Chapter 19: Torque and Rotational Motion

19.1 Why Rotation Requires New Tools

Everything in Chapters 6 through 18 dealt with objects treated as point masses or as objects that translate without rotating. A box sliding down a ramp, a ball launched from a cliff, a car braking on a highway — in all of these, every part of the object moves in the same way. The entire object translates.

But most real objects also rotate. A tire rolls without slipping. A planet spins on its axis while orbiting the Sun. A gymnast executes a somersault. A wrench turns a bolt. In all of these, different parts of the object move differently — points farther from the rotation axis move faster than points closer to it.

To describe and predict rotational motion, we need new quantities. The analogue of position becomes angular position. The analogue of velocity becomes angular velocity. The analogue of acceleration becomes angular acceleration. The analogue of force becomes torque. The analogue of mass becomes moment of inertia. And Newton’s second law \(F = ma\) becomes its rotational counterpart \(\tau = I\alpha\).

This parallelism is not a coincidence. It is the mathematical expression of a deep physical fact: rotational mechanics is not a separate subject from translational mechanics. It is the same mechanics, expressed in the language most natural for rotating systems.

19.2 Translational vs. Rotational Motion

Before introducing new quantities, it is worth making the distinction precise.

Translational (linear) motion occurs when all parts of an object move in the same direction with the same velocity at any given instant. The position of the whole object is described by the position of its center of mass.

Rotational motion occurs when an object spins about an axis. Each particle in the object moves in a circular arc around that axis. Particles at different distances from the axis trace circles of different radii, but all particles complete one revolution in the same time — they all have the same angular velocity.

Combined motion is the most general case: an object translates and rotates simultaneously. A rolling wheel is the classic example. Its center of mass moves forward (translational), while the wheel spins about its own axis (rotational). In this chapter, we focus primarily on pure rotation about a fixed axis. Chapter 19 builds the tools; Chapters 20 and 21 apply them to equilibrium and to objects in complex settings.

19.3 Angular Position and Angular Displacement

When a rigid object rotates about a fixed axis, every point on it traces a circular arc. We describe the object’s orientation using an angular position \(\theta\), measured from a chosen reference line. The reference line is usually chosen to be along the positive \(x\)-axis, and angles are measured counterclockwise as positive (following the standard mathematical convention).

The SI unit of angular position is the radian (rad). One radian is the angle subtended at the center of a circle by an arc length equal to the radius.

\[ \theta = \frac{s}{r} \]

where \(s\) is the arc length and \(r\) is the radius. This definition makes radians dimensionless — the units of arc length divide out. Nonetheless, we write “rad” as a reminder that we are using the radian measure rather than degrees.

Conversion: \(360° = 2\pi \text{ rad}\), so \(1 \text{ rad} = 57.3°\) and \(1° = \pi/180 \text{ rad}\).

The angular displacement is the change in angular position:

\[ \Delta\theta = \theta_f - \theta_i \]

Angular displacement is positive for counterclockwise rotation and negative for clockwise rotation (when viewed from the positive axis direction). Like linear displacement, it can be positive, negative, or zero, and it does not depend on the path — only on the initial and final orientations.

19.4 Angular Velocity

Angular velocity \(\omega\) (lowercase Greek omega) is the rate of change of angular position:

\[ \omega = \frac{\Delta\theta}{\Delta t} \]

For instantaneous angular velocity (at one specific moment), this becomes the limit as \(\Delta t \to 0\) — conceptually the same as the definition of instantaneous linear velocity.

The SI unit of angular velocity is radians per second (rad/s). Angular velocity is positive for counterclockwise rotation and negative for clockwise rotation.

Angular velocity is also related to the period \(T\) (time for one complete rotation) and frequency \(f\) (rotations per second, in Hz):

\[ \omega = \frac{2\pi}{T} = 2\pi f \]

19.5 Angular Acceleration

Angular acceleration \(\alpha\) (lowercase Greek alpha) is the rate of change of angular velocity:

\[ \alpha = \frac{\Delta\omega}{\Delta t} \]

The SI unit is radians per second squared (rad/s²). Angular acceleration is positive when the angular velocity is increasing (speeding up counterclockwise, or slowing down clockwise) and negative when the angular velocity is decreasing.

