Visighta

Rotational Kinematics

9.6 Rolling Kinematics and Constraints Without Slipping

Three Velocities, One Object

A wheel rolls across the floor. It does not skid, it does not slide --- it rolls, the way a bicycle wheel rolls down a road or a bowling ball rolls down a lane.

Watch three points on that wheel. The center moves forward at some steady speed. The top of the wheel moves forward at twice the center's speed. And the bottom --- the contact point, the very spot where rubber meets road --- has zero velocity. It is, for that instant, perfectly still.

One rigid object. Three different velocities at three different points. No part of the wheel has separated from any other part. It is all one piece of metal or rubber, rotating and translating simultaneously. How can a single object have points moving at such wildly different speeds?

The answer is that rolling motion is not pure translation and not pure rotation. It is both, happening at the same time, and the velocities at each point are the sum of these two contributions. This section is about the constraint that ties these two motions together --- a constraint so tight that knowing one completely determines the other.

Prediction

Before you read on: A ball rolls without slipping across a flat surface. The center of the ball moves at speed $v$. What is the velocity of the topmost point of the ball?

(a) $0$ (b) $v$ (c) $2v$

Commit to your answer and your reasoning before continuing.

The Guiding Question

Why does rolling connect linear and angular motion so tightly?

In Sections 9.1 through 9.5, you built the full rotational kinematics framework and explored the analogy between translational and rotational quantities. You learned that $v_t = r\omega$ connects the tangential speed of a point on a rotating body to the angular velocity. But that was a geometric relationship --- it described the speed of a point at radius $r$ on a spinning object.

Rolling introduces something new: a physical constraint. When a wheel rolls along a surface without slipping, translation and rotation are not independent. You cannot spin the wheel faster without also making it move forward faster. You cannot push the center forward faster without also making the wheel spin faster. One motion determines the other, completely. Where does this coupling come from?

Exploration: Watching a Wheel Roll

[Interactive: Rolling Wheel in Slow Motion. A wheel of radius $R$ rolls to the right across a flat surface without slipping. The animation runs in slow motion. Three points are highlighted on the wheel: the center, the topmost point, and the contact point (bottom). At each point, a velocity vector arrow is drawn and updates in real time as the wheel rolls.

The center's velocity vector is a horizontal arrow of constant length. The top's velocity vector is a longer horizontal arrow --- twice the length of the center's. The contact point's velocity vector shrinks to zero at the instant it touches the ground, then grows again as it lifts away.

A "trace" toggle draws the path of a single marked point on the rim as the wheel rolls. This path is a cycloid --- a series of arches, with cusps at the ground.

Guided prompts:

  • "Focus on the contact point. At the instant it touches the ground, what is its velocity? Watch the arrow."
  • "Now focus on the top point. How does its velocity compare to the center's?"
  • "Toggle the trace. What shape does a point on the rim trace out? What happens at the cusps?"
  • "Try increasing the wheel's speed. Do the ratios between the velocities at the three points change, or do they stay the same?"]

Spend time with this visualization, especially the contact point. At the instant a point on the rim touches the ground, its velocity is zero. Not small. Not approximately zero. Exactly zero. And then, as the wheel continues to roll, that point lifts away from the ground and begins to move again, arching upward and forward along the cycloid path until it reaches the top of the wheel, where it travels at twice the center's speed, and then sweeps back down toward the ground to become momentarily still once more.

Pause and think: The contact point has zero velocity, but the wheel is clearly moving. How can part of a moving object be at rest?

The key is to think about what "zero velocity" means for the contact point. That point has two contributions to its motion: the forward translation of the entire wheel, and the backward rotation of the rim at the bottom of the wheel. At the contact point, these two contributions are equal and opposite. They cancel exactly. The forward translation carries the point to the right at $v_{\text{cm}}$. The rotation, at that instant, sweeps the point to the left at the same speed. The result is zero.

Concept Reveal: The No-Slip Constraint

Decomposing rolling motion

The velocity of any point on a rolling wheel can be understood as the sum of two parts:

  1. Translation: Every point on the wheel shares the center-of-mass velocity $\vec{v}_{\text{cm}}$, directed forward.

  2. Rotation about the center: Every point on the wheel also has a velocity due to rotation about the center, with magnitude $r\omega$ (where $r$ is the distance from the center to that point), directed tangent to the circle at that point.

The total velocity of any point is the vector sum of these two contributions.

At the center of the wheel, the rotational contribution is zero (the center is on the axis), so the total velocity is simply $v_{\text{cm}}$ forward.

