10.6 Coupled Translation and Rotation in Rigid-Body Motion

The yo-yo problem

A yo-yo unwinds as it drops. Watch it carefully and you see two things happening at once: the center of mass falls straight down (translation), and the spool spins faster and faster (rotation). Neither motion exists without the other. The string unwinds because the spool falls, and the spool falls slower than it should because energy is going into spinning.

This is new territory. In previous sections, you learned to handle translation ($F = ma$) and rotation ($\tau = I\alpha$) as separate problems. But a yo-yo is not a separate problem. It is one object doing both at once, and the two motions are locked together by the string.

Neither $F = ma$ nor $\tau = I\alpha$ alone tells the whole story. You need both equations, linked by a constraint.

Before you read on: Does a yo-yo fall slower or faster than a freely falling ball of the same mass? Why?

Commit to an answer and a reason before continuing.

If you said slower, you are right --- but the reason is what matters. A freely falling ball converts all of its gravitational potential energy into translational kinetic energy. The yo-yo has to split that energy between translation and rotation. With some energy diverted into spinning, there is less left over for falling. The center of mass accelerates, but not at $g$.

This section is about how to set up and solve problems like this --- problems where translation and rotation are coupled.

The guiding question

What happens when a rigid body both moves through space and spins?

You already have the tools: $F = ma_{\text{cm}}$ governs the center of mass, and $\tau = I\alpha$ governs the rotation. What you need is the bridge between them --- the constraint that says how translational and rotational motion are connected for a particular physical setup.

Exploration: a spool unwinding under gravity

[Interactive: Coupled motion explorer. A spool (drawn as a cylinder with a string wrapped around it) hangs from a fixed support. When released, it unwinds and drops. The display shows three synchronized panels:

1. Free-body diagram panel: The spool with its weight $mg$ acting downward at the center of mass and the string tension $T$ acting upward at the rim. Force arrows are drawn to scale and update in real time.

2. Torque diagram panel: The same spool, now showing the torque produced by each force about the center of mass. The weight acts at the center, so it produces zero torque. The tension acts at the rim at distance $R$, producing a torque $\tau = TR$.

3. Equations panel: Two equations appear side by side:

4. Translation: $mg - T = ma$
5. Rotation: $TR = I\alpha$

Below them, the constraint: $a = R\alpha$

As the animation plays, numerical values fill in. Students can adjust the spool's mass, radius, and moment of inertia (by choosing between a solid cylinder and a hollow cylinder) and watch how the acceleration changes.

Guided prompts appear below the interactive: - "Set the spool to a solid cylinder. Read off the acceleration. Now switch to a hollow cylinder of the same mass and radius. What happens to the acceleration? Why?" - "If you increase the spool radius while keeping mass constant, what happens to the acceleration? Try it." - "Look at the tension $T$. Is it equal to $mg$? Greater? Less? Why does this make physical sense?"]

Spend a few minutes with this. The key observation is that the tension is less than $mg$ --- the spool accelerates downward --- but the acceleration is less than $g$ because the tension partially supports the spool's weight. And the string simultaneously exerts a torque that spins the spool faster.

Concept reveal: two equations, one constraint

Here is the framework for any coupled translation-rotation problem. You write two equations and connect them with a constraint.

Step 1: Translational equation. Apply Newton's second law to the center of mass:

$$\sum F = ma_{\text{cm}}$$

This governs how the center of mass accelerates. It looks exactly like the $F = ma$ you have used since Chapter 5, except now you are careful to track the center of mass specifically.

Step 2: Rotational equation. Apply the rotational form of Newton's second law about the center of mass (or about a fixed axis, if one exists):

$$\sum \tau = I\alpha$$

This governs how the angular velocity changes. It uses the torques computed about the same point you chose for rotation.

Step 3: Constraint. Identify the physical link between translation and rotation. For rolling without slipping or unwinding from a string, the constraint is:

$$a_{\text{cm}} = R\alpha$$

This says that every radian the object rotates corresponds to $R$ meters of translational motion. It is the same rolling constraint you met in Section 9.6, differentiated from the kinematic relation $v_{\text{cm}} = R\omega$.

Step 4: Solve simultaneously. You now have three equations and three unknowns (typically $a$, $\alpha$, and one unknown force like tension or friction). Standard algebra gives you all three.

The power of this method is that it handles the coupling automatically. You do not need to guess how much energy goes into rotation versus translation. The constraint enforces the correct split.