The rotational kinematic equations for constant angular acceleration are direct analogues of the linear kinematic equations from Chapter 8:

\[ \omega_f = \omega_i + \alpha t \]

\[ \theta = \omega_i t + \tfrac{1}{2}\alpha t^2 \]

\[ \omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta \]

\[ \Delta\theta = \tfrac{1}{2}(\omega_i + \omega_f)t \]

These four equations apply only when angular acceleration is constant — the rotational analogue of the condition required for the linear kinematic equations in Chapter 8. Every method, strategy, and caution from Chapter 8 applies here with the substitutions: \(x \to \theta\), \(v \to \omega\), \(a \to \alpha\).

Worked Example 19.1 — Rotational Kinematics

A motor starts from rest and reaches an angular velocity of 240 rad/s in 8.0 s at constant angular acceleration. Find: (a) the angular acceleration; (b) the total angle rotated during the 8.0 s; (c) the number of complete revolutions.

(a) Angular acceleration:

\[ \alpha = \frac{\Delta\omega}{\Delta t} = \frac{240 - 0}{8.0} = 30 \text{ rad/s}^2 \]

(b) Total angle rotated (use equation 2, with \(\omega_i = 0\)):

\[ \Delta\theta = \omega_i t + \tfrac{1}{2}\alpha t^2 = 0 + \tfrac{1}{2}(30)(8.0)^2 = 15 \times 64 = 960 \text{ rad} \]

(c) Number of revolutions:

\[ N = \frac{\Delta\theta}{2\pi} = \frac{960}{2\pi} \approx 152.8 \text{ rev} \]

19.6 Tangential vs. Angular Quantities

A point at distance \(r\) from the rotation axis moves along a circular arc. Its motion can be described either in terms of angular quantities (how many radians per second it sweeps through) or in terms of linear (tangential) quantities (how fast it actually moves along the arc). The two descriptions are linked by the radius \(r\).

Tangential (linear) speed of a point on a rotating object:

\[ v_t = r\omega \]

Tangential acceleration (rate of change of speed along the arc):

\[ a_t = r\alpha \]

Centripetal acceleration (directed inward, toward the axis — this is the acceleration from Chapter 13 required to keep the point moving in a circle):

\[ a_c = \frac{v_t^2}{r} = r\omega^2 \]

The total acceleration of a point on a rotating, accelerating object has two components: \(a_t\) (tangential, along the arc, changing the speed) and \(a_c\) (centripetal, perpendicular to the arc, changing the direction). These add as perpendicular vectors:

\[ a_\text{total} = \sqrt{a_t^2 + a_c^2} \]

Worked Example 19.2 — Tangential and Centripetal Acceleration

A disk of radius 0.50 m accelerates from rest at \(\alpha = 4.0\) rad/s². Find the tangential speed, tangential acceleration, and centripetal acceleration of a point on the rim at \(t = 3.0\) s.

Step 1 — Angular velocity at \(t = 3.0\) s:

\[ \omega = \omega_i + \alpha t = 0 + 4.0 \times 3.0 = 12.0 \text{ rad/s} \]

Step 2 — Tangential speed:

\[ v_t = r\omega = 0.50 \times 12.0 = 6.0 \text{ m/s} \]

Step 3 — Tangential acceleration:

\[ a_t = r\alpha = 0.50 \times 4.0 = 2.0 \text{ m/s}^2 \]

Step 4 — Centripetal acceleration:

\[ a_c = r\omega^2 = 0.50 \times (12.0)^2 = 0.50 \times 144 = 72 \text{ m/s}^2 \]

Step 5 — Total acceleration:

\[ a_\text{total} = \sqrt{(2.0)^2 + (72)^2} = \sqrt{4 + 5184} \approx 72.0 \text{ m/s}^2 \]

At this moment the centripetal component dominates because the disk is spinning fast. The tangential component is relatively small.

19.7 Radius and Rotational Kinematics

The relationship \(s = r\theta\) (arc length equals radius times angle in radians) is the bridge between the rotational kinematic quantities and their linear equivalents. It applies at every level: position, velocity, and acceleration.

\[ s = r\theta \qquad v_t = r\omega \qquad a_t = r\alpha \]

These three relations are all you need to translate between the rotational description (\(\theta\), \(\omega\), \(\alpha\)) and the tangential linear description (\(s\), \(v_t\), \(a_t\)) for any point at radius \(r\) on a rotating body.

A critically important application is rolling without slipping. When a wheel of radius \(r\) rolls without slipping on a flat surface, the contact point is instantaneously at rest, and the center of the wheel moves forward at speed:

\[ v_\text{cm} = r\omega \]

The wheel’s rotation and its translation are locked together. If the wheel spins faster (larger \(\omega\)), the center moves faster (larger \(v_\text{cm}\)) by the same factor. This constraint — \(v_\text{cm} = r\omega\) — connects the rotational and translational descriptions of rolling motion.

