Rotation of a Rigid Body - Vocabulary Flashcards

Chapter 12: Rotation of a Rigid Body

Preview

• Rigid Body: An object whose size and shape remain constant during movement.

• Characterized by its moment of inertia $I$, which is the rotational equivalent of mass.

• Topics include rotation about an axle and rolling without slipping.

• Connects to Section 6.1 on Equilibrium.

Torque

• Torque: The tendency or ability of a force to cause rotation around a pivot point.

• Depends on the magnitude and point of application of the force.

• A longer wrench provides a larger torque.

Torque and Angular Acceleration

• Torque's role in rotation is analogous to force in linear motion.

• Torque $τ$ causes angular acceleration.

• Newton’s Second Law for Rotation: $\alpha = \frac{\tau}{I}$.

• Rotational dynamics is analogous to linear dynamics.

• Connects to Section 6.2 on Newton’s Second Law.

Angular Momentum

• Angular Momentum: The rotational equivalent of linear momentum, representing an object's tendency to continue rotating.

• It is a vector pointing along the rotation axis.

• Used to understand the precession of spinning objects like tops or gyroscopes.

Conservation Laws in Rotational Motion

• Mechanical Energy: Includes rotational kinetic energy, analogous to linear kinetic energy.

• Mechanical energy is conserved in frictionless, rotating systems.

• Angular momentum is conserved in isolated systems.

• Connects to Section 10.4 on Conservation of Energy and Section 11.2 on Conservation of Momentum.

Importance of Rigid-Body Motion

• Understanding rotational motion is crucial for various applications, including windmill turbines, gyroscopes, bicycle/car wheels, rotating molecules, and galaxies.

• Expands understanding of equilibrium by examining conditions under which objects do not rotate.

Types of Motion

• Three basic types of motion exist for a rigid body.

Rotation About the Center of Mass

• An unconstrained object with no net force rotates about its center of mass.

• The center of mass remains motionless, while other points undergo circular motion around it.

Finding the Center of Mass

• For an object made of particles, the center of mass is calculated using:

• $x<em>{cm} = \frac{1}{M} \sum</em>i m<em>i x</em>i$

• $y<em>{cm} = \frac{1}{M} \sum</em>i m<em>i y</em>i$

• Where $m_i$ is the mass of particle $i$.

• Calculating the center of mass is similar to calculating a grade point average; higher mass particles have a greater influence.

Example 12.1: Center of Mass of a Barbell

• A barbell consists of a 500 g ball and a 2.0 kg ball connected by a massless 50-cm-long rod.

• Problem: Find the center of mass and the speed of each ball if they rotate about the center of mass at 40 rpm.

• Model: The barbell is modeled as a rigid body.

• The center of mass is found to be 20% of the way from the 2.0 kg ball to the 0.50 kg ball.

• The tangential velocities are $(v<em>i)</em>t = r_i \omega$, where $\omega$ is in rad/s.

• Conversion: Convert rpm to rad/s using $\omega = (40 \frac{rev}{min})(\frac{2\pi \ rad}{1 \ rev})(\frac{1 \ min}{60 \ s}) = 4.19 \ rad/s$

Finding Center of Mass by Integration

• Divide the object into small cells of mass $dm$.

• The sums become integrals: $x<em>{cm} = \frac{1}{M} \int x dm$ and $y</em>{cm} = \frac{1}{M} \int y dm$

• Replace $dm$ with expressions using $dx$ and $dy$.

• Establish integration limits.

Example 12.2: Center of Mass of a Rod

• Problem: Find the center of mass of a thin, uniform rod of length $L$ and mass $M$, and the tangential acceleration of one tip of a 1.60-m-long rod that rotates about its center of mass with an angular acceleration of 6.0 rad/s^2.

• The rod lies along the x-axis from 0 to $L$, and $y_{cm} = 0$.

• The mass of a small cell is $dm = \frac{M}{L} dx$.

• $x<em>{cm} = \frac{1}{M} \int</em>0^L x dm = \frac{1}{M} \int<em>0^L x \frac{M}{L} dx = \frac{1}{L} \int</em>0^L x dx = \frac{1}{L} [\frac{1}{2}x^2]_0^L = \frac{L}{2}$

• The center of mass is at the center of the rod.

• The tangential acceleration is $a_t = r \alpha = (0.80 \ m)(6.0 \ rad/s^2) = 4.8 \ m/s^2$

Quick Check 12.3

• A baseball bat is cut in half at its center of mass. The hitting end is heavier.

Rotational Energy

• A rotating object has kinetic energy due to the motion of its particles.

• Rotational Kinetic Energy: $K_{rot} = \frac{1}{2} I \omega^2$

• Moment of Inertia: $I = \sum<em>i m</em>i r_i^2$

• Units of moment of inertia are kg*m^2.

• Moment of inertia depends on the axis of rotation; mass farther from the axis contributes more.