At the top of the wheel, the rotational velocity points forward (tangent to the circle, which is horizontal and in the direction of motion at the top). The total velocity is $v_{\text{cm}} + R\omega$ forward.

At the contact point (bottom), the rotational velocity points backward --- opposite to the direction of travel. The total velocity is $v_{\text{cm}} - R\omega$.

The constraint

"Rolling without slipping" means the contact point is instantaneously at rest. The surface does not skid against the ground. Setting the contact point's velocity to zero:

$$v_{\text{cm}} - R\omega = 0$$

$$\boxed{v_{\text{cm}} = R\omega}$$

This single equation is the no-slip constraint. It locks the translational speed of the center of mass to the angular velocity of the wheel. If you know one, you know the other. There is no freedom to choose them independently.

Consequences

With $v_{\text{cm}} = R\omega$ in hand, the velocities at the three special points become:

Point Translational part Rotational part Total velocity
Center $v_{\text{cm}}$ $0$ $v_{\text{cm}}$
Top $v_{\text{cm}}$ $+R\omega = v_{\text{cm}}$ $2v_{\text{cm}}$
Contact (bottom) $v_{\text{cm}}$ $-R\omega = -v_{\text{cm}}$ $0$

The topmost point moves at $2v_{\text{cm}}$. The contact point is at rest. These are not approximations --- they are exact results of the no-slip condition.

Return to the prediction

Go back to your prediction. The topmost point of a ball rolling without slipping at speed $v$ has velocity $2v$. The answer is (c). If this surprised you, trace through the decomposition one more time: the forward translation gives $v$, and the rotation at the top also gives $v$ in the forward direction. The two add.

Velocity of an arbitrary point

For a point on the rim at angle $\theta$ measured from the contact point (counterclockwise), the velocity components are:

$$v_x = v_{\text{cm}} + R\omega \sin\theta = v_{\text{cm}}(1 + \sin\theta)$$

$$v_y = R\omega \cos\theta = v_{\text{cm}} \cos\theta$$

The speed at that point is:

$$v = v_{\text{cm}}\sqrt{(1 + \sin\theta)^2 + \cos^2\theta} = v_{\text{cm}}\sqrt{2(1 + \sin\theta)}$$

At $\theta = 0$ (contact point): $v = v_{\text{cm}}\sqrt{2(1 + 0)} = v_{\text{cm}}\sqrt{2} \cdot 0$ --- wait. Let us be careful. At $\theta = 0$, measured from the bottom, $\sin\theta = 0$ and $\cos\theta = 1$, but the rotational velocity at the bottom points backward, not forward. The formula above uses $\theta$ measured from the bottom going counterclockwise (so the top is at $\theta = \pi/2$). With the correct sign conventions:

At $\theta = -\pi/2$ (the bottom, where rotation is backward): $v_x = v_{\text{cm}} - R\omega = 0$.

At $\theta = \pi/2$ (the top): $v_x = v_{\text{cm}} + R\omega = 2v_{\text{cm}}$.

The key insight is not the formula but the principle: decompose into translation plus rotation, then add as vectors.

The instantaneous axis of rotation

There is an elegant alternative way to see rolling. Instead of decomposing the motion into translation plus rotation about the center, you can describe the entire motion as pure rotation about the contact point. At every instant, the wheel is rotating about the point where it touches the ground. This point is called the instantaneous axis of rotation (or instantaneous center of rotation in 2D).

From this viewpoint, the center is at distance $R$ from the instantaneous axis and moves at speed $R\omega$. The top is at distance $2R$ and moves at speed $2R\omega = 2v_{\text{cm}}$. The contact point is at distance zero and has speed zero. All three results emerge immediately, without needing to add two velocities.

Connection: From Geometry to Constraint

In Section 9.4, you learned that $v_t = r\omega$ connects the tangential speed of a point on a spinning body to its angular velocity. That was a geometric relationship --- a consequence of how arcs relate to angles. It holds for any rotating body, whether it is a spinning top nailed to a table or a freely tumbling asteroid.

Rolling without slipping takes this geometric relationship and promotes it to a physical constraint. The contact point must be at rest, and this requirement forces $v_{\text{cm}} = R\omega$. The relationship is no longer just geometry; it is a restriction on the motion that the physical situation demands. The ground enforces it.

This constraint will prove essential as you move forward. In Chapter 10, when you study the dynamics of rolling objects, the no-slip condition will connect the translational equation ($F_{\text{net}} = ma_{\text{cm}}$) to the rotational equation ($\tau_{\text{net}} = I\alpha$), coupling them through $a_{\text{cm}} = R\alpha$. In Chapter 14, when you analyze rolling on inclined planes and rolling with friction, the constraint will be the link that makes the problem solvable. Without it, translation and rotation would be independent, and rolling problems would require information you do not have.