Connection to prior work

Section 9.6 introduced the rolling constraint kinematically: $v_{\text{cm}} = R\omega$. At that point, you used it to relate velocities. Now we combine it with dynamics. Differentiating the kinematic constraint gives $a_{\text{cm}} = R\alpha$, and this links the force equation to the torque equation.

The translational equation comes from Chapter 5. The rotational equation comes from Section 10.5. The constraint comes from Section 9.6. This section is where all three threads come together.

Worked example: yo-yo falling

A yo-yo of mass $m$ has an inner spool of radius $r$ and a moment of inertia $I = \frac{1}{2}mr^2$ (modeled as a solid cylinder for simplicity). The string unwinds from the inner spool. Find the acceleration of the yo-yo and the tension in the string.

Draw the free-body diagram. Two forces act on the yo-yo: - Weight $mg$ downward, at the center of mass - Tension $T$ upward, at the rim of the inner spool (where the string leaves)

Write the translational equation. Taking downward as positive:

$$mg - T = ma \tag{1}$$

Write the rotational equation. Taking torques about the center of mass, the tension acts at distance $r$ from the center and produces a clockwise torque (consistent with the yo-yo spinning as it unwinds):

$$Tr = I\alpha \tag{2}$$

The weight passes through the center of mass and contributes zero torque.

Apply the constraint. The string unwinds from a spool of radius $r$, so:

$$a = r\alpha \tag{3}$$

Solve. From equation (3): $\alpha = a/r$. Substitute into equation (2):

$$Tr = I \cdot \frac{a}{r} = \frac{1}{2}mr^2 \cdot \frac{a}{r} = \frac{1}{2}mra$$

$$T = \frac{1}{2}ma \tag{4}$$

Substitute equation (4) into equation (1):

$$mg - \frac{1}{2}ma = ma$$

$$mg = \frac{3}{2}ma$$

$$a = \frac{2g}{3}$$

And the tension is:

$$T = \frac{1}{2}m \cdot \frac{2g}{3} = \frac{mg}{3}$$

Check. The acceleration is $\frac{2}{3}g$ --- less than $g$, as expected. The tension is $\frac{1}{3}mg$ --- less than $mg$, so the yo-yo does accelerate downward. The string supports one-third of the weight, and the remaining two-thirds produces the net downward force.

If the moment of inertia were larger (more mass at the rim), the acceleration would be even smaller. More rotational inertia means more of the gravitational work goes into spinning, leaving less for falling.

Faded example: cylinder rolling down a ramp

A solid cylinder of mass $m$ and radius $R$ rolls without slipping down a ramp inclined at angle $\theta$. Find its acceleration.

Step 1: Draw the free-body diagram.

Three forces act on the cylinder: - Weight $mg$ downward (at the center of mass) - Normal force $N$ perpendicular to the ramp surface (at the contact point) - Static friction $f_s$ up the ramp (at the contact point)

Choose coordinates along and perpendicular to the ramp surface.

Step 2: Write the translational equation along the ramp (positive down the ramp):

$$mg\sin\theta - f_s = ma \tag{1}$$

Step 3: Write the rotational equation about the center of mass. The only force producing a torque about the center is friction (weight and normal force both pass through the center of mass or have zero moment arm perpendicular to the axis of rotation):

Fill in the blanks: The friction $f_s$ acts at distance $___$ from the center, producing torque $___$. The rotational equation is $___ = I\alpha$.

Check your answer

Friction acts at distance $R$ from the center, producing torque $f_s R$. The rotational equation is: $$f_s R = I\alpha \tag{2}$$

Step 4: Apply the rolling constraint.

The cylinder rolls without slipping. Write the constraint relating $a$ and $\alpha$.

Check your answer

$$a = R\alpha \tag{3}$$

Step 5: Solve for $a$.

Use the moment of inertia of a solid cylinder ($I = \frac{1}{2}mR^2$) and equations (1)--(3) to eliminate $f_s$ and $\alpha$. Find $a$.

Check your answer

From (3): $\alpha = a/R$. Substitute into (2): $$f_s R = \frac{1}{2}mR^2 \cdot \frac{a}{R} = \frac{1}{2}mRa$$ $$f_s = \frac{1}{2}ma$$ Substitute into (1): $$mg\sin\theta - \frac{1}{2}ma = ma$$ $$mg\sin\theta = \frac{3}{2}ma$$ $$a = \frac{2}{3}g\sin\theta$$ This is exactly $\frac{2}{3}$ of the acceleration of a frictionless sliding block ($g\sin\theta$). The rolling constraint diverts one-third of the gravitational component into rotational kinetic energy.