Calculating Moment of Inertia

• Divide the object into small cells of mass $\Delta m$ and let $\Delta m \rightarrow 0$; $I = \int r^2 dm$

• Moments of inertia for common shapes are tabulated.

Moments of Inertia Table

• Table 12.2 lists moments of inertia for objects with uniform density.

Consequences of Moment of Inertia

• Objects with lower moment of inertia are easier to spin up, while those with higher moment of inertia are harder to spin up.

Quick Check 12.4

• Dumbbell B has the larger moment of inertia because the distance from the axis is more important than mass.

The Parallel-Axis Theorem

• Used to find the moment of inertia about an axis in an unusual position if you know the moment of inertia about a parallel axis through the center of mass.

Example 12.4: The Speed of a Rotating Rod

• Problem: A 1.0-m-long, 200 g rod is hinged at one end and released horizontally. What is the speed of the tip of the rod as it hits the wall?

• Model: Mechanical energy is conserved if the hinge is frictionless.

• The initial conditions are $K<em>i = 0$ and $U</em>i = mg \frac{L}{2}$.

• The final conditions are $U_f = 0$.

• For a rod rotating about one end: $I = \frac{1}{3}ML^2$

• $U<em>i = K</em>f \Rightarrow mg \frac{L}{2} = \frac{1}{2}I \omega^2 = \frac{1}{2} (\frac{1}{3}ML^2) \omega^2$

• $\omega = \sqrt{\frac{3g}{L}} = \sqrt{\frac{3(9.8 \ m/s^2)}{1.0 \ m}} = 5.42 \ rad/s$

• The final speed is $v = \omega L = (5.42 \ rad/s)(1.0 \ m) = 5.42 \ m/s$

Torque

• Torque: The effectiveness of a force at causing rotation, analogous to force in linear motion.

• Torque $τ$ is defined as $τ = rF \sin \phi$.

• SI units of torque are N*m; English units are foot-pounds.

• The ability of a force to cause rotation depends on the magnitude of the force $F$, the distance $r$ from the application point to the pivot, and the angle at which the force is applied.

Torque Sign Convention

• Torque has a sign; it can be positive or negative depending on the direction of rotation.

Moment Arm

• Torque is often calculated using the moment arm or lever arm, $τ = rF \sin \phi$.

Example 12.8: Applying a Torque

• Problem: Luis uses a 20-cm-long wrench to turn a nut. The wrench handle is tilted 30° above the horizontal, and Luis pulls straight down with a force of 100 N. How much torque does Luis exert?

• The tangential component of the force is $F_t = F \sin \phi = (100 \ N) \sin(-30°) = -50 \ N$.

• The torque is $τ = r F_t = (0.20 \ m)(-50 \ N) = -10 \ N \cdot m$.

Gravitational Torque

• The torque due to gravity is found by treating the object as if all its mass is concentrated at the center of mass.

Rotational Dynamics

• A net torque causes an object to have angular acceleration.

• In the absence of a net torque, the object either does not rotate ($\omega = 0$) or rotates with constant angular velocity ($\omega = constant$).

Analogies Between Linear and Rotational Dynamics

Example 12.11: Starting an Airplane Engine

• Problem: An engine has a torque of 60 N*m and drives a 2.0-m-long, 40 kg propeller. How long does it take to reach 200 rpm?

• Model: The propeller is a rigid rod rotating about its center.

• The moment of inertia is $I = \frac{1}{12}ML^2 = \frac{1}{12}(40 \ kg)(2.0 \ m)^2 = 13.3 \ kg \cdot m^2$.

• The angular acceleration is $\alpha = \frac{\tau}{I} = \frac{60 \ N \cdot m}{13.3 \ kg \cdot m^2} = 4.5 \ rad/s^2$.

• The time needed is $t = \frac{\omega_f}{\alpha} = \frac{20.9 \ rad/s}{4.5 \ rad/s^2} = 4.6 \ s$.

Constraints Due to Ropes and Pulleys

• If a rope does not slip as a pulley rotates, the tangential velocity and acceleration of the rim of the pulley must match the motion of the object: $v = r \omega$ and $a = r \alpha$.

Static Equilibrium

• A rigid body is in static equilibrium if there is no net force and no net torque.

• For total equilibrium, there is no net torque about any point.

Quick Check 12.8

• Which object is in static equilibrium?

Quick Check 12.9

• What does the scale read?

Rolling without Slipping

• Rolling is a combination of rotation and translation. The translation of the center of mass and angular velocity are related by $v_{cm} = R \omega$.

• The velocity of a particle is the velocity of the center of mass plus the velocity relative to the center of mass.

• $v<em>{top} = 2v</em>{cm}$, $v_{bottom} = 0$

• The point on the bottom of a rolling object is instantaneously at rest.

Quick Check 12.10

• A wheel rolls without slipping.