19.8 What Torque Is

When you push open a door, the door rotates about its hinges. Push at the edge of the door (far from the hinge) and it swings open easily. Push at a point close to the hinge and it is much harder to open. Push directly at the hinge and the door does not rotate at all, regardless of how hard you push.

The quantity that captures “how effective a force is at causing rotation” is called torque (symbol \(\tau\), lowercase Greek tau). Torque is the rotational analogue of force. Just as a net force causes linear acceleration, a net torque causes angular acceleration.

Torque depends on three things: 1. The magnitude of the force \(F\) 2. The distance from the pivot (rotation axis) to the point where the force is applied, called the moment arm or lever arm \(r\) 3. The angle between the force vector and the line from the pivot to the application point

A large force applied far from the axis, at a right angle to the radius, produces the largest possible torque. The same force applied closer to the axis, or at a non-perpendicular angle, produces less torque.

19.9 Lever Arm and Perpendicular Distance

The lever arm (also called the moment arm) is the perpendicular distance from the rotation axis to the line of action of the force. The line of action is the straight line along which the force vector acts, extended in both directions if necessary.

The lever arm \(d_\perp\) is related to the distance \(r\) from pivot to application point and the angle \(\theta\) between the force and the radius vector by:

\[ d_\perp = r\sin\theta \]

This is the key geometric identity in torque problems. When \(\theta = 90°\) (force perpendicular to the radius), the lever arm equals \(r\) — maximum torque. When \(\theta = 0°\) or \(180°\) (force parallel to the radius, pointing toward or away from the pivot), the lever arm is zero — no torque.

19.10 Calculating Torque

The magnitude of the torque produced by a force \(F\) applied at distance \(r\) from the pivot, at angle \(\theta\) between the force and the radius vector, is:

\[ \tau = rF\sin\theta \]

Equivalently, since the lever arm is \(d_\perp = r\sin\theta\):

\[ \tau = F \cdot d_\perp \]

Both forms are correct and useful. The first is easier to apply when \(r\) and \(\theta\) are given. The second is easier when the perpendicular distance to the line of action has been identified geometrically.

The SI unit of torque is the newton-meter (N·m). Note that this is the same unit as energy (joule = N·m), but torque is not energy — it is a different physical quantity that happens to have the same dimensions. In practice, torques are always written as N·m (never joules) to avoid confusion.

Sign convention for torque: Following the standard convention, torques that tend to produce counterclockwise rotation are positive; torques that tend to produce clockwise rotation are negative. (This matches the sign convention for angles and angular velocity.)

Worked Example 19.3 — Computing Torques

Three forces act on a disk that can rotate about its center: - Force \(F_1\) = 10.0 N applied at \(r_1\) = 0.30 m, perpendicular to the radius (tends to rotate counterclockwise). - Force \(F_2\) = 15.0 N applied at \(r_2\) = 0.20 m, at 40° to the radius (tends to rotate clockwise). - Force \(F_3\) = 8.0 N applied along the radius (\(\theta_3 = 0°\), passing through the center).

Torque from \(F_1\):

\[ \tau_1 = r_1 F_1 \sin 90° = 0.30 \times 10.0 \times 1 = +3.00 \text{ N·m (counterclockwise, positive)} \]

Torque from \(F_2\):

\[ \tau_2 = r_2 F_2 \sin 40° = 0.20 \times 15.0 \times 0.643 = -1.93 \text{ N·m (clockwise, negative)} \]

Torque from \(F_3\):

\[ \tau_3 = r_3 F_3 \sin 0° = 0.25 \times 8.0 \times 0 = 0 \text{ N·m (passes through pivot)} \]

Net torque:

\[ \tau_\text{net} = +3.00 + (-1.93) + 0 = +1.07 \text{ N·m (net counterclockwise)} \]

19.11 Net Torque and Rotational Acceleration

Just as the net force determines translational acceleration (\(F_\text{net} = ma\)), the net torque determines rotational acceleration. Newton’s second law for rotation states:

\[ \tau_\text{net} = I\alpha \]

where: - \(\tau_\text{net}\) is the net torque about the rotation axis (in N·m) - \(I\) is the moment of inertia of the object about that axis (in kg·m²) - \(\alpha\) is the angular acceleration (in rad/s²)

The net torque is the algebraic (signed) sum of all torques acting on the object:

\[ \tau_\text{net} = \sum_i \tau_i = \sum_i r_i F_i \sin\theta_i \]

If the net torque is positive (counterclockwise), the object gains positive angular acceleration (spins faster counterclockwise or slows down if spinning clockwise). If the net torque is zero, the object has zero angular acceleration — it may be rotating, but at constant angular velocity.

