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:
\omegaf = \omegai + \alpha t
Acceleration relationship:
For any initial and final velocities:
\omegaf^2 = \omegai^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{\omegaf - \omegai}{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}
\omegaf = \omegai + \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}
\omegaf = \omegai + \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: sB = RB \theta_B = (3r)(1\text{ rev})
For Gear A: sA = RA \thetaA = (r)\thetaA
Since sA = sB (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{\text{total}} = I{\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 m1 and m2 are at distances r1 and r2 from a pivot, then for equilibrium: r1 m1 g = r2 m2 g, which simplifies to r1 m1 = r2 m2.
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{\text{total}} = KE{\text{translational}} + KE{\text{rotational}} = \frac{1}{2}mv{\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{\text{CoM}} = r\omega \Rightarrow \omega = \frac{v{\text{CoM}}}{r}.
Substituting I and \omega:
KE{\text{total}} = \frac{1}{2}mv{\text{CoM}}^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v{\text{CoM}}}{r}\right)^2 KE{\text{total}} = \frac{1}{2}mv{\text{CoM}}^2 + \frac{1}{4}mv{\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:
Ii \omegai = If \omegaf
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: \omegaf = \omegai + \alpha t
Final angular velocity squared: \omegaf^2 = \omegai^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): r1 m1 = r2 m2
Kinetic Energy for Rolling Without Slipping
Total kinetic energy: KE{\text{total}} = \frac{1}{2}mv{\text{CoM}}^2 + \frac{1}{2}I\omega^2
Total kinetic energy (for solid disk): KE{\text{total}} = \frac{3}{4}mv{\text{CoM}}^2
Angular Momentum Conservation
Conservation of angular momentum: Ii \omegai = If \omegaf
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.