Kinetic Energy of Rolling

• The kinetic energy of a rolling object is $K = \frac{1}{2}Mv<em>{cm}^2 + \frac{1}{2}I</em>{cm} \omega^2$

• Rolling motion is translation of the center of mass plus rotation about the center of mass.

The Angular Velocity Vector

• The magnitude of the angular velocity vector is $\omega$.

• The angular velocity vector points along the axis of rotation in the direction given by the right-hand rule.

The Cross Product of Two Vectors

• The cross product of vectors $\vec{A}$ and $\vec{B}$ with angle $\alpha$ between them is $\vec{C} = \vec{A} \times \vec{B}$.

The Right-Hand Rule

• The cross product is perpendicular to the plane of $\vec{A}$ and $\vec{B}$.

• $\vec{A} \times \vec{B} = - \vec{B} \times \vec{A}$

The Torque Vector

• Torque can also be described as the cross product of the position vector and the force vector: $\vec{\tau} = \vec{r} \times \vec{F}$.

Angular Momentum of a Particle

• The angular momentum vector of a particle is defined as $\vec{L} = \vec{r} \times \vec{p}$.

• Torque causes a particle’s angular momentum to change: $\vec{\tau} = \frac{d\vec{L}}{dt}$.

Angular Momentum of a Rigid Body

• For a rigid body rotating on a fixed axle or about an axis of symmetry, $\vec{L} = I \vec{\omega}$.

• $\vec{\tau}_{net} = \frac{d\vec{L}}{dt}$

Analogies Between Linear and Angular Momentum and Energy

Conservation of Angular Momentum

• The angular momentum of an isolated system is conserved.

• If there is no net torque, $\vec{L}<em>f = \vec{L}</em>i$ where both magnitude and direction are unchanged.

Conservation of Angular Momentum Examples

• An ice skater reduces her moment of inertia $I$ by drawing in her arms, thus increasing her angular speed $\omega$ to conserve angular momentum.

Quick Check 12.11

• Two buckets spin around in a horizontal circle on frictionless bearings, and it starts to rain. The buckets slow down because the angular momentum of the bucket + rain system is conserved.

Preview

• Rigid Body: An object whose size and shape remain constant during movement.

• Characterized by its moment of inertia $I$, which is the rotational equivalent of mass.

• Topics include rotation about an axle and rolling without slipping.

• Connects to Section 6.1 on Equilibrium.

Torque

• Torque: The tendency or ability of a force to cause rotation around a pivot point.

• Depends on the magnitude and point of application of the force.

• A longer wrench provides a larger torque.

Torque and Angular Acceleration

• Torque's role in rotation is analogous to force in linear motion.

• Torque $τ$ causes angular acceleration.

• Newton’s Second Law for Rotation: $\alpha = \frac{\tau}{I}$.

• Rotational dynamics is analogous to linear dynamics.

• Connects to Section 6.2 on Newton’s Second Law.

Angular Momentum

• Angular Momentum: The rotational equivalent of linear momentum, representing an object's tendency to continue rotating.

• It is a vector pointing along the rotation axis.

• Used to understand the precession of spinning objects like tops or gyroscopes.

Conservation Laws in Rotational Motion

• Mechanical Energy: Includes rotational kinetic energy, analogous to linear kinetic energy.

• Mechanical energy is conserved in frictionless, rotating systems.

• Angular momentum is conserved in isolated systems.

• Connects to Section 10.4 on Conservation of Energy and Section 11.2 on Conservation of Momentum.

Importance of Rigid-Body Motion

• Understanding rotational motion is crucial for various applications, including windmill turbines, gyroscopes, bicycle/car wheels, rotating molecules, and galaxies.

• Expands understanding of equilibrium by examining conditions under which objects do not rotate.

Types of Motion

• Three basic types of motion exist for a rigid body.

Rotation About the Center of Mass

• An unconstrained object with no net force rotates about its center of mass.

• The center of mass remains motionless, while other points undergo circular motion around it.

Finding the Center of Mass

• For an object made of particles, the center of mass is calculated using:

• $x<em>{cm} = \frac{1}{M} \sum</em>i m<em>i x</em>i$

• $y<em>{cm} = \frac{1}{M} \sum</em>i m<em>i y</em>i$

• Where $m_i$ is the mass of particle $i$.

• Calculating the center of mass is similar to calculating a grade point average; higher mass particles have a greater influence.

Example 12.1: Center of Mass of a Barbell

• A barbell consists of a 500 g ball and a 2.0 kg ball connected by a massless 50-cm-long rod.

• Problem: Find the center of mass and the speed of each ball if they rotate about the center of mass at 40 rpm.

• Model: The barbell is modeled as a rigid body.

• The center of mass is found to be 20% of the way from the 2.0 kg ball to the 0.50 kg ball.