Worked Example 19.4 — Finding Angular Acceleration

A disk of moment of inertia \(I = 0.80\) kg·m² experiences a net torque of \(\tau_\text{net} = 3.2\) N·m. Find the angular acceleration.

\[ \alpha = \frac{\tau_\text{net}}{I} = \frac{3.2 \text{ N·m}}{0.80 \text{ kg·m}^2} = 4.0 \text{ rad/s}^2 \]

Check: Units: \(\frac{\text{N·m}}{\text{kg·m}^2} = \frac{\text{kg·m}^2/\text{s}^2}{\text{kg·m}^2} = \text{s}^{-2} = \text{rad/s}^2\) ✓

19.12 Moment of Inertia (Conceptual Introduction)

The moment of inertia \(I\) is the rotational analogue of mass. Just as mass measures resistance to translational acceleration (a larger mass requires a larger force to achieve the same acceleration), the moment of inertia measures resistance to rotational acceleration — how hard it is to spin an object faster or slower.

However, the moment of inertia differs from mass in one crucial way: it depends not just on how much mass an object has, but on how that mass is distributed relative to the rotation axis. Mass concentrated far from the axis contributes more to the moment of inertia than the same mass concentrated close to the axis.

For a single point mass \(m\) at distance \(r\) from the axis:

\[ I = mr^2 \]

For a collection of point masses:

\[ I = \sum_i m_i r_i^2 \]

For continuous objects, the sum becomes an integral — a topic handled fully in Volume I of the advanced series. For this book, we use the results of those integrals, listed in Table 19.4 as standard formulas.

19.13 Rotational Analogies to Newton’s Laws

The three laws of Newton that govern translational motion each have a precise rotational counterpart. The analogy is exact — every concept maps directly.

19.14 Common Mistakes in Rotational Analysis

Chapter Summary

Rotational motion is governed by the same physical principles as translational motion, expressed through rotational analogues of every linear quantity and law.

Angular kinematics (constant \(\alpha\)):

\[ \omega_f = \omega_i + \alpha t \qquad \Delta\theta = \omega_i t + \tfrac{1}{2}\alpha t^2 \qquad \omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta \]

Connecting angular and tangential quantities (at radius \(r\)):

\[ s = r\theta \qquad v_t = r\omega \qquad a_t = r\alpha \qquad a_c = r\omega^2 \]

Torque:

\[ \tau = rF\sin\theta = F \cdot d_\perp \]

Units: N·m. Positive = counterclockwise; negative = clockwise.

Rotational Newton’s second law:

\[ \tau_\text{net} = I\alpha \]

Core principles: - Angles must be in radians for all kinematic and dynamic calculations. - Torque depends on the lever arm (perpendicular distance from pivot to line of action), not just the distance \(r\) to the application point. - All points on a rigid body share the same \(\omega\) and \(\alpha\); tangential speed and acceleration scale with \(r\). - Moment of inertia \(I\) depends on both total mass and the distribution of that mass relative to the axis. - The rotational kinematic equations are direct analogues of the linear equations from Chapter 8 — use the same four-step method (list quantities, identify missing variable, select equation, solve).

Key Terms

angular position (\(\theta\))

The orientation of a rotating body measured from a reference line, in radians. Positive counterclockwise.

radian (rad)

The SI unit of angle. One radian is the angle subtended by an arc equal in length to the radius. \(2\pi\) rad = \(360°\).

angular displacement (\(\Delta\theta\))

The change in angular position: \(\Delta\theta = \theta_f - \theta_i\). A vector-like signed quantity.

angular velocity (\(\omega\))

The rate of change of angular position: \(\omega = \Delta\theta/\Delta t\). SI unit: rad/s.

angular acceleration (\(\alpha\))

The rate of change of angular velocity: \(\alpha = \Delta\omega/\Delta t\). SI unit: rad/s².

tangential speed (\(v_t\))

The linear speed of a point on a rotating object along its circular arc: \(v_t = r\omega\).

tangential acceleration (\(a_t\))

The component of acceleration along the arc (changing the speed): \(a_t = r\alpha\).

centripetal acceleration (\(a_c\))

