FORCES & CIRCULAR MOTION

Circular Motion Laboratory

Follow the circle. Release the string. See where the puck goes.

A frictionless table · top view

Ready

Playback

→ v · velocity (m/s)→ T · tension (N)→ a · acceleration (m/s²)⇢ vₓ, vᵧ (m/s)○ release point · pale trail · dashed circle

+x right, +y up; angles and angular velocity are positive counterclockwise. Puck size is schematic. Arrows use separate scales; velocity components share v’s scale. Inward arrows are offset for clarity. Zero vectors have no arrow. Step = 0.05 s.

POSITION (x, y)

SPEED

ANGULAR VELOCITY

TENSION

ACCELERATION

ELAPSED TIME

Motion & tension

Last 30 s · time (s) →

Solid: vₓ · dashed: vᵧ · velocity (m/s)

Tension T (N) · dashed gray: string limit

What the string supplies

Centripetal force is the net inward force required for circular motion. Here, tension supplies it. There is no additional centripetal force or outward force in the table’s stationary frame.

TRY A CHALLENGE

Find the fastest safe circle.

With mass, radius and string strength fixed, find the greatest speed the string supports.

Complete one revolution without breaking, with required tension at 98–100% of the limit. Only speed is adjustable.

Adjust speed, then run a full revolution.

Why?

Constant speed, yet accelerating?

Velocity includes direction. Its magnitude stays constant while its direction changes continuously. That change is acceleration.

Why tangent velocity and inward acceleration?

Velocity follows the instantaneous direction of travel, tangent to the circle. Tension bends the path toward the anchor, producing inward acceleration. Gravity and the table’s normal force cancel vertically.

Why does twice the speed need four times the tension?

T = mv²/r depends on speed squared. Doubling v multiplies v², acceleration and required tension by four.

What does a larger radius change?

At the same speed, the direction changes less sharply on a larger circle. Doubling r halves a = v²/r and T = mv²/r.

What does a heavier puck change?

At fixed speed and radius, acceleration v²/r is unchanged. A heavier puck needs proportionally more tension to achieve that acceleration: T = ma.

Why does release send the puck tangentially?

Ideal release applies no impulse, so position and velocity are continuous. Without tension, the net horizontal force and acceleration are zero. The puck keeps its release velocity in a straight line.

Why does an overloaded string end the circle?

When required tension exceeds the limit, automatic breaking removes the string immediately. The inward force disappears and circular motion ends. An exact match to the limit remains intact. Turning automatic breaking off models an unbreakable ideal constraint, even above the gauge limit.

Why is kinetic energy constant?

Tension is perpendicular to velocity and does no work: T · v = 0. It changes direction without changing speed, so Eₖ = ½mv² stays constant. After release there is no horizontal force and the same energy is retained.