Identify the force that bends a path

Name the interaction supplying the inward force

A particle moves around a horizontal circle on an ideal frictionless support. An ideal cord joins it to the center. In this model, cord tension is the horizontal interaction pulling inward. The upward support and downward weight balance vertically, outside the top-view diagrams.

The phrase centripetal force describes the inward net force required for the curved motion. It is not an additional kind of interaction to draw beside tension. For this setup, the entire horizontal inward force is the tension. Adding “tension plus centripetal force” would count the same role twice.

Other situations can supply inward force through gravity, friction, or several forces together. Identify the actual interactions and sum their inward components. A curved trajectory alone does not tell you which physical interaction supplies that sum. The OpenStax centripetal-force treatment discusses this connection to Newton’s second law.

Calculate the required force before judging feasibility

Uniform circular motion has inward acceleration a = v²/r, where v is speed and r is radius. Newton’s second law then requires inward net force F = ma = mv²/r. The investigation fixes mass m = 2 kg and lets you choose speed and radius independently for a prepared state.

At v = 2 m/s and r = 2 m, acceleration is 2 m/s² and required force is 2 × 2 = 4 N. The numerical force unit is a newton: 1 N = 1 kg·m/s². At the same radius and twice the speed, the required force is 2 × 4²/2 = 16 N. Twice the speed needs four times the force, not twice.

This is the force required by the specified motion. A real cord cannot necessarily supply it: a strength limit, stretch, slackness or attachment failure could invalidate the assumed circle. This model supplies the required force without a strength limit. Selecting a control value is not evidence that a physical device can safely achieve it, and no physical spinning apparatus is required for this lesson.

State what stays fixed when comparing radii

Keep speed at 2 m/s and double radius from 2 to 4 m. Acceleration drops from 2 to 1 m/s², and required force drops from 4 to 2 N. At the same speed, a broader turn needs a smaller change of velocity direction per second. The period doubles because the path is twice as long.

Now compare (speed, radius) pairs (2 m/s, 2 m) and (4 m/s, 4 m). Both have angular rate ω = v/r = 1 rad/s and the same period, about 6.283 s. Their accelerations are 2 and 4 m/s², so required forces are 4 and 8 N. At fixed angular rate, the larger circle requires a greater speed and greater inward force.

Both comparisons are consistent. Substituting v = ωr into F = mv²/r gives F = mω²r. At fixed speed, force varies inversely with radius; at fixed angular rate, it varies directly with radius. The words “larger radius needs less force” are incomplete unless the held-fixed quantity is stated.

The controls choose separate prepared steady states. Changing radius does not simulate reeling in the cord while the particle is already moving, and changing speed does not model the force or energy needed to accelerate it into the selected orbit. A real change during motion requires additional analysis.

Release preserves tangent velocity

The displayed release comparison starts at the rightmost point, position (r, 0), with velocity (0, v). Remove the cord without an impulse while keeping the frictionless horizontal support. The inward force vanishes, so horizontal acceleration becomes zero. The particle continues along the tangent with its current velocity.

After one second, position is (r, v × 1 s). In the default state, it is (2, 2) m with velocity (0, 2) m/s. This is not an outward radial launch to (4, 0) m. Its distance from the original center grows, but that fact does not mean a new outward force was applied. A straight tangent path naturally moves farther from the center.

The release comparison and the retained-cord circle are different physical conditions. The circular-motion formula does not continue to require a mysterious inward force after the specified circular trajectory has ceased. The support remains in place, so this is not a freely falling projectile in a vertical plane.

Keep force, speed and energy roles distinct

During ideal uniform circular motion, the inward force is perpendicular to instantaneous velocity. It changes direction without changing speed. The force’s instantaneous mechanical power is zero because there is no component along the velocity. This does not mean the force is zero, and it does not imply that real bearings or air drag consume no energy.

A tangential component of net force would change speed. If both inward and tangential acceleration occur, v²/r describes only the instantaneous inward part; it is not the total acceleration magnitude. The present investigation intentionally holds speed constant and does not simulate that more general case.

Transfer: A 0.5 kg particle moves at 4 m/s on a 2 m circle. An ideal radial connection can supply at most 6 N inward. Is the specified uniform circle within that limit? What force would be required at 6 m/s, and what is the maximum speed permitted by the stated limit?

Check the inward-force transfer

At 4 m/s the required force is 0.5 × 16 / 2 = 4 N, within the 6 N limit. At 6 m/s it is 0.5 × 36 / 2 = 9 N, beyond the limit. Solving 6 = 0.5v²/2 gives v² = 24 and v ≈ 4.899 m/s. This ideal force-capacity calculation does not certify a real connection or include a safety factor, transients or other failure modes.