Forces, Equilibrium, Momentum, and Collisions: Study Notes

Force and Equilibrium

Torque and Rotational Equilibrium

Rotational equilibrium occurs when the sum of all torques acting on a system is zero, resulting in no angular acceleration. This principle is often applied to problems involving doors, beams, and other rotating bodies.

• Torque (τ): The rotational analog of force, defined as the product of force and the perpendicular distance from the axis of rotation.

• Equation: $\tau = F \cdot r$

• Equilibrium Condition: $\sum \tau = 0$

• Example: Two children push on opposite sides of a door. If one pushes with $F_1 = 175\,\text{N}$ at $r_1 = 0.70\,\text{m}$ and the other at $r_2 = 0.40\,\text{m}$, the force $F_2$ needed for equilibrium is:

$F_1 r_1 = F_2 r_2 \Rightarrow F_2 = \frac{F_1 r_1}{r_2} = \frac{175 \times 0.70}{0.40} = 306.25\,\text{N}$

• Application: Used to determine forces required to keep doors, beams, or levers stationary.

Springs and Hooke's Law

Hooke's Law describes the force exerted by a spring when it is stretched or compressed.

• Hooke's Law: $F = -kx$

• Spring Constant (k): Measures the stiffness of the spring (units: N/m).

• Direction: The force exerted by the spring is always directed toward the equilibrium position.

• Example: For $k = 40\,\text{N/m}$ and $x = 4\,\text{m}$, $F = -40 \times 4 = -160\,\text{N}$ (toward equilibrium).

Statics: Beams and Friction

Uniform Beam Supported by Cable

When analyzing beams supported by cables and walls, both torque and friction must be considered to prevent slipping.

• Static Friction: $f_s = \mu_s N$ where $\mu_s$ is the coefficient of static friction and $N$ is the normal force.

• Torque Balance: Used to determine the minimum distance from the support point to prevent slipping.

• Example: For a beam of length $L = 2.00\,\text{m}$, $\mu_s = 0.560$, and weight $w$, calculate the minimum distance $x$ from point A for equilibrium.

Impulse, Momentum, and Energy

Impulse and Change in Momentum

Impulse is the product of force and the time interval over which it acts, resulting in a change in momentum.

• Impulse (J): $J = F \Delta t = \Delta p$

• Momentum (p): $p = m v$

• Example: A car is hit with $F = 2700\,\text{N}$ for $\Delta t = 0.400\,\text{s}$:

$J = 2700 \times 0.400 = 1080\,\text{N} \cdot \text{s}$

• Direction: The direction of impulse matches the direction of the applied force.

Conservation of Momentum in Collisions

In the absence of external forces, the total momentum of a system remains constant during a collision.

• Conservation Law: $m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f$ (for perfectly inelastic collisions)

• Types of Collisions:

  • Elastic: Both momentum and kinetic energy are conserved.

  • Inelastic: Only momentum is conserved; kinetic energy is not.

  • Perfectly Inelastic: Objects stick together after collision.

• Example: A 1000 kg car moving east at 20 m/s collides with a 1500 kg van moving north at 30 m/s. If they stick together, use vector addition to find the final velocity.

Recoil Velocity (Conservation of Momentum)

When an object (e.g., a cannon) fires a projectile, the recoil velocity can be found using conservation of momentum.

• Equation: $m_\text{cannon} v_\text{cannon} = -m_\text{shell} v_\text{shell}$

• Example: A 2650 kg cannon fires a 16.0 kg shell at 480 m/s at 20° above horizontal. The horizontal component of shell velocity is $v_{\text{shell},x} = 480 \cos(20^\circ)$.

• Recoil velocity: $v_\text{cannon} = -\frac{m_\text{shell} v_{\text{shell},x}}{m_\text{cannon}}$

Momentum, Collisions, Rotations

Ballistic Pendulum and Conservation Laws

A ballistic pendulum is used to measure the speed of a projectile by analyzing the collision and subsequent motion.

• Perfectly Inelastic Collision: The bullet embeds in the block; both move together after impact.

• Conservation of Linear Momentum: $m_\text{bullet} v_\text{bullet} = (m_\text{bullet} + m_\text{block}) v_\text{final}$

• Conservation of Energy (after collision): The combined mass swings upward, converting kinetic energy to potential energy.

• Maximum Vertical Displacement: Use energy conservation: $\frac{1}{2} (m_\text{bullet} + m_\text{block}) v_\text{final}^2 = (m_\text{bullet} + m_\text{block}) g h$

Springs and Friction

When a mass is pressed against a spring and released, it can slide across a surface, losing energy to friction.

• Work Done by Spring: $W = \frac{1}{2} k x^2$

• Work Done by Friction: $W_f = \mu_k m g d$

• Maximum Displacement: Set spring work equal to friction work to solve for $d$.

Rotational Motion and Angular Momentum

Angular momentum is conserved in the absence of external torques. This principle is used in problems involving rotating doors and objects.

• Angular Momentum (L): $L = I \omega$ where $I$ is the moment of inertia and $\omega$ is angular velocity.

• Torque: $\tau = I \alpha$

• Example: Throwing a ball at a door to just barely close it, considering the door's moment of inertia and frictional torque.

Summary Table: Types of Collisions

Type of Collision	Momentum Conserved?	Kinetic Energy Conserved?	Objects Stick Together?
Elastic	Yes	Yes	No
Inelastic	Yes	No	No
Perfectly Inelastic	Yes	No	Yes

Additional info: These notes expand on the original questions by providing definitions, formulas, and context for each topic, ensuring a self-contained study guide for exam preparation.