Rotational Dynamics

Chapter Overview

• Rotational Physics is omnipresent in everyday life and can be observed in various contexts from amusement rides to celestial rotations.

Angular Displacement

• Definition: Angular displacement, denoted by the symbol $heta$, refers to the change in angular position as an object moves around a circle. It is the angle through which an object has rotated about a specified axis.

• Measurement: Typically measured in radians.

  • Conversion: 360 degrees is equivalent to $2 ext{π}$ radians, representing one complete revolution around a circle.

  • Important Fact: 1 radian is approximately 57.3 degrees.

Practice with Radians

• Conversion examples:

  • Convert 52° to radians:

    $heta = 52° imes\frac{2 ext{π}}{360°} = 0.91 ext{ rad}$

  • Convert 2 revolutions to radians:

    $2 ext{ rev} = 2 imes 2 ext{π rad} = 12.6 ext{ rad}$

  • Convert 0.75 rad to degrees:

    $0.75 ext{ rad} imes\frac{360°}{2 ext{π}} = 43°$

  • Note: Keep $2 ext{π}$ together in calculations for clarity.

Rotational Velocity

• Definition: Average rotational velocity, denoted as $\omega$, is defined as the rotational distance covered over a period of time.

• Formula:

  $\omega =\frac{\Delta\theta}{t}$

• Characteristics:

  • Similar to linear velocity, it shows how rotational position varies with time.

  • Every point in a rotating rigid body shares the same rotational velocity.

Rotational Acceleration

• Definition: Rotational acceleration, denoted as $\alpha$, measures how rotational velocity changes over time.

• Formula:

  $\alpha =\frac{\Delta\omega}{t}$

• Describes how the rate of rotation alters over time.

Comparisons of Motion Quantities

• Linear vs. Rotational Quantities:

  • Linear Distance: $s$ (meters)

  • Linear Velocity: $v$ (m/s)

  • Linear Acceleration: $a$ (m/s²)

  • Rotational Distance: $heta$ (radians)

  • Rotational Velocity: $\omega$ (rad/s)

  • Rotational Acceleration: $\alpha$ (rad/s²)

Relationship Between Linear and Angular Motion

• Equations reflecting relationships between linear ($s, v, a$) and angular ($heta, \omega, \alpha$) quantities:

  • For linear motion:

    $s = v_0t +\frac{1}{2}at^2$

  • For angular motion:

    $heta = \omega_0t +\frac{1}{2}\alpha t^2$

• Final angular velocity:

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

• Acceleration relationship:

  • For any initial and final velocities:

    $\omega<em>f^2 = \omega</em>i^2 + 2\alpha heta$

Problems and Applications

Example Problem 1: Merry-Go-Round

• Problem: Calculate the approximate angular velocity ($\omega$) and period ($T$) of a merry-go-round that revolves at 10 rpm.

• Solution:

  • Convert rpm to rad/s:

    $\omega = 10\frac{\text{rev}}{\text{min}} \times \frac{2\pi \text{ rad}}{1\text{ rev}} \times \frac{1\text{ min}}{60\text{ s}} = \frac{20\pi}{60}\frac{\text{rad}}{\text{s}} = \frac{\pi}{3}\frac{\text{rad}}{\text{s}} \approx 1.05\frac{\text{rad}}{\text{s}}$

  • Calculate the period ($T$):

    $T = \frac{2\pi}{\omega} = \frac{2\pi}{\frac{\pi}{3}} = 6\text{ s}$

Example Problem 2: Ultra Centrifuge

• Details:

  • Accelerates from rest to 100 krpm in 2 minutes.

  • Calculate angular acceleration ($\alpha$) in rad/s².

    • Initial angular velocity $\omega_i = 0$

    • Final angular velocity $\omega_f = 100\text{ krpm} = 100 \times 10^3 \frac{\text{rev}}{\text{min}} \times \frac{2\pi \text{ rad}}{1\text{ rev}} \times \frac{1\text{ min}}{60\text{ s}} \approx 10472 \frac{\text{rad}}{\text{s}}$

    • Time $t = 2\text{ min} = 120\text{ s}$

    • $\alpha = \frac{\Delta\omega}{t} = \frac{\omega<em>f - \omega</em>i}{t} = \frac{10472 - 0}{120} \approx 87.27\frac{\text{rad}}{\text{s}^2}$

  • Find linear acceleration for a point 9.50 cm from the axis.

    • Radius $r = 9.50\text{ cm} = 0.095\text{ m}$

    • Tangential linear acceleration $a_t = r\alpha = (0.095\text{ m}) \times (87.27\frac{\text{rad}}{\text{s}^2}) \approx 8.29\frac{\text{m}}{\text{s}^2}$

  • Determine angular velocity after 2 seconds when it slows at a rate of $3.00 \times 10^3 \text{ rad/s²}$.