Independent problem: Atwood machine with a massive pulley

This is a classic problem that combines everything from this section. Try it on your own before checking.

A pulley of mass $M$, radius $R$, and moment of inertia $I = \frac{1}{2}MR^2$ (solid disk) is mounted on a frictionless axle. A light, inextensible string passes over the pulley without slipping. Masses $m_1$ and $m_2$ hang from the two ends, with $m_1 > m_2$.

Find the acceleration of the masses and the tensions $T_1$ and $T_2$ on each side of the string.

Hints: - The string does not slip on the pulley, so $a = R\alpha$. - The tensions on the two sides of the string are not equal (unlike the massless-pulley version from Chapter 5). The difference in tensions provides the net torque on the pulley. - You need three equations: translational Newton's second law for $m_1$, translational Newton's second law for $m_2$, and rotational Newton's second law for the pulley.

Check your answer

**For mass $m_1$** (taking downward as positive since $m_1$ descends): $$m_1 g - T_1 = m_1 a \tag{1}$$ **For mass $m_2$** (taking upward as positive since $m_2$ ascends): $$T_2 - m_2 g = m_2 a \tag{2}$$ **For the pulley** (net torque from the two tensions): $$(T_1 - T_2)R = I\alpha = \frac{1}{2}MR^2 \cdot \frac{a}{R} = \frac{1}{2}MRa$$ $$T_1 - T_2 = \frac{1}{2}Ma \tag{3}$$ Add equations (1) and (2): $$(m_1 - m_2)g - (T_1 - T_2) = (m_1 + m_2)a$$ Substitute (3): $$(m_1 - m_2)g - \frac{1}{2}Ma = (m_1 + m_2)a$$ $$(m_1 - m_2)g = \left(m_1 + m_2 + \frac{M}{2}\right)a$$ $$a = \frac{(m_1 - m_2)g}{m_1 + m_2 + \frac{M}{2}}$$ Notice what the pulley does: it adds $\frac{M}{2}$ to the effective inertia of the system. The system accelerates as if there were extra mass resisting the motion. When $M = 0$ (massless pulley), this reduces to the standard Atwood result $a = \frac{(m_1 - m_2)g}{m_1 + m_2}$. The tensions are found by substituting $a$ back into equations (1) and (2): $$T_1 = m_1(g - a), \qquad T_2 = m_2(g + a)$$ And indeed $T_1 > T_2$, which is what creates the net torque on the pulley.

Practice

Layer 1: Concrete

A solid cylinder of mass $m = 4.0$ kg and radius $R = 0.10$ m rolls without slipping down a ramp inclined at $\theta = 30°$. Using the coupled-equations method from this section, find:

(a) The acceleration of the cylinder.

(b) The friction force acting on the cylinder.

Check your answer

Using the result derived in this section for a solid cylinder ($I = \frac{1}{2}mR^2$): **(a)** The acceleration is: $$a = \frac{2}{3}g\sin\theta = \frac{2}{3}(9.8)(0.50) = 3.27 \text{ m/s}^2$$ Compare this to a sliding block on the same ramp: $a_{\text{slide}} = g\sin\theta = 4.9 \text{ m/s}^2$. The rolling cylinder accelerates at only two-thirds of the sliding value. **(b)** From the rotational equation, $f_s = \frac{1}{2}ma$: $$f_s = \frac{1}{2}(4.0)(3.27) = 6.5 \text{ N}$$ This is the static friction force required to maintain rolling without slipping. It is *not* the maximum static friction --- it is exactly the friction needed to enforce the constraint $a = R\alpha$.

Layer 2: Pattern

A solid cylinder ($I = \frac{1}{2}mR^2$) and a hollow cylinder ($I = mR^2$) have the same mass and radius. Both roll without slipping down the same ramp.

(a) Find the acceleration of each.

(b) Which reaches the bottom first?

(c) Does the result depend on mass or radius?

Check your answer

The general result for a cylinder with moment of inertia $I = cmR^2$ rolling down a ramp is: $$a = \frac{g\sin\theta}{1 + c}$$ where $c$ is the geometric factor in the moment of inertia. **(a)** For the solid cylinder, $c = \frac{1}{2}$: $$a_{\text{solid}} = \frac{g\sin\theta}{1 + \frac{1}{2}} = \frac{2}{3}g\sin\theta$$ For the hollow cylinder, $c = 1$: $$a_{\text{hollow}} = \frac{g\sin\theta}{1 + 1} = \frac{1}{2}g\sin\theta$$ **(b)** The solid cylinder has the larger acceleration, so it reaches the bottom first. This makes physical sense: the hollow cylinder has more of its mass at the rim, giving it a larger moment of inertia. More rotational inertia means more energy goes into spinning and less into translating. **(c)** The acceleration depends on $c$ (the mass distribution) and $\theta$ (the ramp angle), but *not* on $m$ or $R$. A tiny solid marble and a massive solid drum would reach the bottom at the same time, as long as both are solid cylinders rolling without slipping. This is the rotational analog of the fact that all objects in free fall have the same acceleration regardless of mass.

