11.6 Energy and Angular Momentum in Rolling and Rotating Systems

Four Tools, One Problem

A ball rolls down a ramp. How fast is it going at the bottom?

You could solve this with Newton's second law for translation: draw the FBD, decompose gravity along the ramp, find the friction force, solve $\sum F = ma$ for the translational acceleration, then use kinematics.

You could solve it with Newton's second law for rotation: find the torque from friction about the center, apply $\sum \tau = I\alpha$, relate $\alpha$ to $a$ through the rolling constraint, then use kinematics.

You could solve it with energy conservation: set the gravitational potential energy at the top equal to the total kinetic energy at the bottom --- translational plus rotational --- and solve for $v$.

You could even ask whether angular momentum conservation applies (it does not here, because there is a net external torque, but knowing when to rule out a method is just as important as knowing when to use one).

That is four possible approaches for a single problem. The physics does not care which one you choose --- the answer is the same. But the effort, the algebra, and the information you extract are very different. This section is about making that choice deliberately.

Prediction

Before you read on: A solid ball and a hollow ball have the same mass $m$ and the same radius $R$. Both start from rest at the top of the same ramp and roll without slipping to the bottom.

Which one reaches the bottom first? Is it close, or is it a decisive lead? Commit to a prediction and a rough estimate of the difference before continuing.

The Guiding Question

When several methods --- forces, torques, energy, angular momentum --- are all available, how do you decide which one to use?

This is not a new question. Section 7.7 asked the same thing when you first had two tools (forces vs. energy) for translational problems. The answer there was: energy methods are fastest when the question asks for speed or distance; force methods are necessary when the question asks for time, acceleration, or a specific force value.

Now the toolkit has doubled. You have translational forces, rotational torques, energy conservation (with both translational and rotational kinetic energy), and angular momentum conservation. The decision is more complex, but the logic is the same: each method keeps some information and discards some information. The best method is the one whose outputs match what the problem asks for.

Exploration: The Same Rolling Problem, Three Ways

Here are two rolling problems. For each one, three solution approaches are outlined. Read all three, then compare.

Problem A: Speed at the Bottom of a Ramp

A solid sphere of mass $m$ and radius $R$ starts from rest at the top of a ramp of height $h$ and rolls without slipping to the bottom. Find the speed of the center of mass at the bottom.

Approach 1: Force + torque.

Draw the FBD. Forces: gravity $mg$ (downward), normal force $N$ (perpendicular to ramp), static friction $f$ (up the ramp, at the contact point --- this is what provides the torque for rolling).

Translation along the ramp:

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

Rotation about the center of mass:

$$fR = I\alpha$$

Rolling constraint: $a = R\alpha$. For a solid sphere, $I = \frac{2}{5}mR^2$. Substituting:

$$f = I\alpha / R = \frac{2}{5}mR^2 \cdot \frac{a}{R^2} = \frac{2}{5}ma$$

Plug into the translational equation:

$$mg\sin\theta - \frac{2}{5}ma = ma \implies a = \frac{5}{7}g\sin\theta$$

Now use kinematics. The ramp length is $L = h/\sin\theta$:

$$v^2 = 2aL = 2 \cdot \frac{5}{7}g\sin\theta \cdot \frac{h}{\sin\theta} = \frac{10}{7}gh$$

$$v = \sqrt{\frac{10}{7}gh}$$

That took six equations and required knowing the moment of inertia, setting up two Newton's law equations, applying the rolling constraint, solving a system, and then using kinematics. It works. It gives you the acceleration, the friction force, and the speed. But it was a lot of algebra.

Approach 2: Energy conservation.

At the top: $KE = 0$, $U = mgh$. At the bottom: $U = 0$, and the total kinetic energy is translational plus rotational:

$$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

Rolling constraint: $\omega = v/R$. For a solid sphere, $I = \frac{2}{5}mR^2$:

$$KE = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mR^2\right)\left(\frac{v}{R}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$$

Energy conservation:

$$mgh = \frac{7}{10}mv^2 \implies v = \sqrt{\frac{10}{7}gh}$$

Three equations. No force decomposition. No friction. No ramp angle. No kinematics. The same answer, half the work.

Approach 3: Angular momentum.

Angular momentum about any point is $L = I\omega$ (for the spin) plus $mv_{\text{cm}} \times r$ (for the orbital motion of the center of mass). Is angular momentum conserved here?

