Visighta

Rotational Kinematics

9.4 Arc Length, Tangential Speed, and Centripetal Acceleration

The Bridge Between Two Descriptions

A vinyl record plays at 33 1/3 RPM. Every point on the record --- the outer groove, the inner groove, the label --- sweeps through the same angle every second. The angular velocity is the same everywhere. Sections 9.1 through 9.3 built an entire kinematics around this angular description, mirroring the translational kinematics of Chapters 1 and 2.

But the needle at the outer edge of the record covers more distance per revolution than a needle near the center. It moves faster, in the ordinary translational sense. Same angular velocity, different translational speed. The angular description is universal across the body; the translational description depends on where you are.

How are these two descriptions connected? The answer is a single quantity that has been quietly waiting in the background: the radius. This section makes that connection explicit. The relationships you will find are not new physical laws --- they are pure geometry, consequences of what it means to move along a circular arc. But they are the essential bridge between the rotational world of Sections 9.1--9.3 and the translational world of Chapters 1--3.

Prediction

Before you read on: A point on the rim of a wheel has a tangential speed of 12 m/s. The wheel's radius is 0.3 m.

(a) What is the angular velocity of the wheel?

(b) What is the tangential speed of a point halfway between the rim and the center (at radius 0.15 m)?

Commit to numerical answers for both before continuing.

Exploration: Radius as the Converter

[Interactive: Spinning Wheel with Arc Readouts. A wheel of adjustable radius spins at an adjustable angular velocity $\omega$. A highlighted arc is drawn from a reference line to a marked point on the wheel, with the arc length $s$ displayed numerically. Three readouts are shown at the marked point's location: arc length $s$, tangential speed $v_t$, and centripetal acceleration $a_c$. The student can drag the marked point to any radius on the wheel (from the center to the rim). Two sliders control the radius $r$ and the angular velocity $\omega$ independently. As the student adjusts either slider, all three readouts update in real time. A second marked point at a different radius can be toggled on for comparison.]

Work through the following guided steps.

Step 1: Set the angular velocity to some moderate value and place the marked point at the rim. Note the tangential speed and centripetal acceleration. Now drag the marked point to half the radius. What happened to the tangential speed? What happened to the centripetal acceleration?

Step 2: Return the point to the rim. Now double the radius of the wheel while keeping $\omega$ fixed. What happens to the tangential speed? What happens to the centripetal acceleration?

Step 3: Turn on the second marked point and place it at a different radius. Both points share the same $\omega$ (they are on the same rigid wheel). Compare their tangential speeds and centripetal accelerations. What pattern do you see?

If you explored carefully, here is what you should have noticed:

  1. Tangential speed is proportional to radius. At fixed $\omega$, doubling the radius doubles $v_t$. Halving the radius halves $v_t$. The outer edge always moves fastest.

  2. Centripetal acceleration is also proportional to radius --- at fixed $\omega$. Doubling the radius doubles $a_c$. This may be surprising if you remember $a_c = v^2/r$ from Section 3.7, where larger $r$ meant smaller $a_c$. The difference is what you hold fixed: $\omega$ or $v$.

  3. Angular velocity is the same everywhere on the wheel. The two marked points always share the same $\omega$. The radius converts this shared angular quantity into different translational quantities at each location.

Concept Reveal: Three Geometric Relationships

Every relationship in this section follows from one foundational fact about circles: an arc of a circle subtends an angle equal to the arc length divided by the radius, provided the angle is measured in radians. This is the definition of the radian, the reason we insisted on radians back in Section 9.1.

Arc length

Consider a point at radius $r$ from the axis. When the wheel rotates through an angle $\theta$ (in radians), the point traces an arc. The length of that arc is:

$$s = r\theta$$

This is not a derived result --- it is essentially the definition of radians restated. One radian is the angle that cuts an arc equal to the radius. Two radians cut an arc equal to twice the radius. The relationship is perfectly linear.

For two points at different radii on the same wheel, both experience the same $\theta$, but the one at larger $r$ traces a longer arc. The radius acts as a scaling factor, converting a shared angular displacement into a radius-dependent arc length.

Tangential speed

Now take a time derivative. If $s = r\theta$ and $r$ is constant (the point stays at the same radius on a rigid body), then:

$$\frac{ds}{dt} = r\frac{d\theta}{dt}$$

The left side is the rate at which arc length accumulates --- this is the speed along the circular path, the tangential speed $v_t$. The right side contains $d\theta/dt$, which is the angular velocity $\omega$. So:

$$v_t = r\omega$$

This is the bridge equation. It says: the translational speed of a point on a spinning body equals the radius times the angular velocity. Every point on the wheel shares the same $\omega$, but the tangential speed scales linearly with $r$. The outer edge moves fastest. The center ($r = 0$) does not move at all.