Layer 3: Structure

A block of mass $m$ slides (frictionlessly) down a ramp at angle $\theta$, accelerating at $a = g\sin\theta$. A solid cylinder of the same mass rolls without slipping down the same ramp, accelerating at $a = \frac{2}{3}g\sin\theta$.

Why does the rolling object accelerate slower than the sliding object on the same ramp? Explain in terms of where the energy goes, and connect your explanation to the role of the constraint equation.

Check your answer

Both objects experience the same gravitational force component down the ramp: $mg\sin\theta$. For the sliding block, this entire force goes into translational acceleration. There is nothing else to accelerate. For the rolling cylinder, the situation is different. Gravity still pulls the center of mass down the ramp, but friction at the contact point produces a torque that spins the cylinder. The rolling constraint $a = R\alpha$ demands that the translational and rotational accelerations match --- the cylinder cannot translate without also rotating. This means the gravitational potential energy released as the cylinder descends must be split between translational kinetic energy ($\frac{1}{2}mv^2$) and rotational kinetic energy ($\frac{1}{2}I\omega^2$). Since some energy is "diverted" into rotation, less is available for translation, and the center of mass accelerates more slowly. The constraint is what enforces this energy split. Without it, the translational and rotational motions would be independent, and there would be no reason for the cylinder to accelerate any differently from the block. The constraint couples the two equations and reduces the translational acceleration.

Layer 4: Debug

A student is solving for the acceleration of a cylinder rolling down a ramp. They correctly write the translational equation:

$$mg\sin\theta - f_s = ma$$

and the rotational equation:

$$f_s R = I\alpha$$

But they forget the constraint $a = R\alpha$. They now have two equations and three unknowns ($a$, $\alpha$, and $f_s$).

What goes wrong? Can they still solve the problem? What does the missing constraint represent physically?

Check your answer

Without the constraint, the student has two equations but three unknowns. The system is underdetermined --- there is no unique solution. They could pick any value of $f_s$ and find *some* pair $(a, \alpha)$ that satisfies both equations, but those values would not correspond to rolling without slipping. For example, if they set $f_s = 0$, they would get $a = g\sin\theta$ (free sliding, no friction) and $\alpha = 0$ (no rotation). This describes a sliding block, not a rolling cylinder. If they set $f_s = mg\sin\theta$, they would get $a = 0$ (no translation) and a large $\alpha$ (spinning in place). Neither describes the physical situation. The constraint $a = R\alpha$ is what makes this a *rolling* problem. It says that the translational and rotational motions are not independent --- every meter the center of mass moves corresponds to exactly $1/R$ radians of rotation. Physically, it enforces the no-slip condition at the contact point: the contact point has zero velocity relative to the ground, which is what rolling means. Without the constraint, the student has written the correct force and torque equations but has not specified *what kind of motion* the cylinder is doing. The constraint is the third equation that pins down the physics.

Reflection

Think about the constraint equation $a = R\alpha$ and the role it plays in this section.

Is the constraint adding an equation to the system, or is it reducing the number of unknowns?

Both descriptions are mathematically correct --- adding an equation to the system is equivalent to reducing the degrees of freedom. But which way of thinking about it helps you more? When you set up a coupled problem, do you think of the constraint as giving you one more equation to work with, or as telling you that $a$ and $\alpha$ are really the same variable in disguise?

There is no single right answer here. But developing a clear mental picture of what constraints do will serve you well in every coupled problem you encounter.

Looking Ahead

You now have the complete toolkit for rigid-body dynamics: forces and torques, moments of inertia, Newton's second law in both translational and rotational form, and the constraints that couple them. In this section, you wrote three equations and solved them simultaneously. It worked --- but you may have noticed that the algebra, while straightforward, can get involved.

In the next section, you will learn how to make that algebra shorter. Section 10.7 covers choosing axes and simplifying rotational dynamics problems --- the rotational analog of the coordinate-system strategy you used in Chapter 6 for ramps. A well-chosen axis can eliminate unknown forces from the torque equation entirely, cutting the number of equations you need to solve. The physics does not change, but the work does.