Gravity exerts a torque about most axes. The normal force exerts a torque about most axes. Friction exerts a torque. There is no obvious axis about which all external torques vanish. Angular momentum is not conserved in this problem.

This approach does not apply. Recognizing that quickly is itself a skill.

[Interactive: Three-Method Comparison. The ramp problem is shown with a rolling sphere. Three panels display the force/torque solution, the energy solution, and the angular momentum assessment side by side. Each panel shows the equations used, the number of algebraic steps, and highlights what information each method provides (acceleration, friction force, speed) or does not provide. A summary bar at the bottom reads: "Energy: 3 steps, gives speed. Force/torque: 6 steps, gives speed + acceleration + friction. Angular momentum: does not apply." Students can toggle between a solid sphere, hollow sphere, and solid cylinder to see how the moment of inertia changes each solution but does not change which method is shortest.]

Problem B: Turntable Collision

A disk of moment of inertia $I_1$ spins freely at angular velocity $\omega_0$. A second disk of moment of inertia $I_2$, initially at rest, is dropped onto it from above. The two disks lock together and spin at a new angular velocity. Find $\omega_f$.

Approach 1: Force + torque.

The contact between the disks produces friction torques that are large, brief, and unknown in detail. You would need to model the friction force as a function of time, integrate it, and solve coupled differential equations for the angular velocities of each disk. This is extremely difficult --- the force is not given, and it depends on the relative sliding speed, which changes during the process.

This approach is technically possible but practically awful.

Approach 2: Energy conservation.

Is kinetic energy conserved? The disks slide against each other during the collision. Friction between them converts kinetic energy to thermal energy. So $KE_f \neq KE_i$. Energy conservation does not directly give $\omega_f$.

You could use energy methods if you knew how much energy was lost, but you do not. Energy does not help here.

Approach 3: Angular momentum.

No external torques act on the two-disk system (the axle is frictionless, and the internal friction between the disks is an internal torque). Angular momentum is conserved:

$$L_i = L_f$$ $$I_1\omega_0 = (I_1 + I_2)\omega_f$$ $$\omega_f = \frac{I_1}{I_1 + I_2}\omega_0$$

One equation. One line of algebra. Done.

Pause and compare: In Problem A, energy conservation was the clear winner. In Problem B, angular momentum conservation was the only practical method. What determined the winner in each case?

In Problem A, energy was conserved (no slipping means no energy loss to friction), but angular momentum was not (external torques from gravity and friction). In Problem B, angular momentum was conserved (no external torques), but energy was not (internal friction dissipated KE). Each method works precisely when its conservation condition is met and the others are not.

Concept Reveal: The Rolling Energy Formula and Method Selection

Total Kinetic Energy for Rolling Without Slipping

When an object rolls without slipping, its motion has two components: the center of mass translates, and the object rotates about its center of mass. The total kinetic energy is the sum:

$$KE_{\text{total}} = \frac{1}{2}mv_{\text{cm}}^2 + \frac{1}{2}I_{\text{cm}}\omega^2$$

The rolling constraint $v_{\text{cm}} = R\omega$ lets you write everything in terms of a single variable. Using $\omega = v_{\text{cm}}/R$:

$$KE_{\text{total}} = \frac{1}{2}mv_{\text{cm}}^2 + \frac{1}{2}I_{\text{cm}}\frac{v_{\text{cm}}^2}{R^2} = \frac{1}{2}\left(m + \frac{I_{\text{cm}}}{R^2}\right)v_{\text{cm}}^2$$

This is a single expression in $v_{\text{cm}}$. The term $I_{\text{cm}}/R^2$ captures how much of the kinetic energy goes into rotation. For a solid sphere ($I = \frac{2}{5}mR^2$), it adds $\frac{2}{5}m$. For a hollow sphere ($I = \frac{2}{3}mR^2$), it adds $\frac{2}{3}m$. The more of the mass that is far from the center, the more energy goes into spinning, and the less is available for translating.

Answering the Prediction

Now return to the opening prediction. A solid ball ($I = \frac{2}{5}mR^2$) and a hollow ball ($I = \frac{2}{3}mR^2$) roll down the same ramp from the same height.