The component of acceleration directed inward toward the axis (changing the direction): \(a_c = r\omega^2 = v_t^2/r\).

torque (\(\tau\))

The rotational effect of a force; the product of force, lever arm, and \(\sin\theta\): \(\tau = rF\sin\theta\). SI unit: N·m. The rotational analogue of force.

lever arm (moment arm)

The perpendicular distance from the rotation axis to the line of action of the force. Equal to \(r\sin\theta\).

line of action

The infinite straight line along which a force vector acts, extended in both directions.

moment of inertia (\(I\))

The rotational analogue of mass; measures resistance to angular acceleration; depends on total mass and its distribution relative to the axis: \(I = \sum m_i r_i^2\). SI unit: kg·m².

rotational Newton’s second law

\(\tau_\text{net} = I\alpha\); the net torque on an object equals its moment of inertia times its angular acceleration.

Concept Review Questions

Answer in complete sentences. No calculation required.

1. Every linear quantity in mechanics has a rotational analogue. List five pairs of analogous quantities and state what each member of the pair measures.

2. Why must angles be expressed in radians rather than degrees when using the relations \(v_t = r\omega\) and \(s = r\theta\)? What goes wrong if degrees are used?

3. A large disk and a small disk are both rotating at the same angular velocity. Which has the greater tangential speed at its rim? Explain using \(v_t = r\omega\).

4. Explain why a force applied directly through the pivot point of a rotating object produces zero torque, regardless of the magnitude of the force.

5. A wrench is used to tighten a bolt. Explain in terms of lever arm why pushing perpendicular to the wrench handle produces more torque than pushing at an angle.

6. Two objects have the same mass: a solid cylinder and a hollow cylinder of the same outer radius. Which has the larger moment of inertia? Explain physically why, without using the formulas.

7. A figure skater spins and then pulls her arms inward. Her moment of inertia decreases. What happens to her angular velocity? What physical principle governs this? (You do not need to know angular momentum formally — reason from \(\tau = I\alpha\) and the absence of external torques.)

8. State Newton’s three laws in their rotational form. For each one, give a physical example that illustrates it.

9. What is the difference between angular velocity and tangential velocity? Can two points on the same rotating rigid body have different angular velocities? Can they have different tangential velocities?

10. Explain why the moment of inertia of a rod rotating about its end (\(I = \frac{1}{3}mL^2\)) is greater than the moment of inertia of the same rod rotating about its center (\(I = \frac{1}{12}mL^2\)).

Angular Motion Problems

Use radians throughout. Show all unit conversions.

1. Convert the following to radians per second: (a) 33⅓ rpm, (b) 1500 rpm, (c) 0.5 rev/s, (d) 120°/s.

2. A wheel starts from rest and reaches 240 rpm in 8.0 s with constant angular acceleration. Find: (a) \(\omega_f\) in rad/s; (b) \(\alpha\); (c) total angle rotated; (d) number of revolutions.

3. A disk rotates at \(\omega_i = 25.0\) rad/s and decelerates at \(\alpha = -5.0\) rad/s². (a) When does it stop? (b) How many radians does it rotate before stopping? (c) How many revolutions?

4. A bicycle wheel (radius 0.35 m) accelerates from rest at \(\alpha = 2.0\) rad/s² for 4.0 s. Find: (a) final angular velocity; (b) tangential speed of the rim at \(t = 4.0\) s; (c) centripetal acceleration of a point on the rim at \(t = 4.0\) s; (d) total distance traveled by a point on the rim.

5. Two points are on a rotating disk: Point A is at \(r_A = 0.10\) m and Point B is at \(r_B = 0.30\) m. The disk rotates at \(\omega = 8.0\) rad/s. Find: (a) \(v_t\) for each point; (b) \(a_c\) for each point; (c) the angle (in degrees) swept by each point in 2.0 s.

6. A motor spins up from rest to 3000 rpm in 12.0 s. A grinding wheel of radius 0.15 m is attached to the motor shaft. Find: (a) angular acceleration; (b) tangential acceleration of the wheel’s rim; (c) centripetal acceleration of the rim at \(t = 12.0\) s; (d) total angle rotated in 12.0 s.

Torque Problems

Include the sign of each torque (+ for counterclockwise, − for clockwise).

1. A force of 25.0 N is applied at the end of a 0.60 m wrench. Find the torque if the force is applied: (a) perpendicular to the wrench; (b) at 60° to the wrench; (c) parallel to the wrench.

2. A 40.0 N force is applied at a point 0.50 m from a pivot. (a) What angle produces a torque of 15.0 N·m? (b) What angle produces zero torque?