• The tangential velocities are $(v<em>i)</em>t = r_i \omega$, where $\omega$ is in rad/s.

• Conversion: Convert rpm to rad/s using $\omega = (40 \frac{rev}{min})(\frac{2\pi \ rad}{1 \ rev})(\frac{1 \ min}{60 \ s}) = 4.19 \ rad/s$

Finding Center of Mass by Integration

• Divide the object into small cells of mass $dm$.

• The sums become integrals: $x<em>{cm} = \frac{1}{M} \int x dm$ and $y</em>{cm} = \frac{1}{M} \int y dm$

• Replace $dm$ with expressions using $dx$ and $dy$.

• Establish integration limits.

Example 12.2: Center of Mass of a Rod

• Problem: Find the center of mass of a thin, uniform rod of length $L$ and mass $M$, and the tangential acceleration of one tip of a 1.60-m-long rod that rotates about its center of mass with an angular acceleration of 6.0 rad/s^2.

• The rod lies along the x-axis from 0 to $L$, and $y_{cm} = 0$.

• The mass of a small cell is $dm = \frac{M}{L} dx$.

• $x<em>{cm} = \frac{1}{M} \int</em>0^L x dm = \frac{1}{M} \int<em>0^L x \frac{M}{L} dx = \frac{1}{L} \int</em>0^L x dx = \frac{1}{L} [\frac{1}{2}x^2]_0^L = \frac{L}{2}$

• The center of mass is at the center of the rod.

• The tangential acceleration is $a_t = r \alpha = (0.80 \ m)(6.0 \ rad/s^2) = 4.8 \ m/s^2$

Quick Check 12.3

• A baseball bat is cut in half at its center of mass. The hitting end is heavier.

Rotational Energy

• A rotating object has kinetic energy due to the motion of its particles.

• Rotational Kinetic Energy: $K_{rot} = \frac{1}{2} I \omega^2$

• Moment of Inertia: $I = \sum<em>i m</em>i r_i^2$

• Units of moment of inertia are kg*m^2.

• Moment of inertia depends on the axis of rotation; mass farther from the axis contributes more.

Calculating Moment of Inertia

• Divide the object into small cells of mass $\Delta m$ and let $\Delta m \rightarrow 0$; $I = \int r^2 dm$

• Moments of inertia for common shapes are tabulated.

Moments of Inertia Table

• Table 12.2 lists moments of inertia for objects with uniform density.

Consequences of Moment of Inertia

• Objects with lower moment of inertia are easier to spin up, while those with higher moment of inertia are harder to spin up.

Quick Check 12.4

• Dumbbell B has the larger moment of inertia because the distance from the axis is more important than mass.

The Parallel-Axis Theorem

• Used to find the moment of inertia about an axis in an unusual position if you know the moment of inertia about a parallel axis through the center of mass.

Example 12.4: The Speed of a Rotating Rod

• Problem: A 1.0-m-long, 200 g rod is hinged at one end and released horizontally. What is the speed of the tip of the rod as it hits the wall?

• Model: Mechanical energy is conserved if the hinge is frictionless.

• The initial conditions are $K<em>i = 0$ and $U</em>i = mg \frac{L}{2}$.

• The final conditions are $U_f = 0$.

• For a rod rotating about one end: $I = \frac{1}{3}ML^2$

• $U<em>i = K</em>f \Rightarrow mg \frac{L}{2} = \frac{1}{2}I \omega^2 = \frac{1}{2} (\frac{1}{3}ML^2) \omega^2$

• $\omega = \sqrt{\frac{3g}{L}} = \sqrt{\frac{3(9.8 \ m/s^2)}{1.0 \ m}} = 5.42 \ rad/s$

• The final speed is $v = \omega L = (5.42 \ rad/s)(1.0 \ m) = 5.42 \ m/s$

Torque

• Torque: The effectiveness of a force at causing rotation, analogous to force in linear motion.

• Torque $τ$ is defined as $τ = rF \sin \phi$.

• SI units of torque are N*m; English units are foot-pounds.

• The ability of a force to cause rotation depends on the magnitude of the force $F$, the distance $r$ from the application point to the pivot, and the angle at which the force is applied.

Torque Sign Convention

• Torque has a sign; it can be positive or negative depending on the direction of rotation.

Moment Arm

• Torque is often calculated using the moment arm or lever arm, $τ = rF \sin \phi$.

Example 12.8: Applying a Torque

• Problem: Luis uses a 20-cm-long wrench to turn a nut. The wrench handle is tilted 30° above the horizontal, and Luis pulls straight down with a force of 100 N. How much torque does Luis exert?

• The tangential component of the force is $F_t = F \sin \phi = (100 \ N) \sin(-30°) = -50 \ N$.

• The torque is $τ = r F_t = (0.20 \ m)(-50 \ N) = -10 \ N \cdot m$.

