CLASSICAL-MECHANICS · Interactive Practice | Unit 5 · Video 1

Why Circular Motion Always Accelerates Inward

IKey Formulas

Formula	Name	Description
$\vec{v} \perp \vec{r}$	Tangential velocity	Velocity is always tangent to the circle (perpendicular to the radius)
$\vec{a} = \dfrac{d\vec{v}}{dt}$	Acceleration	Rate of change of the velocity vector (changes if length or direction changes)
$\vec{a} = \vec{a}_r + \vec{a}_t$	Acceleration split	A centripetal (radial, inward) part and a tangential part
$F_r = m\,a_r$	Radial Newton's 2nd Law	Inward (centripetal) force produces the inward (centripetal) acceleration

For constant speed, the tangential part vanishes and only the inward centripetal acceleration remains.

IIInteractive Visualizations

Visualization 1 — Velocity Direction Never Stops Turning

At constant speed the velocity keeps its length but changes direction at every point on the circle.

Position θ 45°

💡 A vector whose direction changes is itself a changing vector — so even at perfectly constant speed, a turning velocity means there is an acceleration.

Visualization 2 — Change in Velocity Points Inward

Subtracting two nearby velocities gives a change vector $\Delta\vec{v} = \vec{v}_2 - \vec{v}_1$ that aims straight at the center.

Angular gap 60°

💡 This inward change in velocity is the centripetal acceleration (Latin for center-seeking).

Visualization 3 — Splitting Acceleration into Two Parts

Any acceleration in circular motion splits into an inward centripetal part and an along-motion tangential part.

Tangential aₜ 1/2

💡 An inward acceleration always demands an inward net force: $F_r = m\,a_r$ .

IIIQuiz Questions

Question 1

A ball is whirled on a string at a perfectly constant speed in a horizontal circle. At any instant, in which direction does its velocity vector point?

✅ Correct! Velocity rides tangent to the circle at every instant.

❌ Not quite. Constant speed is not zero speed — the ball is still moving, so its velocity is nonzero.

❌ Not quite. Radial directions describe acceleration/force, not the velocity. The velocity is tangent.

Show solution

Solution:

The velocity is always tangent to the circle, pointing along the direction of travel at each instant. If you cut the string, the ball flies off along this tangent line — not along the radius. Constant speed means the velocity's length is fixed, but its direction still points tangentially and keeps changing as the ball goes around.

Question 2

An object moves around a circle at constant speed. A classmate says: "Constant speed means the velocity is constant, so the acceleration must be zero."

Is this reasoning correct?

✅ Correct! Constant speed but changing direction still means nonzero acceleration.

❌ Not quite. Speed (a scalar) is constant, but velocity (a vector) is not — its direction changes.

Show solution

Solution:

The statement is false. Acceleration is the rate of change of the velocity vector, and a vector changes if either its length or its direction changes. Here the length (speed) is constant, but the direction is continuously turning. A changing velocity is an acceleration — the centripetal acceleration, pointing toward the center.

Question 3

You build the change-in-velocity vector $\Delta\vec{v} = \vec{v}_2 - \vec{v}_1$ for two nearby instants of circular motion at constant speed. As the time interval shrinks toward zero, in which direction does $\Delta\vec{v}$ (and hence the acceleration) point?

✅ Correct! Delta v points to the center — that is the centripetal acceleration.

❌ Not quite. That is the common 'centrifugal' intuition, but the actual acceleration points inward.

❌ Not quite. Build the subtraction triangle: delta v ends up pointing inward, not along the tangent.

Show solution

Solution:

Place $\vec{v}_1$ and $\vec{v}_2$ tail-to-tail and draw the connecting arrow from the tip of $\vec{v}_1$ to the tip of $\vec{v}_2$ — that is $\Delta\vec{v}$ . Translating it back onto the circle shows it pointing toward the center. This inward acceleration is the centripetal acceleration (center-seeking). The outward feeling of being "thrown out" is not an acceleration of the object — the real acceleration is inward.

Question 4

A car drives around a circular track and is also speeding up as it goes. Which statement best describes its acceleration, and what does the radial component of Newton's Second Law say?

✅ Correct! Speeding up adds a tangential part; the radial component still gives F_r = m a_r.

❌ Not quite. That is the constant-speed case. Speeding up also adds a tangential acceleration.

❌ Not quite. Turning always needs an inward (centripetal) acceleration, and speeding up adds a tangential one.

Show solution

Solution:

When the car both turns and speeds up, the acceleration has two perpendicular parts:

a centripetal part pointing inward (it changes which way the car goes), and
a tangential part pointing along the motion (it changes how fast it goes).

Newton's Second Law is a vector equation, so it holds component by component. Projecting onto the radial (inward) direction gives the boxed result:

$F_r = m\,a_r$

The inward (radial) force equals the mass times the centripetal acceleration. An inward acceleration always demands an inward (centripetal) force.

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