Equilibrium, Torque, and Rotational Dynamics Study Notes

Equilibrium Conditions in Physics

Static Equilibrium

Static equilibrium occurs when an object is at rest and remains at rest, meaning all the forces and torques acting on it are balanced. This is a fundamental concept in mechanics, especially when analyzing structures and systems that do not move.

• Translational Equilibrium: The sum of all forces in both the x and y directions must be zero.

• Rotational Equilibrium: The sum of all torques about any axis must be zero.

Equilibrium Conditions:

• $\sum F_x = 0$

• $\sum F_y = 0$

• $\sum \tau = 0$

Choosing the Axis: For torque calculations, the axis can be chosen to simplify the problem, often at a point where unknown forces act.

Torque and Lever Arm

Definition of Torque

Torque is a measure of the tendency of a force to rotate an object about an axis. It depends on the magnitude of the force, the distance from the axis (lever arm), and the angle at which the force is applied.

• Formula: $\tau = r F \sin \theta$

• Lever Arm: The perpendicular distance from the axis of rotation to the line of action of the force.

Example: Seesaw Problem

• Two children sit on opposite sides of a seesaw. To remain stationary, the torques produced by their weights about the pivot must balance.

• Given: 60g N and 25g N forces at distances x and 2.0 m, respectively.

• Equilibrium condition: $60g \cdot x = 25g \cdot 2.0$

• Solve for x: $x = \frac{25g \cdot 2.0}{60g} = \frac{50}{60} = 0.83$ m

Group Problem Solving: Balancing Torques

When additional weights are added, the position must be adjusted to maintain balance.

• Girl weighs 200 N, receives a 40 N bag (total 240 N).

• Original lever arm: 3 m, original force: 200 N.

• New lever arm $x$ must satisfy: $(\text{lever arm})_{\text{new}} \times F_{\text{new}} = (\text{lever arm})_{\text{old}} \times F_{\text{old}}$

• Calculation: $x \times 240 = 3 \times 200 \implies x = \frac{600}{240} = 2.5$ m

Rotational Dynamics and Applications

Rotational Equilibrium Example

Consider a plank held horizontally by a rope and a hinge, with a box placed at a distance x from the hinge. The tension in the rope must not exceed a maximum value.

• Plank mass: 10 kg, length: 1.0 m, box mass: 10 kg.

• Maximum tension: 58 N.

• Set up torque equations about the hinge to solve for the maximum x before the rope breaks.

• Torque due to plank: $\tau_{\text{plank}} = (\text{mass}_{\text{plank}} \cdot g) \cdot (\text{distance to center of mass})$

• Torque due to box: $\tau_{\text{box}} = (\text{mass}_{\text{box}} \cdot g) \cdot x$

• Torque due to tension: $\tau_{T} = T \cdot L \cdot \sin \theta$

• Set $\sum \tau = 0$ and solve for x.

Conservation of Angular Momentum

When two rotating disks interact (e.g., a disk dropped onto a rotating turntable), angular momentum is conserved if no external torque acts.

• Initial angular momentum: $L_i = I_1 \omega_1$

• Final angular momentum: $L_f = (I_1 + I_2) \omega_f$

• Conservation: $I_1 \omega_1 = (I_1 + I_2) \omega_f$

• Solve for final angular velocity: $\omega_f = \frac{I_1 \omega_1}{I_1 + I_2}$

• Example: $I_1 = 3 \text{kg m}^2$, $I_2 = 1 \text{kg m}^2$, $\omega_1 = 4 \text{rad/s}$

• $\omega_f = \frac{3 \times 4}{3 + 1} = 3$ rad/s

Changing Radius in Rotational Motion

When the radius of rotation changes (e.g., a block attached to a cord pulled through a hole), angular momentum is conserved if no external torque acts.

• Initial angular momentum: $L_i = m r_i^2 \omega_i$

• Final angular momentum: $L_f = m r_f^2 \omega_f$

• Conservation: $m r_i^2 \omega_i = m r_f^2 \omega_f$

• Solve for new angular speed: $\omega_f = \omega_i \left( \frac{r_i}{r_f} \right)^2$

• Example: $m = 0.025$ kg, $r_i = 0.3$ m, $\omega_i = 1.5$ rad/s, $r_f = 0.1$ m

• $\omega_f = 1.5 \left( \frac{0.3}{0.1} \right)^2 = 1.5 \times 9 = 13.5$ rad/s

Change in Kinetic Energy:

• Initial KE: $KE_i = \frac{1}{2} m r_i^2 \omega_i^2$

• Final KE: $KE_f = \frac{1}{2} m r_f^2 \omega_f^2$

• Calculate the difference: $\Delta KE = KE_f - KE_i$

Rotational Dynamics: Pulley Systems

Acceleration of Masses in a Pulley System

In systems with pulleys and masses, analyze forces and torques to determine acceleration.

• Free-body diagrams: Draw forces acting on each mass and the pulley.

• Equations: Use Newton's second law for linear and rotational motion.

• For the pulley: $\sum \tau = I \alpha$

• For the masses: $m a = T - m g$ (for each mass, with appropriate direction)

• Relate angular acceleration $\alpha$ to linear acceleration $a$: $a = R \alpha$

Example: Masses $3m$ and $m$ connected over a pulley of radius $R$ and mass $M$.

Summary Table: Equilibrium and Rotational Dynamics

Concept	Condition/Equation	Example/Application
Translational Equilibrium	$\sum F_x = 0$, $\sum F_y = 0$	Stationary plank, seesaw
Rotational Equilibrium	$\sum \tau = 0$	Balancing torques on seesaw
Torque	$\tau = r F \sin \theta$	Lever arm problems
Conservation of Angular Momentum	$L_i = L_f$	Disk dropped on turntable
Rotational Kinetic Energy	$KE = \frac{1}{2} I \omega^2$	Block on cord, changing radius
Pulley System Acceleration	$\sum \tau = I \alpha$, $a = R \alpha$	Masses over pulley

Additional info: These notes expand on the provided slides and problems, adding definitions, formulas, and example calculations for clarity and completeness. The content covers topics from chapters on rotational motion, dynamics of rotational motion, and applications of Newton's laws, relevant for college-level physics.