Gravitational Torque

• The torque due to gravity is found by treating the object as if all its mass is concentrated at the center of mass.

Rotational Dynamics

• A net torque causes an object to have angular acceleration.

• In the absence of a net torque, the object either does not rotate ($\omega = 0$) or rotates with constant angular velocity ($\omega = constant$).

Analogies Between Linear and Rotational Dynamics

Example 12.11: Starting an Airplane Engine

• Problem: An engine has a torque of 60 N*m and drives a 2.0-m-long, 40 kg propeller. How long does it take to reach 200 rpm?

• Model: The propeller is a rigid rod rotating about its center.

• The moment of inertia is $I = \frac{1}{12}ML^2 = \frac{1}{12}(40 \ kg)(2.0 \ m)^2 = 13.3 \ kg \cdot m^2$.

• The angular acceleration is $\alpha = \frac{\tau}{I} = \frac{60 \ N \cdot m}{13.3 \ kg \cdot m^2} = 4.5 \ rad/s^2$.

• The time needed is $t = \frac{\omega_f}{\alpha} = \frac{20.9 \ rad/s}{4.5 \ rad/s^2} = 4.6 \ s$.

Constraints Due to Ropes and Pulleys

• If a rope does not slip as a pulley rotates, the tangential velocity and acceleration of the rim of the pulley must match the motion of the object: $v = r \omega$ and $a = r \alpha$.

Static Equilibrium

• A rigid body is in static equilibrium if there is no net force and no net torque.

• For total equilibrium, there is no net torque about any point.

Quick Check 12.8

• Which object is in static equilibrium?

Quick Check 12.9

• What does the scale read?

Rolling without Slipping

• Rolling is a combination of rotation and translation. The translation of the center of mass and angular velocity are related by $v_{cm} = R \omega$.

• The velocity of a particle is the velocity of the center of mass plus the velocity relative to the center of mass.

• $v<em>{top} = 2v</em>{cm}$, $v_{bottom} = 0$

• The point on the bottom of a rolling object is instantaneously at rest.

Quick Check 12.10

• A wheel rolls without slipping.

Kinetic Energy of Rolling

• The kinetic energy of a rolling object is $K = \frac{1}{2}Mv<em>{cm}^2 + \frac{1}{2}I</em>{cm} \omega^2$

• Rolling motion is translation of the center of mass plus rotation about the center of mass.

The Angular Velocity Vector

• The magnitude of the angular velocity vector is $\omega$.

• The angular velocity vector points along the axis of rotation in the direction given by the right-hand rule.

The Cross Product of Two Vectors

• The cross product of vectors $\vec{A}$ and $\vec{B}$ with angle $\alpha$ between them is $\vec{C} = \vec{A} \times \vec{B}$.

The Right-Hand Rule

• The cross product is perpendicular to the plane of $\vec{A}$ and $\vec{B}$.

• $\vec{A} \times \vec{B} = - \vec{B} \times \vec{A}$

The Torque Vector

• Torque can also be described as the cross product of the position vector and the force vector: $\vec{\tau} = \vec{r} \times \vec{F}$.

Angular Momentum of a Particle

• The angular momentum vector of a particle is defined as $\vec{L} = \vec{r} \times \vec{p}$.

• Torque causes a particle’s angular momentum to change: $\vec{\tau} = \frac{d\vec{L}}{dt}$.

Angular Momentum of a Rigid Body

• For a rigid body rotating on a fixed axle or about an axis of symmetry, $\vec{L} = I \vec{\omega}$.

• $\vec{\tau}_{net} = \frac{d\vec{L}}{dt}$

Analogies Between Linear and Angular Momentum and Energy

Conservation of Angular Momentum

• The angular momentum of an isolated system is conserved.

• If there is no net torque, $\vec{L}<em>f = \vec{L}</em>i$ where both magnitude and direction are unchanged.

Conservation of Angular Momentum Examples

• An ice skater reduces her moment of inertia

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Chapter 12: Rotation of a Rigid Body

Preview

• Rigid Body: An object whose size and shape remain constant during movement.

  • Defined by consistent mass distribution relative to each other.

  • Essential for simplified dynamics; deformation is negligible.

• Characterized by its moment of inertia $I$, which is the rotational equivalent of mass.

  • Represents resistance to rotational acceleration.

  • Depends on mass distribution and axis of rotation.

• Topics include rotation about an axle and rolling without slipping.

  • Rotation about an axle involves fixed axis motion.

  • Rolling without slipping combines rotation and translation, where the point of contact is instantaneously at rest.

• Connects to Section 6.1 on Equilibrium.

  • Equilibrium is when net force and net torque are zero, ensuring no translational or rotational acceleration.

Torque

• Torque: The tendency or ability of a force to cause rotation around a pivot point.

  • Not just force but also its distance from the pivot.

  • Measured in Newton-meters (N·m).

• Depends on the magnitude and point of application of the force.

  • Greater force or longer lever arm gives larger torque.

  • Torque is maximized when force is perpendicular to the lever arm.

• A longer wrench provides a larger torque.

  • Easier to turn bolts with longer handles due to increased torque.

Torque and Angular Acceleration

• Torque's role in rotation is analogous to force in linear motion.

  • Just as force causes linear acceleration, torque causes angular acceleration.

• Torque $τ$ causes angular acceleration.

  • Greater torque results in greater angular acceleration if inertia is constant.

  • The relationship is quantitative, described by Newton's second law for rotation.

• Newton’s Second Law for Rotation: $\alpha = \frac{\tau}{I}$.

  • Angular acceleration is directly proportional to net torque and inversely proportional to moment of inertia.

  • Analogous to $a = F/m$ in linear motion.

• Rotational dynamics is analogous to linear dynamics.

  • Concepts like work, energy, and power have rotational equivalents.

  • Analyzing rotational systems can parallel linear systems.

• Connects to Section 6.2 on Newton’s Second Law.

  • Newton's Second Law forms the basis for understanding both linear and rotational dynamics.

Angular Momentum

• Angular Momentum: The rotational equivalent of linear momentum, representing an object's tendency to continue rotating.

  • Defined as $L = I \omega$ for rigid objects.

  • It is a vector quantity; its direction matters.

• It is a vector pointing along the rotation axis.

  • Direction determined by the right-hand rule.

  • Indicates the orientation of rotation.

• Used to understand the precession of spinning objects like tops or gyroscopes.

  • Explains why spinning objects maintain stability.

  • Precession is the slow change in the orientation of the rotation axis.

Conservation Laws in Rotational Motion

• Mechanical Energy: Includes rotational kinetic energy, analogous to linear kinetic energy.

  • Total mechanical energy is the sum of translational kinetic, rotational kinetic, and potential energy.

  • Rotational kinetic energy: $\frac{1}{2}I\omega^2$

• Mechanical energy is conserved in frictionless, rotating systems.

  • If no external torques do work, total mechanical energy remains constant.

  • Initial energy equals final energy: $K<em>i + U</em>i = K<em>f + U</em>f$

• Angular momentum is conserved in isolated systems.

  • If no net external torque acts, angular momentum remains constant.

  • $I<em>i \omega</em>i = I<em>f \omega</em>f$

• Connects to Section 10.4 on Conservation of Energy and Section 11.2 on Conservation of Momentum.

  • These conservation laws are fundamental principles in physics.

Importance of Rigid-Body Motion

• Understanding rotational motion is crucial for various applications, including windmill turbines, gyroscopes, bicycle/car wheels, rotating molecules, and galaxies.

  • Wind turbines convert wind energy into rotational kinetic energy into electrical energy.

  • Gyroscopes maintain orientation due to angular momentum conservation.

  • Molecular rotations affect material properties and chemical reactions.

• Expands understanding of equilibrium by examining conditions under which objects do not rotate.

  • Static equilibrium requires both net force and net torque to be zero.

Types of Motion

• Three basic types of motion exist for a rigid body.

  1. Translation: Movement in a straight line without rotation.

  2. Rotation: Movement about a fixed axis.

  3. General Motion: Combination of translation and rotation.

Rotation About the Center of Mass

• An unconstrained object with no net force rotates about its center of mass.

  • The axis of rotation naturally passes through the center of mass.

• The center of mass remains motionless, while other points undergo circular motion around it.

  • Simplifies analysis, as the center of mass's motion describes the overall translation.

Finding the Center of Mass

• For an object made of particles, the center of mass is calculated using:

  • Weighted average of the positions of all particles.

• $x<em>{cm} = \frac{1}{M} \sum</em>i m<em>i x</em>i$

  • X-coordinate of the center of mass.

  • Sum of each particle's mass times its x-coordinate, divided by total mass.

• $y<em>{cm} = \frac{1}{M} \sum</em>i m<em>i y</em>i$

  • Y-coordinate of the center of mass.

  • Sum of each particle's mass times its y-coordinate, divided by total mass.

• Where $m*i$ is the mass of particle $i$.

  • Each particle contributes to the overall center of mass.

• Calculating the center of mass is similar to calculating a grade point average; higher mass particles have a greater influence.

  • The location of the center of mass shifts toward more massive particles.

Example 12.1: Center of Mass of a Barbell

• A barbell consists of a 500 g ball and a 2.0 kg ball connected by a massless 50-cm-long rod.

  • Simplified model to illustrate center of mass calculation.

• Problem: Find the center of mass and the speed of each ball if they rotate about the center of mass at 40 rpm.

  • Calculating both position and velocity adds complexity.

• Model: The barbell is modeled as a rigid body.

  • Simplifies analysis assuming constant shape.

• The center of mass is found to be 20% of the way from the 2.0 kg ball to the 0.50 kg ball.

  • Close to heavy ball, as expected.

• The tangential velocities are $(v<em>i)</em>t = r_i \omega$, where $\\omega$ is in rad/s.

  • Relates tangential speed to distance from the center and angular velocity.

• Conversion: Convert rpm to rad/s using $\omega = (40 \frac{rev}{min})(\frac{2\pi \ rad}{1 \ rev})(\frac{1 \ min}{60 \ s}) = 4.19 \ rad/s$

  • Crucial step for correct unit handling.

Finding Center of Mass by Integration

• Divide the object into small cells of mass $dm$.

  • Infinitesimal mass elements.

• The sums become integrals: $x<em>{cm} = \frac{1}{M} \int x dm$ and $y</em>{cm} = \frac{1}{M} \int y dm$

  • Extend center of mass calculation to continuous objects.

• Replace $dm$ with expressions using $dx$ and $dy$.

  • Relate mass elements to spatial coordinates.

• Establish integration limits.

  • Define boundaries of the object.

Example 12.2: Center of Mass of a Rod

• Problem: Find the center of mass of a thin, uniform rod of length $L$ and mass $M$, and the tangential acceleration of one tip of a 1.60-m-long rod that rotates about its center of mass with an angular acceleration of 6.0 rad/s^2.

  • Two-part problem testing center of mass and rotational kinematics.

• The rod lies along the x-axis from 0 to $L$, and $y_{cm} = 0$.

  • Simplifies the geometry.

• The mass of a small cell is $dm = \frac{M}{L} dx$.

  • Relates mass to length.

• $x<em>{cm} = \frac{1}{M} \int</em>0^L x dm = \frac{1}{M} \int<em>0^L x \frac{M}{L} dx = \frac{1}{L} \int</em>0^L x dx = \frac{1}{L} [\frac{1}{2}x^2]_0^L = \frac{L}{2}$

  • Step-by-step integration.

• The center of mass is at the center of the rod.

  • As expected for a uniform rod.

• The tangential acceleration is $a_t = r \alpha = (0.80 \ m)(6.0 \ rad/s^2) = 4.8 \ m/s^2$

  • Applies rotational kinematic equation.

Quick Check 12.3

• A baseball bat is cut in half at its center of mass. The hitting end is heavier.

Rotational Energy

• A rotating object has kinetic energy due to the motion of its particles.

  • Each particle has kinetic energy; summing them gives rotational kinetic energy.

• Rotational Kinetic Energy: $K_{rot} = \frac{1}{2} I \omega^2$

  • Kinetic energy due to rotation.

• Moment of Inertia: $I = \sum<em>i m</em>i r_i^2$

  • Sum of mass times the square of the distance from the axis of rotation for each particle.

• Units of moment of inertia are kg*m^2.

• Moment of inertia depends on the axis of rotation; mass farther from the axis contributes more.

Calculating Moment of Inertia

• Divide the object into small cells of mass $\Delta m$ and let $\Delta m \rightarrow 0$; $I = \int r^2 dm$

  • Transition from discrete to continuous mass distribution.

• Moments of inertia for common shapes are tabulated.

  • Simplifies calculations for standard geometric shapes.

Moments of Inertia Table

• Table 12.2 lists moments of inertia for objects with uniform density.

Consequences of Moment of Inertia

• Objects with lower moment of inertia are easier to spin up, while those with higher moment of inertia are harder to spin up.

Quick Check 12.4

• Dumbbell B has the larger moment of inertia because the distance from the axis is more important than mass.

The Parallel-Axis Theorem

• Used to find the moment of inertia about an axis in an unusual position if you know the moment of inertia about a parallel axis through the center of mass.

Example 12.4: The Speed of a Rotating Rod

• Problem: A 1.0-m-long, 200 g rod is hinged at one end and released horizontally. What is the speed of the tip of the rod as it hits the wall?

• Model: Mechanical energy is conserved if the hinge is frictionless.

• The initial conditions are $K<em>i = 0$ and $U</em>i = mg \frac{L}{2}$.

• The final conditions are $U_f = 0$.

• For a rod rotating about one end: $I = \frac{1}{3}ML^2$

• $U<em>i = K</em>f \Rightarrow mg \frac{L}{2} = \frac{1}{2}I \omega^2 = \frac{1}{2} (\frac{1}{3}ML^2) \omega^2$

• $\omega = \sqrt{\frac{3g}{L}} = \sqrt{\frac{3(9.8 \ m/s^2)}{1.0 \ m}} = 5.42 \ rad/s$

• The final speed is $v = \omega L = (5.42 \ rad/s)(1.0 \ m) = 5.42 \ m/s$

Torque

• Torque: The effectiveness of a force at causing rotation, analogous to force in linear motion.

• Torque $τ$ is defined as $τ = rF \sin \phi$.

• SI units of torque are N*m; English units are foot-pounds.

• The ability of a force to cause rotation depends on the magnitude of the force $F$, the distance $r$ from the application point to the pivot, and the angle at which the force is applied.

Torque Sign Convention

• Torque has a sign; it can be positive or negative depending on the direction of rotation.

Moment Arm

• Torque is often calculated using the moment arm or lever arm, $τ = rF \sin \phi$.

Example 12.8: Applying a Torque

• Problem: Luis uses a 20-cm-long wrench to turn a nut. The wrench handle is tilted 30° above the horizontal, and Luis pulls straight down with a force of 100 N. How much torque does Luis exert?

• The tangential component of the force is $F_t = F \sin \phi = (100 \ N) \sin(-30°) = -50 \ N$.

• The torque is $τ = r F_t = (0.20 \ m)(-50 \ N) = -10 \ N \cdot m$.

Gravitational Torque

• The torque due to gravity is found by treating the object as if all its mass is concentrated at the center of mass.

Rotational Dynamics

• A net torque causes an object to have angular acceleration.

• In the absence of a net torque, the object either does not rotate ($\omega = 0$) or rotates with constant angular velocity ($\omega = constant$).

Analogies Between Linear and Rotational Dynamics

Example 12.11: Starting an Airplane Engine

• Problem: An engine has a torque of 60 N*m and drives a 2.0-m-long, 40 kg propeller. How long does it take to reach 200 rpm?

• Model: The propeller is a rigid rod rotating about its center.

• The moment of inertia is $I = \frac{1}{12}ML^2 = \frac{1}{12}(40 \ kg)(2.0 \ m)^2 = 13.3 \ kg \cdot m^2$.

• The angular acceleration is $\alpha = \frac{\tau}{I} = \frac{60 \ N \cdot m}{13.3 \ kg \cdot m^2} = 4.5 \ rad/s^2$.

• The time needed is $t = \frac{\omega_f}{\alpha} = \frac{20.9 \ rad/s}{4.5 \ rad/s^2} = 4.6 \ s$.

Constraints Due to Ropes and Pulleys

• If a rope does not slip as a pulley rotates, the tangential velocity and acceleration of the rim of the pulley must match the motion of the object: $v = r \omega$ and $a = r \alpha$.

Static Equilibrium

• A rigid body is in static equilibrium if there is no net force and no net torque.

• For total equilibrium, there is no net torque about any point.

Quick Check 12.8

• Which object is in static equilibrium?

Quick Check 12.9

• What does the scale read?

Rolling without Slipping

• Rolling is a combination of rotation and translation. The translation of the center of mass and angular velocity are related by $v_{cm} = R \omega$.

• The velocity of a particle is the velocity of the center of mass plus the velocity relative to the center of mass.

• $v<em>{top} = 2v</em>{cm}$, $v_{bottom} = 0$

• The point on the bottom of a rolling object is instantaneously at rest.

Quick Check 12.10

• A wheel rolls without slipping.

Kinetic Energy of Rolling

• The kinetic energy of a rolling object is $K = \frac{1}{2}Mv<em>{cm}^2 + \frac{1}{2}I</em>{cm} \omega^2$

• Rolling motion is translation of the center of mass plus rotation about the center of mass.

The Angular Velocity Vector

• The magnitude of the angular velocity vector is $\omega$.

• The angular velocity vector points along the axis of rotation in the direction given by the right-hand rule.

The Cross Product of Two Vectors

• The cross product of vectors $\vec{A}$ and $\vec{B}$ with angle $\alpha$ between them is $\vec{C} = \vec{A} \times \vec{B}$.

The Right-Hand Rule

• The cross product is perpendicular to the plane of $\vec{A}$ and $\vec{B}$.

• $\vec{A} \times \vec{B} = - \vec{B} \times \vec{A}$

The Torque Vector

• Torque can also be described as the cross product of the position vector and the force vector: $\vec{\tau} = \vec{r} \times \vec{F}$.

Angular Momentum of a Particle

• The angular momentum vector of a particle is defined as $\vec{L} = \vec{r} \times \vec{p}$.

• Torque causes a particle’s angular momentum to change: $\vec{\tau} = \frac{d\vec{L}}{dt}$.

Angular Momentum of a Rigid Body

• For a rigid body rotating on a fixed axle or about an axis of symmetry, $\vec{L} = I \vec{\omega}$.

• $\vec{\tau}_{net} = \frac{d\vec{L}}{dt}$

Analogies Between Linear and Angular Momentum and Energy

Conservation of Angular Momentum

• The angular momentum of an isolated system is conserved.

• If there is no net torque, $\vec{L}f = \vec{L}i$ where both magnitude and direction are unchanged.

Conservation of Angular Momentum Examples

• An ice skater reduces her moment of inertia $I$ by drawing in her arms, thus increasing her