For the solid ball:

$$mgh = \frac{1}{2}\left(m + \frac{2m}{5}\right)v^2 = \frac{7}{10}mv^2 \implies v_{\text{solid}} = \sqrt{\frac{10gh}{7}}$$

For the hollow ball:

$$mgh = \frac{1}{2}\left(m + \frac{2m}{3}\right)v^2 = \frac{5}{6}mv^2 \implies v_{\text{hollow}} = \sqrt{\frac{6gh}{5}}$$

The solid ball is faster. It reaches the bottom first. The ratio of speeds is:

$$\frac{v_{\text{solid}}}{v_{\text{hollow}}} = \sqrt{\frac{10/7}{6/5}} = \sqrt{\frac{50}{42}} = \sqrt{\frac{25}{21}} \approx 1.091$$

The solid ball is about 9% faster. That is noticeable but not dramatic. The key insight: the outcome depends only on how $I$ compares to $mR^2$, not on the mass $m$, the radius $R$, or the ramp height $h$. A bowling ball and a marble of the same shape would tie.

Notice that this entire analysis used energy. The force/torque approach would have required solving for the acceleration of each ball separately and then comparing kinematics. The energy approach got the answer in two lines per ball.

When to Use Which Method

Here is the decision framework, extended from Section 7.7 to include rotational tools:

Method	What it requires	What it provides	Use when...
Forces ($\sum F = ma$)	All forces, their directions	Translational acceleration, specific forces	You need $a$, time, or a specific force
Torques ($\sum \tau = I\alpha$)	All torques, moment of inertia	Angular acceleration, specific torques	You need $\alpha$, angular timing, or a specific torque
Energy conservation	Conservative forces, work by nonconservative forces	Speeds, distances	You need speed or distance, not time or specific forces
Angular momentum conservation	Net external torque = 0	Final angular velocities after reconfiguration or collision	Internal changes only, no external torques

The methods are not competitors. They are complementary tools. The expert's job is to read the problem, identify what is being asked, and match it to the method that provides that output most directly.

Connection: The Same Lesson at a Higher Level

In Section 7.7, you faced this question for the first time: forces or energy? The answer was that energy methods bypass force details and give speed directly, while force methods give acceleration and time. You learned to pause before computing and ask what the problem actually wants.

This section extends that lesson to include rotational tools. The toolkit is now four methods deep. But the logic of choosing has not changed. Each method is a different way of processing $\sum \vec{F} = m\vec{a}$ (and its rotational counterpart $\sum \vec{\tau} = I\vec{\alpha}$). Each integrates the equations differently --- over displacement (energy), over time (impulse/angular impulse), or not at all (direct force/torque analysis). Each integration preserves some information and discards some.

The structural analogy is exact:

Translational	Rotational
$KE = \frac{1}{2}mv^2$	$KE_{\text{rot}} = \frac{1}{2}I\omega^2$
$W = \int F\,dx$	$W = \int \tau\,d\theta$
$\vec{p} = m\vec{v}$	$\vec{L} = I\vec{\omega}$
$\int F\,dt = \Delta p$	$\int \tau\,dt = \Delta L$
If $\sum F_{\text{ext}} = 0$: $p$ conserved	If $\sum \tau_{\text{ext}} = 0$: $L$ conserved

For rolling objects, both columns are active simultaneously. The total kinetic energy combines both rows. The total angular momentum may include both orbital and spin contributions. The real challenge is not learning the formulas --- you already know them from Sections 11.1 through 11.5. The real challenge is knowing which formula to reach for in a given problem.

Metacognition: The Expert's Toolkit

You now have four tools: forces, torques, energy, and angular momentum. Expert problem solving is largely about choosing the right tool --- and that choice is a skill worth practicing explicitly.

Here is a practice that will accelerate your development: before solving any problem, write one sentence stating which method you will use and why. Not "I will use energy because it is easier" --- that is a wish, not a reason. Instead: "I will use energy because the problem asks for speed, not time, and energy is conserved because the ball rolls without slipping on a surface where friction does no work."

If you cannot write that sentence, you do not yet understand the problem well enough to start solving it. The sentence forces you to engage with the problem's structure before engaging with its algebra. This is the single most important habit that separates novice and expert problem solvers.

Spaced Retrieval

Before moving to practice, test your recall from earlier in this chapter and from Chapter 7.

Recall prompt 1: What is the total kinetic energy of a rolling object? Write the formula and explain each term. (Section 11.1)

Recall prompt 2: When is angular momentum conserved? State the condition precisely. (Section 11.5)

Recall prompt 3: In Section 7.7, you learned that energy methods cannot provide what piece of information? (Section 7.7)

Practice Layers

Layer 1: Concrete --- Rolling Down a Ramp Using Energy

Problem 1. A solid cylinder of mass $m = 4\,\text{kg}$ and radius $R = 0.10\,\text{m}$ starts from rest at the top of a ramp of height $h = 2.0\,\text{m}$ and rolls without slipping to the bottom. The moment of inertia of a solid cylinder is $I = \frac{1}{2}mR^2$.