    • Assuming it slows from its maximum speed calculated above:

    • Initial angular velocity $\omega_i = 10472\frac{\text{rad}}{\text{s}}$

    • Angular acceleration $\alpha = -3.00 \times 10^3\frac{\text{rad}}{\text{s}^2}$ (negative because it slows down)

    • Time $t = 2\text{ s}$

    • $\omega<em>f = \omega</em>i + \alpha t = 10472 + (-3.00 \times 10^3)(2) = 10472 - 6000 = 4472\frac{\text{rad}}{\text{s}}$

  • Calculate time to fully stop.

    • Initial angular velocity $\omega_i = 10472\frac{\text{rad}}{\text{s}}$

    • Final angular velocity $\omega_f = 0$

    • Angular acceleration $\alpha = -3.00 \times 10^3\frac{\text{rad}}{\text{s}^2}$

    • $\omega<em>f = \omega</em>i + \alpha t \Rightarrow 0 = 10472 + (-3.00 \times 10^3)t$

    • $t = \frac{-10472}{-3.00 \times 10^3} \approx 3.49\text{ s}$

Gears

• Functionality: Gears transfer motion through interlocked teeth.

• Key Concept: The linear and rotational velocities at the points where gear teeth meet must be equal, underpinning the mechanics of gear systems.

Example Problem: Gears

• Gear A has radius $r$, and Gear B has radius $3r$. If Gear B makes one full rotation, Gear A rotates a number of times determined by their radii relationship.

• Solution:

  • The linear distance moved by the teeth at the point of contact is the same for both gears.

  • For Gear B: $s<em>B = R</em>B \theta_B = (3r)(1\text{ rev})$

  • For Gear A: $s<em>A = R</em>A \theta<em>A = (r)\theta</em>A$

  • Since $s<em>A = s</em>B$ (no slipping), $(r)\theta_A = (3r)(1\text{ rev})$

  • $\theta_A = \frac{3r}{r} (1\text{ rev}) = 3\text{ revolutions}$

  • So, Gear A rotates 3 times.

Moment of Inertia

• Definition: Moment of inertia (I), or rotational inertia, quantifies how mass is distributed in relation to the axis of rotation.

• Formula: $I = mr^2$ - where $m$ is mass and $r$ is the distance from the axis.

• Describes resistance to changes in rotational motion.

• For simple objects, the rotational inertia is represented as $I = mr^2$.

Examples of Moments of Inertia

• Various geometrical shapes and their corresponding moments of inertia can be defined, primarily given on an equation sheet for problem-solving.

Demonstration: Spinning Faster

• Discussion on human body positioning (arms out vs. arms in) and its effect on rotational speed, demonstrating the conservation of angular momentum where mass distribution affects spin.

Example Problem 3: Merry-Go-Round

• Given rotational inertia ($I = 800\text{ kg} \cdot \text{m²}$) and radius (2.0 m), calculate total rotational inertia of a child situated at the edge (40 kg treated as a point mass).

• Solution:

  • Moment of inertia of the merry-go-round: $I_{\text{MG}} = 800\text{ kg} \cdot \text{m²}$

  • Moment of inertia of the child (as a point mass): $I_{\text{child}} = mr^2 = (40\text{ kg})(2.0\text{ m})^2 = 40 \times 4 = 160\text{ kg} \cdot \text{m²}$

  • Total rotational inertia: $I<em>{\text{total}} = I</em>{\text{MG}} + I_{\text{child}} = 800\text{ kg} \cdot \text{m²} + 160\text{ kg} \cdot \text{m²} = 960\text{ kg} \cdot \text{m²}$

Balance and Center of Mass

• Concept of Balance: To be in a balanced state means no rotational acceleration occurs. All forces and torques acting on the system are equal in their effects.

• Example: Balancing a baseball bat at point P highlights the concept of center of mass.

Torque

• Definition: Torque ($\tau$), the rotational analog of force, represents the force causing an object to rotate around an axis.

• Formula:

  $\tau = rF_{\perp}$

• Key elements: distance from axis and effective force perpendicular to the lever arm.

Examples and Problems in Torque

• Analyze problems like opening a door:

  • Given door length ($L$) and applied force ($F$), calculate torque.

  • Solution: If force is applied perpendicularly at distance $r$ from the hinge (e.g., $r=L$ at the edge), the torque is: $\tau = rF_{\perp} = LF$

  • If applied at an angle $\theta$: $\tau = LF\sin(\theta)$

  • Units: Newton-meters ($\text{N} \cdot \text{m}$).

• A mobile construction problem involves mass distribution along a rigid rod to maintain horizontal balance.

  • Solution: For horizontal balance, the net torque about any pivot must be zero. If two masses $m<em>1$ and $m</em>2$ are at distances $r<em>1$ and $r</em>2$ from a pivot, then for equilibrium: $r<em>1 m</em>1 g = r<em>2 m</em>2 g$, which simplifies to $r<em>1 m</em>1 = r<em>2 m</em>2$.