Practice Layers

Layer 1: Concrete --- From Angular to Linear and Back

Problem 1. A bicycle wheel has radius $R = 0.35$ m and rolls without slipping. The wheel makes 2.0 revolutions per second. What is the speed of the center of the wheel?

Check your answer First, convert to angular velocity: $$\omega = 2.0 \text{ rev/s} \times 2\pi \text{ rad/rev} = 4\pi \text{ rad/s} \approx 12.6 \text{ rad/s}$$ Apply the no-slip constraint: $$v_{\text{cm}} = R\omega = (0.35)(4\pi) = 1.4\pi \approx 4.4 \text{ m/s}$$ The bicycle moves forward at about 4.4 m/s, or roughly 16 km/h.

Problem 2. A bowling ball of radius $R = 0.11$ m rolls without slipping at $v_{\text{cm}} = 6.0$ m/s. What is its angular velocity?

Check your answer $$\omega = \frac{v_{\text{cm}}}{R} = \frac{6.0}{0.11} \approx 54.5 \text{ rad/s}$$ That is about 8.7 revolutions per second. The ball is spinning rapidly, even though its forward speed may not seem extreme.

Layer 2: Pattern --- Velocity at Any Point on the Wheel

Problem 3. A wheel of radius $R$ rolls without slipping with center-of-mass speed $v_{\text{cm}}$. Find the speed of the following points:

(a) The point at the very top of the wheel.

(b) The point at the very bottom (contact point).

(c) A point at the same height as the center, on the front of the wheel (the "3 o'clock" position if the wheel rolls to the right).

(d) A point at the same height as the center, on the back of the wheel (the "9 o'clock" position).

Check your answer For each point, add the translational velocity (forward at $v_{\text{cm}}$) and the rotational velocity (tangent to the circle, magnitude $R\omega = v_{\text{cm}}$): **(a) Top:** Translation gives $v_{\text{cm}}$ forward. Rotation at the top also gives $v_{\text{cm}}$ forward. Total: $2v_{\text{cm}}$ forward. **(b) Bottom (contact point):** Translation gives $v_{\text{cm}}$ forward. Rotation at the bottom gives $v_{\text{cm}}$ backward. Total: $0$. **(c) Front ("3 o'clock"):** Translation gives $v_{\text{cm}}$ forward (horizontal). Rotation at this point gives $v_{\text{cm}}$ straight up (tangent to the circle at the front). These two are perpendicular, so the total speed is: $$v = \sqrt{v_{\text{cm}}^2 + v_{\text{cm}}^2} = v_{\text{cm}}\sqrt{2}$$ The direction is 45 degrees above horizontal. **(d) Back ("9 o'clock"):** Translation gives $v_{\text{cm}}$ forward. Rotation at this point gives $v_{\text{cm}}$ straight down. Again perpendicular: $$v = v_{\text{cm}}\sqrt{2}$$ The direction is 45 degrees below horizontal. Notice the symmetry: the front and back points have the same speed, but their velocity vectors tilt in opposite vertical directions.

Layer 3: Structure --- Why Zero Velocity at the Contact Point?

Problem 4. A student says: "I understand the math --- $v_{\text{cm}} - R\omega = 0$ at the bottom. But I do not understand physically why the contact point must have zero velocity. Can you explain it without equations?"

Explain to this student, in physical terms, why "rolling without slipping" requires the contact point to be instantaneously at rest.

Check your answer Think about what "no slipping" means physically. Slipping is when the surface of the wheel slides against the surface of the ground --- when there is relative motion between the two surfaces in contact. If the contact point on the wheel had any velocity, it would be moving relative to the ground at that point. That is sliding. That is slipping. For the wheel to roll without slipping, the part of the wheel that touches the ground must, at that instant, be moving at the same velocity as the ground. The ground is not moving. So the contact point must not be moving either. Think of it this way: at the instant a point on the rim touches the ground, it is doing exactly what a point on the ground does --- sitting still. The wheel "plants" each point on the rim momentarily, then peels it away. Like a tank tread laying itself down and picking itself up. The piece touching the ground is at rest; the rest of the wheel moves around it. This is why the instantaneous-axis-of-rotation picture is so natural for rolling. At each instant, the wheel really is rotating about the contact point. The contact point is the pivot. A pivot does not move.

Layer 4: Debug --- Friction and Rolling

Problem 5. A student argues: "If the contact point has zero velocity, then there is no sliding. If there is no sliding, there is no friction. So a rolling ball experiences no friction."