(a) Find the speed of the center of mass at the bottom using energy conservation.

(b) What fraction of the total kinetic energy at the bottom is rotational?

(c) If the cylinder were replaced by a hoop ($I = mR^2$) of the same mass and radius, would it be faster or slower at the bottom?

Check your answer

**(a)** Energy conservation: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. Using $I = \frac{1}{2}mR^2$ and $\omega = v/R$: $$mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mR^2\right)\frac{v^2}{R^2} = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$$ $$v = \sqrt{\frac{4gh}{3}} = \sqrt{\frac{4(9.8)(2.0)}{3}} = \sqrt{26.1} = 5.11\,\text{m/s}$$ **(b)** Rotational KE: $\frac{1}{2}I\omega^2 = \frac{1}{4}mv^2$. Total KE: $\frac{3}{4}mv^2$. Fraction that is rotational: $$\frac{\frac{1}{4}mv^2}{\frac{3}{4}mv^2} = \frac{1}{3}$$ One-third of the kinetic energy is rotational. This fraction depends only on the shape (the ratio $I/mR^2$), not on the mass, radius, or ramp height. **(c)** For a hoop, $I = mR^2$: $$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)\frac{v^2}{R^2} = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$$ $$v = \sqrt{gh} = \sqrt{(9.8)(2.0)} = 4.43\,\text{m/s}$$ The hoop is slower (4.43 m/s vs. 5.11 m/s). It has a larger moment of inertia relative to $mR^2$, so more energy goes into rotation and less into translation. Half of the hoop's kinetic energy is rotational, compared to one-third for the cylinder.

Layer 2: Pattern --- Predict the Best Method Before Solving

For each problem below, predict which method (forces/torques, energy, or angular momentum) will be the most efficient. Write your prediction and reasoning before checking the answer. Do not solve the problems fully --- the goal is to practice method selection.

Problem 2a. A ball rolls without slipping down a ramp. Find the acceleration of the ball's center of mass.

Problem 2b. A figure skater pulls her arms in, reducing her moment of inertia from $3.0\,\text{kg}\cdot\text{m}^2$ to $1.2\,\text{kg}\cdot\text{m}^2$. Her initial angular velocity is $2.0\,\text{rad/s}$. Find her final angular velocity.

Problem 2c. A solid sphere rolls without slipping along a horizontal surface and then up a ramp. Find the maximum height it reaches.

Check your answer

**2a. Forces and torques win.** The problem asks for *acceleration*, not speed or distance. Energy conservation can give you the speed at the bottom but not the acceleration during the roll. You need $\sum F = ma$ along the ramp and $\sum \tau = I\alpha$ about the center, combined with the rolling constraint $a = R\alpha$, to find $a$ directly. **2b. Angular momentum wins.** No external torques act on the skater (the ice exerts negligible friction torque about the vertical axis). Her moment of inertia changes, so angular momentum conservation gives $\omega_f$ immediately: $I_1\omega_1 = I_2\omega_2$, so $\omega_f = (3.0/1.2)(2.0) = 5.0\,\text{rad/s}$. Energy is *not* conserved (her muscles do work), so energy methods would require knowing the work done by her muscles. Force/torque methods would require modeling internal forces. Angular momentum is the only clean approach. **2c. Energy wins.** The problem asks for a *height* (a distance), not an acceleration or time. Energy conservation connects the initial kinetic energy (translational + rotational) to the final potential energy at maximum height. One equation, one unknown. Force/torque methods would require analyzing two phases (flat surface, then ramp) with separate FBDs. Angular momentum is not conserved (gravity exerts a torque). **The pattern:** match the method to the question. Acceleration? Forces/torques. Speed or distance with no energy loss? Energy. Internal reconfiguration with no external torques? Angular momentum.

Layer 3: Structure --- Why Energy Makes Rolling Problems Easier

Problem 3. Answer these questions from your understanding of how the methods work, not by solving a specific problem.

(a) When a ball rolls without slipping down a ramp, static friction acts at the contact point. Does static friction do work? Why or why not?

(b) Given your answer to (a), why can you use energy conservation for rolling-without-slipping problems even though friction is present?