Rotational Motion and Newton’s Second Law

• Connection with Newton’s 2nd Law: The relationship states that if linear acceleration is proportional to the sum of forces, rotational acceleration is proportional to the sum of torques:

  • $\alpha \propto \tau$

• Where mass is substituted by moment of inertia in rotational dynamics.

Statics

• Definition: Statics is the study of systems at rest or in uniform motion (not accelerating either linearly or rotationally).

• It is essential in multiple fields, including structural engineering and architecture.

Conditions for Equilibrium

• The equilibrium condition necessitates balance in both linear forces and torques acting on an object:

  • Ensure total torque about any chosen axis equals zero for rotational equilibrium.

Statics Problems Methodology

• Recommendations for solving statics problems:

  • Start with a Free Body Diagram (FBD).

  • Carefully designate a coordinate system and axis of rotation.

  • Solve for forces or torques effectively, going back and forth between the two as necessary.

Cantilever Problem

• Calculation of forces acting on a uniform cantilevered beam must factor its weight and position.

Rolling Without Slipping

• Key Concept: In rolling without slipping, the point of contact between the rolling object and the surface does not slide, utilizing static friction to facilitate motion.

• Important equations relate translation and rotation:

  • $x_{\text{CoM}} = r\theta$

  • $v_{\text{CoM}} = r\omega$

  • $a_{\text{CoM}} = r\alpha$

Total Kinetic Energy

Example Problem 4

• A solid disk wheel of mass 1.0 kg rolls without slipping at a speed of 6.0 m/s. Calculate the total kinetic energy:

• Solution:

  • Total kinetic energy for rolling without slipping is the sum of translational and rotational kinetic energy:

    $KE<em>{\text{total}} = KE</em>{\text{translational}} + KE<em>{\text{rotational}} = \frac{1}{2}mv</em>{\text{CoM}}^2 + \frac{1}{2}I\omega^2$

  • For a solid disk, the moment of inertia is $I = \frac{1}{2}mr^2$.

  • For rolling without slipping, $v<em>{\text{CoM}} = r\omega \Rightarrow \omega = \frac{v</em>{\text{CoM}}}{r}$.

  • Substituting $I$ and $\omega$:

    $KE<em>{\text{total}} = \frac{1}{2}mv</em>{\text{CoM}}^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v<em>{\text{CoM}}}{r}\right)^2$ $KE</em>{\text{total}} = \frac{1}{2}mv<em>{\text{CoM}}^2 + \frac{1}{4}mv</em>{\text{CoM}}^2 = \frac{3}{4}mv_{\text{CoM}}^2$

  • Given: $m = 1.0\text{ kg}$ and $v_{\text{CoM}} = 6.0\text{ m/s}$

  • $KE_{\text{total}} = \frac{3}{4} (1.0\text{ kg})(6.0\text{ m/s})^2 = \frac{3}{4} (36) = 27\text{ J}$

Angular Momentum

• Fundamental Principle: The total angular momentum remains conserved if no net torque is acting on the system:

  • Relation is established:

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

Formulas in Rotational Physics

• Angular Displacement

  • Conversion: $2\pi\text{ radians} = 360°$

• Rotational Velocity

  • Average rotational velocity: $\omega = \frac{\Delta\theta}{t}$

• Rotational Acceleration

  • Rotational acceleration: $\alpha = \frac{\Delta\omega}{t}$

• Kinematic Equations for Rotational Motion

  • Angular displacement: $heta = \omega_0t +\frac{1}{2}\alpha t^2$

  • Final angular velocity: $\omega<em>f = \omega</em>i + \alpha t$

  • Final angular velocity squared: $\omega<em>f^2 = \omega</em>i^2 + 2\alpha heta$

• Relationship Between Linear and Angular Motion

  • Center of Mass displacement: $x_{\text{CoM}} = r\theta$

  • Center of Mass velocity: $v_{\text{CoM}} = r\omega$

  • Center of Mass acceleration: $a_{\text{CoM}} = r\alpha$

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

• Moment of Inertia

  • For a point mass: $I = mr^2$

• Torque

  • Torque: $\tau = rF_{\perp}$

  • Torque at an angle: $\tau = LF\sin(\theta)$

  • Static equilibrium condition (for two masses): $r<em>1 m</em>1 = r<em>2 m</em>2$

• Kinetic Energy for Rolling Without Slipping

  • Total kinetic energy: $KE<em>{\text{total}} = \frac{1}{2}mv</em>{\text{CoM}}^2 + \frac{1}{2}I\omega^2$

  • Total kinetic energy (for solid disk): $KE<em>{\text{total}} = \frac{3}{4}mv</em>{\text{CoM}}^2$

• Angular Momentum Conservation

  • Conservation of angular momentum: $I<em>i \omega</em>i = I<em>f \omega</em>f$

Turntable Demonstration

• An experiment demonstrates the change in angular velocity upon adding a disk to a turntable and questions the ratio after the interaction, exploring fundamental physics principles in rotational dynamics.