Centripetal acceleration

A point moving at speed $v_t$ along a circular path of radius $r$ is constantly changing direction. Section 3.7 showed, through the velocity-vector-change construction, that this requires an inward acceleration of magnitude:

$$a_c = \frac{v_t^2}{r}$$

Now substitute $v_t = r\omega$:

$$a_c = \frac{(r\omega)^2}{r} = r\omega^2$$

This gives a second form of the centripetal acceleration --- expressed in terms of the angular velocity rather than the tangential speed:

$$\boxed{a_c = \frac{v_t^2}{r} = r\omega^2}$$

Both forms are correct. They are the same quantity written in two different languages. But they lead to opposite-sounding statements about how $a_c$ depends on $r$, which is a source of real confusion. We will confront that directly in the practice problems.

Summary of the bridge equations

Angular quantity Bridge Translational quantity
$\theta$ (rad) $s = r\theta$ $s$ (m)
$\omega$ (rad/s) $v_t = r\omega$ $v_t$ (m/s)
$\omega$ (rad/s) $a_c = r\omega^2$ $a_c$ (m/s$^2$)

These are geometric identities, not dynamical laws. They do not tell you why something spins or what force is responsible. They tell you how to convert between two equivalent descriptions of the same motion. The angular description is shared by the whole body; the translational description varies with radius. The radius is the conversion factor.

Return to the Prediction

Now revisit your prediction. A point on the rim has $v_t = 12$ m/s at $r = 0.3$ m.

From $v_t = r\omega$:

$$\omega = \frac{v_t}{r} = \frac{12}{0.3} = 40 \text{ rad/s}$$

This angular velocity belongs to the entire wheel. At half the radius ($r = 0.15$ m):

$$v_t = r\omega = (0.15)(40) = 6 \text{ m/s}$$

Half the radius, half the tangential speed. The angular velocity is unchanged; only the translational quantity depends on where you look.

Connection: Section 3.7 Revisited

In Section 3.7, you derived $a_c = v^2/r$ from the geometry of a rotating velocity vector. You drew adjacent velocity vectors, placed them tail-to-tail, and found that the change $\Delta\vec{v}$ always pointed inward. The derivation used arc-length reasoning on a circle of radius $v$ (the speed circle in velocity space), arriving at $a_c = v^2/r$.

Here you see the same result through a different lens. The rotational description gives $a_c = r\omega^2$ directly, and substituting $v_t = r\omega$ recovers $a_c = v_t^2/r$. The two derivations are not independent --- they are the same geometry viewed from two vantage points. Section 3.7 worked entirely in translational language. This section works in rotational language and converts at the end. They must agree, and they do.

The fact that you can arrive at the same formula by two routes is not a coincidence. It is a sign that the geometry is robust --- the centripetal acceleration formula is locked in by the structure of circular motion itself, regardless of which language you use to describe it.

Spaced Retrieval

Before moving to practice, test your recall of earlier material.

Recall prompt 1: What is the direction of centripetal acceleration, and why? Trace the answer back to the velocity-vector-change construction in Section 3.7.

Recall prompt 2: In Section 9.1, why did we insist on measuring angles in radians rather than degrees? How does that choice show up in the bridge equations of this section?

Recall prompt 3: Section 9.2 gave the kinematic equation $\omega = \omega_0 + \alpha t$ for constant angular acceleration. If you multiply both sides by $r$, what translational equation do you get?

Practice

Layer 1: Concrete -- Converting Between Angular and Translational Quantities

Problem 1. A bicycle wheel has a radius of 0.35 m and rotates at 10 rad/s. Find the tangential speed and the centripetal acceleration of a point on the rim.

Check your answer $$v_t = r\omega = (0.35)(10) = 3.5 \text{ m/s}$$ $$a_c = r\omega^2 = (0.35)(10)^2 = (0.35)(100) = 35 \text{ m/s}^2$$ Alternatively, $a_c = v_t^2/r = (3.5)^2/0.35 = 12.25/0.35 = 35$ m/s$^2$. Both forms give the same answer, as they must.

Problem 2. A ceiling fan blade is 0.8 m long and spins at 120 RPM. What is the tangential speed of the tip of the blade? What is the tangential speed of a point 0.2 m from the center?

Check your answer First convert to rad/s: $\omega = 120 \text{ RPM} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = 4\pi$ rad/s $\approx 12.57$ rad/s. At the tip ($r = 0.8$ m): $$v_t = (0.8)(4\pi) = 3.2\pi \approx 10.1 \text{ m/s}$$ At $r = 0.2$ m: $$v_t = (0.2)(4\pi) = 0.8\pi \approx 2.51 \text{ m/s}$$ The tip moves four times faster than the point at one-quarter the radius. Both share the same $\omega$; the radius makes the difference.

Layer 2: Pattern -- Comparing Points on the Same Wheel

Problem 3. A merry-go-round spins at constant angular velocity. Child A sits at the outer edge, at radius $R$. Child B sits at radius $R/3$. Compare their tangential speeds and centripetal accelerations.

Check your answer Both children share the same $\omega$. Tangential speed: $v_A = R\omega$ and $v_B = (R/3)\omega = v_A/3$. Child A moves three times faster. Centripetal acceleration: $a_A = R\omega^2$ and $a_B = (R/3)\omega^2 = a_A/3$. Child A experiences three times the centripetal acceleration. At constant $\omega$, both $v_t$ and $a_c$ scale linearly with radius. The outer edge always has the largest values of both.