(c) The force/torque approach to rolling requires solving two coupled equations (translation and rotation) with the rolling constraint. The energy approach requires one equation. What information does the force/torque approach provide that the energy approach does not?

(d) Why does the energy approach make the ramp angle irrelevant for finding the speed at the bottom?

Check your answer

**(a)** Static friction does no work. The contact point is instantaneously at rest (that is what "rolling without slipping" means), so the displacement of the point where the force acts is zero. No displacement means no work, regardless of the force magnitude. **(b)** Since static friction does no work, no mechanical energy is lost to friction. Only gravity does work, and gravity is conservative. Therefore mechanical energy is conserved: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. The presence of friction does not violate energy conservation here because the friction force does not transfer energy out of the mechanical system. **(c)** The force/torque approach gives the *acceleration* of the center of mass ($a = \frac{g\sin\theta}{1 + I/(mR^2)}$) and the *friction force* ($f = \frac{mg\sin\theta}{1 + mR^2/I}$). From the acceleration, you can find the time to reach the bottom, the velocity at any intermediate point, and whether the friction required for rolling exceeds $\mu_s N$ (which would mean the object slips instead of rolling). The energy approach gives none of this. **(d)** Energy conservation says $mgh = \frac{1}{2}(m + I/R^2)v^2$. The height $h$ is the vertical drop, which is the same regardless of the ramp angle. A steeper ramp gives a larger component of gravity along the ramp (larger acceleration) but a shorter ramp length, and these exactly cancel. The energy method never introduces the ramp angle in the first place --- it works directly with the vertical height --- so the cancellation is automatic rather than algebraic.

Layer 4: Debug --- The Missing Rotational KE Term

Problem 4. A student solves a rolling-without-slipping problem for a solid sphere ($I = \frac{2}{5}mR^2$) descending a ramp of height $h$. The student writes:

$$mgh = \frac{1}{2}mv^2$$ $$v = \sqrt{2gh}$$

The student forgot the rotational kinetic energy term.

(a) What is the correct answer?

(b) By what percentage does the student's answer overestimate the speed?

(c) For a hoop ($I = mR^2$), the student makes the same mistake. Is the percentage error larger or smaller than for the sphere? Why?

(d) The student says: "The difference is small, so it doesn't matter." Under what circumstances would the error actually matter?

Check your answer

**(a)** The correct energy equation is $mgh = \frac{7}{10}mv^2$, so $v_{\text{correct}} = \sqrt{\frac{10gh}{7}}$. **(b)** The student's answer: $v_{\text{student}} = \sqrt{2gh}$. The ratio: $$\frac{v_{\text{student}}}{v_{\text{correct}}} = \sqrt{\frac{2gh}{10gh/7}} = \sqrt{\frac{14}{10}} = \sqrt{1.4} \approx 1.183$$ The student overestimates the speed by about 18.3%. That is not small. **(c)** For a hoop: $mgh = mv^2$, so $v_{\text{correct}} = \sqrt{gh}$. The student's answer is still $\sqrt{2gh}$. The ratio is $\sqrt{2} \approx 1.414$, an overestimate of about 41.4%. The error is **larger** for the hoop because the hoop has a larger moment of inertia relative to $mR^2$. A larger fraction of the energy goes into rotation, so ignoring the rotational term misses a larger share of the total kinetic energy. For the hoop, half the energy is rotational. Ignoring half the kinetic energy is a major error. **(d)** The error matters whenever the speed prediction feeds into a subsequent calculation. If you are predicting whether a ball clears a gap at the bottom of a ramp, an 18% speed overestimate translates to a 40% overestimate of range (since range scales as $v^2$ for projectile motion). In engineering, an 18% error in speed means a 40% error in kinetic energy, which could mean the difference between a safe design and a failure. The error also matters conceptually: the student's model says the ball behaves like a sliding block, which misses the entire physics of rolling.

Reflection

After this chapter, how do you decide which conservation law to use?

Think about the problems you solved in this section. In Problem A (rolling down a ramp), energy was the right tool because you wanted speed and energy was conserved. In Problem B (turntable collision), angular momentum was the right tool because you wanted the final angular velocity and there were no external torques. In the skater problem (Practice 2b), angular momentum was again the right tool --- but energy was not conserved, even though nothing seemed to be lost. (The skater's muscles added energy.)