Is this correct? Identify the error and explain what role friction actually plays in rolling.

Check your answer The student is wrong, and the error is a common one. The student is confusing *kinetic* friction (which requires sliding) with *static* friction (which does not). It is true that there is no sliding at the contact point, so there is no *kinetic* friction. But there can still be *static* friction. Static friction acts between surfaces that are in contact but not sliding relative to each other. It is the same force that prevents your shoes from sliding on the floor when you walk, or that lets you push off the ground to start running. For a ball rolling on a flat surface at constant speed, static friction may indeed be zero (or negligibly small) --- there is nothing trying to make the contact point slip, so no friction is needed. But consider a ball rolling down a ramp. Gravity tries to accelerate the center of mass down the slope. Without friction, the ball would slide without rotating (like a block on a frictionless ramp). Static friction at the contact point provides the torque needed to spin the ball, and also modifies the translational acceleration so that the no-slip constraint $a_{\text{cm}} = R\alpha$ is maintained. Friction is essential for rolling in the first place --- it is what couples the rotation to the translation. So: no kinetic friction (because no sliding), but static friction can be present and is often crucial.

Layer 5: Creation --- When the Constraint Breaks

Problem 6. A ball rolls without slipping along a horizontal table and then reaches the edge. It rolls off and becomes a projectile. At what point does the rolling constraint $v_{\text{cm}} = R\omega$ break? What happens to the translational and rotational motions after the ball leaves the table?

Check your answer The no-slip constraint $v_{\text{cm}} = R\omega$ requires a surface for the wheel to roll on. It is the ground that enforces the constraint: static friction at the contact point couples translation and rotation. Once the ball leaves the edge of the table and becomes airborne, there is no contact point. No surface, no constraint. At the instant the ball leaves the table, it has some $v_{\text{cm}}$ and some $\omega$ that satisfy $v_{\text{cm}} = R\omega$. After that instant, the two motions decouple: - **Translation:** The center of mass follows a parabolic projectile trajectory under gravity, just like any other projectile (Section 3.4). $v_x$ remains constant; $v_y$ increases downward at rate $g$. - **Rotation:** With no friction and no torque (ignoring air resistance), the angular velocity $\omega$ remains constant. The ball keeps spinning at the same rate it had when it left the table. So $v_{\text{cm}}$ changes (because gravity accelerates the center of mass) while $\omega$ stays constant. The constraint $v_{\text{cm}} = R\omega$ no longer holds. Translation and rotation are now independent, decoupled by the absence of the surface that linked them. When the ball eventually hits the ground again, the constraint may be re-established --- though typically there is a brief period of skidding as friction brings the translational and rotational speeds back into agreement.

Reflection

How does one constraint lock together two descriptions of motion?

Throughout this chapter, you built parallel frameworks for translation and rotation. Separate quantities, separate equations, separate intuitions. They looked like two independent languages for describing the same physical world.

Rolling without slipping is where those two languages are forced to speak in unison. The single equation $v_{\text{cm}} = R\omega$ removes one degree of freedom from the system. You can no longer choose the translational speed and the angular velocity independently --- fixing one fixes the other. Differentiate the constraint and you get $a_{\text{cm}} = R\alpha$: the translational and angular accelerations are locked together too.

This is the power of a constraint. It does not add new physics. It reduces the number of things you need to know. A rolling wheel looks complicated --- every point has a different velocity, the motion is a combination of translation and rotation --- but the constraint means the entire motion is determined by a single number: either $v_{\text{cm}}$ or $\omega$. Everything else follows.

Chapter-End Retrieval

This is the final section of Chapter 9. Before moving on, close your notes and test your recall of the entire chapter. Answer from memory.

1. What are the rotational analogs of position $x$, velocity $v$, and acceleration $a$? Write their definitions as derivatives.

2. What are the constant-angular-acceleration kinematic equations? How do they relate to the translational ones from Chapter 2?

3. When angular acceleration is not constant, how do you find $\omega(t)$ and $\theta(t)$? What mathematical operation replaces the kinematic formulas?

4. How are arc length, tangential speed, and centripetal acceleration related to angular quantities? Write the three key relationships involving radius $r$.

5. Where does the translational-rotational analogy work perfectly, and where does it break down?

6. What does "rolling without slipping" mean mathematically? What is the velocity of the contact point, and why?

7. A wheel rolls without slipping. What is the speed of the topmost point relative to the ground? Explain why, using the decomposition into translation and rotation.

Take your time. If any answer feels uncertain, revisit the relevant section before continuing to Chapter 10.