Problem 4. Two wheels spin at the same angular velocity $\omega$. Wheel 1 has radius $R$; Wheel 2 has radius $2R$. For a point on the rim of each wheel, compare the tangential speed and the centripetal acceleration.

Check your answer Tangential speed: $v_1 = R\omega$, $v_2 = 2R\omega = 2v_1$. Wheel 2's rim point moves twice as fast. Centripetal acceleration: $a_1 = R\omega^2$, $a_2 = 2R\omega^2 = 2a_1$. Wheel 2's rim point has twice the centripetal acceleration. Again, at fixed $\omega$, everything scales with $r$.

Layer 3: Structure -- The Radius Puzzle

Problem 5. Here is a question that confuses many students: centripetal acceleration increases with $r$ when $\omega$ is fixed, but decreases with $r$ when $v$ is fixed. How can the same quantity go in opposite directions?

Check your answer The resolution is that you cannot change the radius while keeping *both* $\omega$ and $v$ fixed. The constraint $v = r\omega$ means that if you change $r$, at least one of the other two must change. **Case 1: Fixed $\omega$ (same wheel, different radii).** Then $v = r\omega$ increases with $r$. The centripetal acceleration is $a_c = r\omega^2$, which increases with $r$. Farther from the center means faster tangential speed, which means more direction change per unit time, which means more acceleration. **Case 2: Fixed $v$ (different situations engineered to have the same speed).** Then $\omega = v/r$ decreases with $r$. The centripetal acceleration is $a_c = v^2/r$, which decreases with $r$. At the same speed, a gentler curve (larger $r$) requires less inward acceleration to maintain the path. The two cases describe physically different setups. In Case 1, you are comparing different locations on one spinning body. In Case 2, you are comparing different circular paths that happen to be traversed at the same speed. There is no contradiction --- just two different questions with two different answers.

Layer 4: Debug -- Choosing the Right Velocity

Problem 6. A ball of radius 0.05 m rolls without slipping across a floor at a center-of-mass speed of 2 m/s. A student wants to find the centripetal acceleration of a point on the ball's surface (relative to the ball's center). The student writes:

$$a_c = \frac{v^2}{r} = \frac{(2)^2}{0.05} = 80 \text{ m/s}^2$$

Is this correct? When does using the center-of-mass velocity in $a_c = v^2/r$ give the wrong answer?

Check your answer The student used $v = 2$ m/s, but this is the velocity of the center of mass relative to the ground. The relevant velocity for computing the centripetal acceleration of a point on the surface *relative to the ball's center* is the tangential speed of that point in the rotating frame --- which is $v_t = r\omega$. For rolling without slipping, $v_{\text{cm}} = R\omega$, so $\omega = v_{\text{cm}}/R = 2/0.05 = 40$ rad/s. The tangential speed of a surface point relative to the center is: $$v_t = R\omega = (0.05)(40) = 2 \text{ m/s}$$ In this particular case, $v_t$ happens to equal $v_{\text{cm}}$ because for rolling without slipping, the tangential speed of a rim point equals the center-of-mass speed (this is the rolling constraint from Section 9.6). So the student's numerical answer of 80 m/s$^2$ is actually correct --- by coincidence of the constraint. But the reasoning is fragile. If the ball were spinning freely (not rolling), or if it were rolling on a surface that itself moves, the center-of-mass velocity and the tangential speed of the rim point would differ. In general, $a_c = v^2/r$ requires $v$ to be the tangential speed of the point in question relative to the center of its circular path, not the velocity of the object's center of mass. Using the wrong $v$ gives the wrong $a_c$ whenever the two velocities differ.

Reflection

What role does radius play in connecting the rotational and translational worlds?

In Sections 9.1--9.3, every quantity was angular: $\theta$, $\omega$, $\alpha$. These quantities are the same for every point on a rigid body --- they describe the body as a whole. In Chapters 1--3, every quantity was translational: $s$, $v$, $a$. These describe the motion of individual points.

The radius is the bridge between these two worlds. Multiply an angular quantity by $r$, and you get the corresponding translational quantity at that particular radius. This is not a deep physical law --- it is geometry, built into the definition of the radian. But it is the relationship that makes angular descriptions useful for predicting what actually happens at specific locations on a rotating body.

Think about why a vinyl record sounds the same whether the needle is near the outer edge or near the center. The angular velocity is the same, so the groove pattern passes at the same angular rate. But the tangential speed is different --- and the groove spacing must be adjusted to compensate. The geometry of $v_t = r\omega$ is not just a formula; it is a design constraint that record engineers had to respect.

Looking Ahead

You now have two parallel sets of kinematic tools --- translational and rotational --- and the bridge equations that connect them through the radius. Section 9.5 consolidates this work into a complete analogy table, laying out every translational-rotational pair side by side. More importantly, it asks where the analogy breaks down. The translational-rotational parallel is a powerful scaffold, but like all scaffolds, it has edges. Knowing where those edges are is just as important as knowing how to use the scaffold itself.