The decision framework is not complicated, but it requires discipline:

What does the problem ask for? (Speed, distance, angular velocity, acceleration, time, a force?)

Is energy conserved? (Are the only forces conservative? Does friction do work? Do internal forces change the system's KE?)

Is angular momentum conserved? (Is the net external torque zero about some axis?)

Which method's outputs match what the problem asks for?

If you can answer these four questions before writing any equations, you will almost always choose the right method on the first try.

Chapter Summary: Rotational Work, Energy, and Angular Momentum

This chapter built the rotational counterparts of the energy and momentum tools you learned in Chapters 7 and 8 --- and showed how translational and rotational versions combine in real systems.

Section 11.1: Rotational kinetic energy. A spinning object carries kinetic energy $KE_{\text{rot}} = \frac{1}{2}I\omega^2$, structurally identical to $\frac{1}{2}mv^2$. For objects that both translate and rotate, the total kinetic energy is the sum: $KE_{\text{total}} = \frac{1}{2}mv_{\text{cm}}^2 + \frac{1}{2}I_{\text{cm}}\omega^2$.

Section 11.2: Work done by torques. Torques do work through angular displacement: $W = \int \tau\,d\theta$. For constant torque, $W = \tau\Delta\theta$. This parallels $W = \int F\,dx$ for translational work and leads to the rotational work-energy theorem: $W_{\text{net}} = \Delta KE_{\text{rot}}$.

Section 11.3: Angular momentum. For a rigid body spinning about a fixed axis, $L = I\omega$. For a particle, $\vec{L} = \vec{r} \times m\vec{v}$. Angular momentum is the rotational analog of linear momentum $\vec{p} = m\vec{v}$.

Section 11.4: Angular impulse. A torque applied over time changes angular momentum: $\int \tau\,dt = \Delta L$. This is the rotational analog of the impulse-momentum theorem $\int F\,dt = \Delta p$.

Section 11.5: Conservation of angular momentum. When no net external torque acts on a system, its total angular momentum is conserved: $L_i = L_f$. When a system's moment of inertia changes (skater pulling arms in, collapsing star), angular velocity adjusts to keep $L$ constant.

Section 11.6: Method selection for rolling and rotating systems. For rolling without slipping, $KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ with $v = R\omega$. Energy methods bypass force details and give speed directly. Force/torque methods give acceleration and specific forces. Angular momentum methods handle collisions and reconfigurations where no external torques act. The best method depends on what the problem asks for.

The hero concept of this chapter is the conservation architecture: the same pattern (work $\to$ energy, impulse $\to$ momentum, conservation when external influence is zero) repeats in both translational and rotational settings. The two domains are not separate theories --- they are two instances of the same structure, and for rolling objects, they operate simultaneously.

Chapter-End Retrieval

Close your notes. Answer these from memory.

1. What is rotational kinetic energy? Write the formula and identify what each variable represents.

2. What is angular momentum for a rigid body rotating about a fixed axis? Write the formula.

3. When is angular momentum conserved? State the condition precisely.

4. How do translational and rotational kinetic energies combine for a rolling object? Write the total kinetic energy and explain the rolling constraint that links $v$ and $\omega$.

5. A solid sphere and a hollow sphere of the same mass and radius roll down the same ramp. Which is faster at the bottom? Why?

6. For a rolling-without-slipping problem, when would you choose energy conservation over force/torque analysis? When would you choose force/torque analysis over energy?

After you have attempted all six, check your answers against the chapter summary above.

Looking Ahead

This chapter completes your rotational dynamics toolkit. You now have the rotational versions of every major concept from translational dynamics: Newton's second law for rotation ($\sum \tau = I\alpha$, Chapter 10), rotational kinetic energy and work ($\frac{1}{2}I\omega^2$ and $W = \int\tau\,d\theta$), angular momentum and angular impulse ($L = I\omega$ and $\Delta L = \int\tau\,dt$), and conservation of angular momentum.

Chapter 12 shifts the focus to a specific and practically important case: static equilibrium. When a rigid body is not accelerating and not rotating, both $\sum \vec{F} = \vec{0}$ and $\sum \vec{\tau} = \vec{0}$ must hold simultaneously. These two conditions together govern the design of bridges, buildings, cranes, and every structure that is meant to stay still under load. The physics is a special case of what you already know --- but the problem-solving challenges are rich, because statics problems demand careful geometric reasoning and strategic axis choice. The setup matters more than the algebra, and Section 10.7's lesson about choosing axes will prove essential.