2. Kinetic friction from the road on the tires
  3. The normal force from the road
  4. The engine's forward push
Show the answer

The tires do not slide sideways, so the road's sideways grip is static friction, pointing toward the center of the curve. It is the only horizontal force pointing inward.

  • Correct: Static friction from the road on the tires: Right: static friction toward the center.
  • Kinetic friction from the road on the tires: Kinetic friction would mean the tires skid sideways.
  • The normal force from the road: On a flat road the normal force is vertical; it balances the weight.
  • The engine's forward push: A forward force along the road cannot turn the car; the turning force must point sideways, toward the center.

Part 9 · Summary

Summary

Circular motion at constant speed is accelerated motion: a_c = v²/r toward the center, with v = 2πr/T. Real forces (tension, friction, normal force, gravity) must add to a net force mv²/r toward the center; for circular orbits this gives v = √(GM/r) and T² ∝ r³.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections