Chapter 12: Rotation of a Rigid Body

Rigid Body:
An extended object with a fixed size and shape during its motion. Rigid bodies are modeled as collections of particles that are bound together through rigid, massless rods, representing interactions akin to atomic bonds. Understanding rigid body motion requires rethinking conventional definitions of mass and deriving new equations for kinetic energy and momentum.

Properties:

• Rigid bodies retain their shape and size, meaning they cannot be compressed, stretched, or deformed under applied forces.

• Throughout any motion, all points within the rigid body exhibit the same angular velocity ($\omega$) and angular acceleration ($\alpha$).

• This modeling is valid only under conditions where the rigid body remains intact; deformable bodies require different modeling approaches that account for internal stresses and strains.

Types of Motion:

1. Translational Motion: Movement along a path without rotation (e.g., moving in a straight line).

2. Rotational Motion: Circular movement around an axis (e.g., a spinning wheel).

3. Combined Motion: A mixture of both translational and rotational motion (e.g., a rolling ball).

Review: Rotational Kinematics

• Angular Displacement ($\theta$): The angle through which a point or line has been rotated in a specified sense about an axis; measured in radians, with counter-clockwise being the positive direction from the +x axis.

• Angular Velocity ($\omega$): The rate of change of angular displacement over time, defined by the equation:
  $\omega = \frac{d\theta}{dt}$
  Unit for angular velocity: radians per second (rad/s).

• Angular Acceleration ($\alpha$): The rate of change of angular velocity over time, given by:
  $\alpha = \frac{d^2\theta}{dt^2}$
  Unit: radians per second squared (rad/s$^{2}$). Both $\omega$ and $\alpha$ remain constant across all points in a rigid body during motion.

Sign Conventions:

• Positive $\theta$ is designated for counter-clockwise rotations from the +x axis.

• Positive $\omega$ corresponds to counter-clockwise motion; negative $\omega$ indicates clockwise motion.

• The sign of $\alpha$ depends on the current direction of $\omega$ and whether the object is accelerating either positively (speeding up) or negatively (slowing down). For instance, if a wheel is speeding up with a positive $\omega$, then \alpha > 0, whereas if the wheel is slowing down, \alpha < 0.

Relationships Linking Instantaneous Velocity and Acceleration to Rotational Values:

• Tangential velocity: $v_t = r\omega$, where $r$ is the radius.

• Radial velocity: $v_r = 0$, indicating points in a rigid body do not move inward or outward.

• Tangential acceleration: $a_t = r\alpha$.

• Radial (centripetal) acceleration: $a<em>r = \frac{v</em>t^2}{r} = r\omega^2$.

Total Velocity and Acceleration:

• Overall velocity and acceleration are vector sums combining tangential and radial components, represented as:
  $\vec{v} = \vec{v}<em>t + \vec{v}</em>r$
  $\vec{a} = \vec{a}<em>t + \vec{a}</em>r$.

Kinematic Equations for Rotational Motion (constant $\alpha$):

1. $\theta<em>f = \theta</em>0 + \omega_i t + \frac{1}{2} \alpha t^2$

2. $\omega<em>f = \omega</em>i + \alpha t$

3. $\omega<em>f^2 = \omega</em>i^2 + 2 \alpha \Delta \theta$
   These equations assume constant angular acceleration, becoming invalid if $\alpha$ varies during motion.

Center of Mass

• The center of mass represents the mass-weighted central point of an object. An unconstrained object with no net external force tends to rotate about its center of mass while maintaining its translational motion.

• In a finite particle system, the center of mass can be calculated by summing the products of each particle's mass and its displacement, then dividing by the total mass, allowing the center of mass to shift toward larger mass particles.

• For rigid bodies or continuous objects, integration methods are employed to find the center of mass:
  $x<em>{CM} = \frac{1}{M} \int x dm$ $y</em>{CM} = \frac{1}{M} \int y dm$.
  This method accounts for the distribution of mass across the object, ensuring precise calculations necessary for understanding motion dynamics and stability.

Rotational Energy

• Origin: Arises from the motion of particles within an object rotating about an axis.

• Calculation: Defined by looking at the motion of individual particles; the total kinetic energy is the sum of the energies of these particles.

• Setup: Consider an object rotating counterclockwise on an axle (not necessarily at the center of mass). Highlight three points on the object, each moving with tangential velocity $v_{t,i}$; velocities differ due to differing radii $r$’s.

• Kinetic Energy of Particle $i$:
  $K<em>i = \frac{1}{2} m</em>i v<em>{i,t}^2 = \frac{1}{2} m</em>i r_i \omega^2$. Here, $\omega$ is constant.

• Total Kinetic Energy:
  $K<em>{tot} = K</em>1 + K<em>2 + … = \frac{1}{2} m</em>1 r<em>1^2 \omega^2 + \frac{1}{2} m</em>2 r<em>2^2 \omega^2 + …$ Thus, $K</em>{tot} = \sum \frac{1}{2} m<em>i r</em>i^2 \omega^2 = \omega^2 \sum \frac{1}{2} m<em>i r</em>i^2$.

• Moment of Inertia ($I$):
  $I = \sum m<em>i r</em>i^2$. Units are kg$\cdot$m$^{2}$.

• Total Kinetic Energy of the Object:
  $K_{tot} = \frac{1}{2} I \omega^2$; uses the same kinetic energy formula as before but presented differently.

• Gravitational Potential Energy: Depends on the location of its center of mass. If the object is rotating about an axle not located at its center of mass, the center of mass may move up and down, and the change in potential energy is given by:
  $\Delta U<em>g = M g \Delta y</em>{cm}$.

• Total Energy:
  $E<em>{tot} = K</em>{tot} + \Delta U<em>g = \frac{1}{2} I \omega^2 + M g \Delta y</em>{cm}$.

Moment of Inertia

• Facts About Moment of Inertia:

  • Rotational equivalent of mass (also known as inertial mass).

  • Dependent on the axis of rotation; not an inherent property of the object.

  • Variation with size and shape of the object is significant; common shapes have known moments of inertia listed in tables, while others require calculations.

  • A larger moment of inertia results in greater difficulty in rotating the object, analogous to how a large mass is hard to accelerate.

• Calculating Moment of Inertia of a Rigid Body:

  • Nontrivial; requires knowledge of total mass, mass distribution, shape, and location of the rotation axis.

  • For an object rotating CCW about an axle not at its center of mass, divide the object into small pieces and integrate over all the pieces, but this is not performed in this class!

• In this class, moment of inertia calculations come in two forms:

  • System of particles – calculated using sums.

  • Rigid body – utilize expressions from a table.

Parallel-Axis Theorem

• When: Useful when computing moment of inertia is difficult because the object does not rotate about its center of mass.

• Simplification: One can simplify the process if the moment of inertia about the center of mass, $I_{CM}$, is known, along with the distance $d$ away from the center of mass.

• Theorem:
  For rotation about an axis parallel to an axis through the center of mass, we derive:
  $I = I_{CM} + Md^2$.

Torque

• Definition: Torque causes rotational motion and is the result of force application. The effectiveness of a force to cause rotation is known as torque.

• Effectiveness Depends On:

  • The magnitude of the force applied.

  • The distance between the pivot point and where the force is applied.

  • The angle at which the force is applied. Only the component perpendicular to the movement will influence the torque.

• Definition Using Vector Cross Product:
  $\vec{\tau} = \vec{r} \times \vec{F}$
  Where $\vec{F}$ is the applied force, and $\vec{r}$ is the vector from the pivot to the point where the force is applied. Torque is a vector quantity with magnitude and direction. The unit of torque is the newton-meter, $N \cdot m$, which does not simplify to joules since torque is not a form of energy.

• Magnitude and Direction:

  • Magnitude:
    $\tau = rF \sin{\phi},$ where $\phi$ is the angle between the two vectors.

  • If $\vec{r}$ and $\vec{F}$ are in the xy-plane, the torque points along z.

  • $\vec{\tau}$ out of the page = positive torque → CCW rotation (\omega > 0)

  • $\vec{\tau}$ into the page = negative torque → CW rotation (\omega < 0).

• Moment Arm or Lever Arm:
  The line along which the force acts is an imaginary line; the distance of the lever arm, $d$, separates the line of action from the pivot point. The relationship can be shown as:
  $\sin{(180 - \phi)} = \sin{\phi} = \frac{d}{r} \implies d = r \sin{\phi}$. Thus the magnitude of torque is:
  $\tau = Fd$.

• Gravitational Torque:
  Will exert a torque on an object on a pivot, inducing rotation. This can be expressed as:
  $\tau<em>{net} = \sum |\vec{\tau</em>i}|$. To find the net gravitational torque, divide an object into equal pieces, calculate the torque on each, and sum. The torque on each piece is given by:
  $\tau<em>i = -m</em>i g x<em>i$; so, $\tau</em>{net} = -g \sum m<em>i x</em>i$. The net torque equation results in:
  $\tau<em>{net} = -Mgx</em>{CM}$.

Rotational Dynamics

• Torque: Causes angular acceleration, $\alpha$. The net torque is given by:
  $\tau = \sum \tau_i$.

• Force Components:

  • Radial component: $F_r$ is the centripetal force and does not contribute to torque; it does not affect $\alpha$.

  • Tangential component: $F_t$ produces torque, altering $\alpha$.

• For one particle, this is written as:
  $F<em>{t,i} = m</em>i a<em>{t,i} = m</em>i r<em>i \alpha$. Since $\alpha$ is consistent, no $i$ is necessary. The torque from this force yields: $\tau</em>i = r<em>i F</em>{t,i} = r<em>i m</em>i r<em>i \alpha = m</em>i r<em>i^2 \alpha$. Thus, the total torque sums as: $\tau</em>{net} = \sum m<em>i r</em>i^2 \alpha = \alpha \sum m<em>i r</em>i^2$.
  This leads to:
  $\tau_{net} = \alpha I$

• To solve for $\alpha$, we have:
  $\alpha = \frac{\tau}{I}$, applying Newton's 2nd Law for Rotation.

• If the net torque on an object is zero, the object may either be stationary ($\omega = 0$) or rotating at a constant angular velocity ($\omega = const$).

Comparisons Between Linear and Rotational Motion

Problem Solving - Rotational Dynamics

1. Model the object as a rigid body.

2. Create a pictorial representation to clarify the situation, define coordinates, symbols, and list known information.

3. Identify the axis around which the object rotates.

4. Assess forces and their distances from the axis; drawing a free-body diagram is usually beneficial.

5. Determine torques induced by the forces and their signs.

6. Use Newton's second law for rotational motion mathematically:
   $\tau_{net} = I\alpha$

7. Find the moment of inertia from Table 12.2 or calculate if necessary via integral or parallel-axis theorem.

8. Apply rotational kinematics to determine angles and angular velocities.

9. Validate that your results meet unit consistency, significant figures, and answer the question logically.

Constraints due to Ropes and Pulleys

• When a rope passes over a pulley without slipping, the velocity and acceleration of the rope match those of a point on the rim of the pulley:

• $v<em>{rope} = v</em>{rim} = \omega R$

• $a<em>{rope} = a</em>{rim} = \alpha R$

• Any object attached to the rope must obey the same constraints.

Constant Torque

• For objects where the net torque is constant:

• Model the object as a rigid body with constant angular acceleration.

• Consider constraints from ropes and pulleys.

• Use Newton's second law:
  $\tau<em>{net} = I\alpha$ $\alpha = \frac{\tau</em>{net}}{I}$

• Apply kinematics of constant angular acceleration.

• Note: This model is invalid if the torque is variable.

Static Equilibrium

• An object with no net force or net torque is in static equilibrium.

• Previously established rules still apply here, as we include torque and rotation.

• From Newton’s second law:

• Zero net force implies no linear acceleration.

• Zero net torque indicates no angular acceleration.

• Caution: for a rigid body in total static equilibrium, it experiences no net torque about any point within the object.

Static Equilibrium Model

• This applies to extended objects at rest.

• Model the object as a rigid body with no acceleration.

• Mathematically:

  • No net force: $F<em>{net} = \Sigma F</em>i = 0$

  • No net torque: $\tau<em>{net} = \Sigma \tau</em>i = 0$

• The torque equals zero about every point; thus, use any convenient point as the pivot.

• Limitations arise if either forces or torques are unbalanced.

Balance and Stability

• Upon tilting an object counter-clockwise (CCW), three scenarios can happen:

1. The object returns to its original position (natural rotation shifts to clockwise).

2. The object continues to fall even further CCW and tips over.

3. The object remains in its tilted position (balanced).

Tipping Point:

• The balance scenario depends on the center of mass (CM) being inside, outside, or at the edge of the support base. The tipping point occurs when rotation exceeds a critical angle.

Rolling Motion

• Rolling encompasses both rotation about an axis and translation.

• Example: A round object with radius $R$ rolls on a flat surface, utilizing sufficient static friction to prevent slipping.

• To analyze the motion, focus on a point on the rim:

• During one complete revolution, the horizontal displacement of its center of mass is equal to its circumference:
  $\Delta x<em>{cm} = 2\pi R$ Thus, the velocity of the center of mass can be expressed as: $v</em>{CM} = \frac{\Delta x<em>{CM}}{\Delta t} \implies \frac{\Delta x</em>{CM}}{v<em>{CM} \Delta t} = v</em>{CM} T$
  Through these perspectives, we derive the rolling constraint:
  $v_{CM} = \omega R$

• The rolling constraint narrates the motion physics, leading to:
  $\vec{r} = \vec{r}<em>{CM} + \vec{r}</em>{i,rel}$
  $\vec{v} = \vec{v}<em>{CM} + \vec{v}</em>{i,rel}$

• The rotational energy follows as:
  $K<em>{rot} = \frac{1}{2} I</em>{CM} \omega^2$

• The translational energy is in terms of $v<em>{CM}$: $K</em>{CM} = \frac{1}{2} Mv<em>{CM}^2 = \frac{1}{2} MR^2 \omega^2$. Then, the total kinetic energy equation is: $K</em>{tot} = K<em>{rot} + K</em>{CM} = \frac{1}{2} I_{CM} \omega^2 + \frac{1}{2} M R^2 \omega^2$.

Rotational Motion – Including Vectors

• Angular velocity ($\omega$) points along the rotation axis; its direction is determined using the right-hand rule: Fingers curl in rotation direction, thumb points along $\omega$.

• Angular acceleration ($\vec{\alpha}$) is the rate of change of $\omega$; its direction is parallel to or opposite $\omega$, based on whether the object accelerates or decelerates.

Vectors – The Cross Product

• The dot product ($\vec{A} \cdot \vec{B}$) returns a scalar result: $C = A \cdot B = \vec{A} B \cos{\theta}$. This value peaks as vectors become parallel and equals zero when they are perpendicular.

• Example: Work is defined as $W = \vec{F} \cdot \Delta \vec{s}$.

• Conversely, the cross product yields a vector: $C = A \times B = A B \sin{\alpha}$. The magnitude and direction follow the right-hand rule.

The Torque Vector

• We define torque as a vector formed by a cross product: $\vec{\tau} = \vec{r} \times \vec{F}$. Applying the right-hand rule will help determine its direction.

Angular Momentum

• Linear momentum is defined as: $\vec{p} = m \vec{v}$. Conservation law upholds that total momentum remains constant, $\vec{p}<em>i = \vec{p}</em>f$.

• The angular momentum of a rotating object can be expressed as: $L \equiv \vec{r} \times \vec{p}$. The unit is kg⋅m$^{2}$/s, signifying that angular momentum depends on the choice of origin. At each point along the path, the angle between $\vec{r}$ and $\vec{p}$ is 90°, thus yielding:
  $L = mr^2\omega.$

Angular Momentum – Newton’s 2nd Law

• Like the linear scenario, Newton’s 2nd Law can be expressed in terms of angular momentum:
  $\frac{dL}{dt} = \frac{d}{dt}(\vec{r} \times \vec{p}) = \frac{d\vec{r}}{dt} \times \vec{p} + \vec{r} \times \frac{d\vec{p}}{dt} = \vec{v} \times \vec{p} + \vec{r} \times \vec{F}.$

Conservation of Angular Momentum

• In a no-net-torque environment, the derivative of angular momentum results in zero:
  $\frac{dL}{dt} = 0.$

• The Law of Conservation of Angular Momentum articulates that an isolated system’s angular momentum remains consistent, preserving both magnitude and direction over time.

• Analogous to how linear momentum relates to velocity, we can express angular momentum concerning angular velocity as follows:
  $L = I\omega,$ relevant for rotation around a fixed axle.

Comparison of Angular and Linear Motion
| Rotational Quantities | Linear Quantities |
| K{rot} = $\frac{1}{2} I \omega^2$ | $$K{CM} = \frac{