3. Three forces act on a wheel of radius 0.25 m that is free to rotate about its center: \(F_1 = 12\) N applied tangentially counterclockwise; \(F_2 = 20\) N applied tangentially clockwise; \(F_3 = 15\) N applied radially outward. Find: (a) the torque from each force; (b) the net torque.

4. A torque of 8.0 N·m is required to open a jar lid. The lid has a radius of 4.0 cm. What force must be applied tangentially to the edge of the lid?

5. A beam of length 3.0 m can rotate about its left end. Forces are applied as follows: \(F_1 = 50\) N downward at 1.0 m from the left end; \(F_2 = 80\) N downward at 2.5 m from the left end; \(F_3 = 60\) N upward at 3.0 m from the left end. Find the net torque about the left end.

6. A disk of moment of inertia \(I = 0.50\) kg·m² experiences a net torque of 4.0 N·m counterclockwise. (a) Find \(\alpha\). (b) Starting from rest, what is \(\omega\) after 3.0 s? (c) How many revolutions in 3.0 s?

Lever Arm Exercises

For each, identify the pivot, find the lever arm using the perpendicular-distance method, and calculate the torque.

1. A 30.0 N force is applied to a door at 0.90 m from the hinge. Find the torque if the force is: (a) perpendicular to the door; (b) at 45° to the door surface.

2. The moment arm of gravity on a sign hanging from a horizontal rod: the sign hangs 0.40 m from the pivot, and its weight is 25.0 N acting straight down. What is the torque of gravity about the pivot?

3. A wrench of length 0.25 m is used to turn a bolt. A 90.0 N force is applied at the end of the wrench at 70° to the wrench axis. (a) Find the torque. (b) Find the perpendicular lever arm directly.

4. A 500 N person sits 1.2 m from the center of a seesaw. Another person sits 0.8 m from the center on the other side. What force must the second person exert to balance the torques? (Consider both sides — does the answer have to be their weight, or just the force they exert on the seesaw?)

5. A flagpole of mass 15.0 kg and length 4.0 m is horizontal and attached at one end to a wall. Its weight acts at its center of mass (2.0 m from the wall). A 100 N downward force is applied at the free end. Find the total torque about the wall attachment.

Mixed Rotational Analysis

1. A solid disk (mass 3.0 kg, radius 0.20 m) is initially at rest. A tangential force of 4.5 N is applied at its rim. Find: (a) torque; (b) moment of inertia; (c) angular acceleration; (d) angular velocity after 6.0 s; (e) tangential speed of rim at \(t = 6.0\) s.

2. A hollow cylinder (mass 2.0 kg, radius 0.15 m, \(I = mR^2\)) is mounted on a horizontal axle and is initially at rest. A 3.0 N tangential force is applied at the rim. (a) Find \(\tau\), \(I\), and \(\alpha\). (b) After 4.0 s, find \(\omega\) and \(v_t\) at the rim. (c) How does your answer for \(\alpha\) compare with a solid disk of the same mass and radius? Why is it different?

3. Two forces are applied to a rod of length 2.0 m that is pinned at its left end: \(F_1 = 20\) N downward at 0.50 m from the pin, and \(F_2 = 15\) N upward at 1.80 m from the pin. The rod has moment of inertia \(I = \frac{1}{3}mL^2\) with \(m = 4.0\) kg. Find: (a) the net torque about the pin; (b) the angular acceleration of the rod.

4. (Rolling wheel) A bicycle wheel (mass 1.5 kg, radius 0.33 m, \(I = mR^2\) for a thin ring) rolls without slipping at \(v_\text{cm} = 6.0\) m/s. Find: (a) \(\omega\); (b) \(KE_\text{translational} = \frac{1}{2}mv^2\); (c) \(KE_\text{rotational} = \frac{1}{2}I\omega^2\); (d) total kinetic energy.

5. (Conceptual) A yo-yo (treated as a solid disk) falls from rest on its string. As it falls, the string unwinds from the axle (small radius \(r\)) while the disk has larger outer radius \(R\). The yo-yo accelerates both translationally (its center moves down) and rotationally (it spins). Describe qualitatively: does the yo-yo fall with the same acceleration as an object in free fall? Would you expect a larger or smaller acceleration? Explain your reasoning without calculating.

Chapter 19 Checkpoint

Before moving to Chapter 20, confirm you can do each of the following without consulting your notes.

Continue to Chapter 20: Static Equilibrium →