🔑:## Step 1: Define the problem and identify key componentsWe have a uniform chain of length L hanging from a horizontal table, with a portion of its length l hanging off the table. The system is in equilibrium. To find the Lagrangian, we need to consider the kinetic energy (T) and potential energy (U) of the system.## Step 2: Determine the kinetic energy of the systemThe kinetic energy of the system can be considered as the sum of the kinetic energies of the hanging portion and the portion on the table. However, since the system is in equilibrium, the kinetic energy due to motion of the chain as a whole is zero. Thus, we primarily focus on the potential energy for the equilibrium condition.## Step 3: Calculate the potential energy of the systemThe potential energy (U) of the hanging portion of the chain can be calculated by considering the weight of the chain and its height. The weight of the chain hanging off the table is proportional to its length (l) and the density of the chain (ρ). The potential energy of the hanging portion is given by (U = frac{1}{2} rho g l^2), where g is the acceleration due to gravity.## Step 4: Consider the potential energy of the portion on the tableThe portion of the chain on the table does not contribute to the potential energy in terms of height, as it is at the same level. Thus, the potential energy of the system is primarily due to the hanging portion.## Step 5: Introduce the concept of LagrangianThe Lagrangian (L) of a system is defined as the difference between the kinetic energy (T) and the potential energy (U), (L = T - U). However, since the system is in equilibrium and we're focusing on the potential energy for the calculation of the Lagrangian in this context, we recognize that the kinetic energy term is not contributing to the equilibrium condition directly.## Step 6: Discuss the implications of introducing a friction forceIntroducing a friction force would affect the equilibrium condition by opposing the motion of the chain. However, in the calculation of the Lagrangian for the equilibrium state, frictional forces are typically considered as non-conservative forces and do not directly affect the potential energy calculation. They would, however, influence the dynamics of the system if it were not in equilibrium.## Step 7: Formulate the Lagrangian for the systemGiven that the kinetic energy is not contributing to the equilibrium condition and focusing on the potential energy, the Lagrangian for the system in equilibrium, considering only the potential energy of the hanging portion, is (L = 0 - frac{1}{2} rho g l^2), simplifying to (L = -frac{1}{2} rho g l^2).The final answer is: boxed{-frac{1}{2} rho g l^2}

🔑:## Step 1: Understand the relationship between torque, angular acceleration, and moment of inertia.The torque τ applied to the wheel causes it to rotate. The relationship between torque, angular acceleration α, and the moment of inertia I of the wheel is given by τ = Iα. For a solid wheel, I = (1/2)mr^2, where m is the mass of the wheel and r is its radius.## Step 2: Apply the torque equation to find the angular acceleration.Given τ = Iα and I = (1/2)mr^2, we substitute I into the torque equation to get τ = (1/2)mr^2α. Solving for α gives α = τ / ((1/2)mr^2) = 2τ / (mr^2).## Step 3: Relate angular acceleration to linear acceleration.For an object rolling without slipping, the linear acceleration a of the axle is related to the angular acceleration α by the equation a = αr. This is because the distance the wheel travels in one rotation (its circumference, 2πr) is equal to the distance its axle would move if it were not rotating, implying a direct relationship between the rotational motion and the linear motion.## Step 4: Derive the expression for linear acceleration.Substituting α from Step 2 into the relationship a = αr gives a = (2τ / (mr^2))r = 2τ / (mr).The final answer is: boxed{frac{2tau}{mr}}

🔑:A delightful question about the fascinating world of super-hydrophobic surfaces!Super-hydrophobic surfaces are characterized by their ability to repel water, with contact angles greater than 150°. This property is achieved through the creation of micro- and nano-structured surfaces that increase the surface roughness, leading to a significant reduction in the solid-liquid interfacial energy. When a water droplet comes into contact with a super-hydrophobic surface, it forms a spherical shape, minimizing its contact area with the surface.Now, let's dive into the principles governing the adhesion between dust particles and water droplets on super-hydrophobic surfaces.Adhesion between dust particles and water droplets:The adhesion between dust particles and water droplets is primarily governed by the following forces:1. Van der Waals forces: These are weak intermolecular forces that arise from the interaction between temporary dipoles in the dust particles and water molecules. The van der Waals force (FvdW) between a dust particle and a water droplet can be estimated using the Hamaker constant (A) and the radius of the dust particle (R) and water droplet (r):FvdW = (A * R * r) / (6 * (R + r)^2)2. Electrostatic forces: These forces can arise from the interaction between charged dust particles and water droplets. However, on super-hydrophobic surfaces, the electrostatic forces are typically negligible due to the low dielectric constant of the surface.3. Capillary forces: These forces arise from the interaction between the water droplet and the dust particle through the formation of a liquid bridge. The capillary force (Fcap) can be estimated using the surface tension of water (γ), the radius of the dust particle (R), and the contact angle (θ) between the water droplet and the dust particle:Fcap = 2 * π * γ * R * cos(θ)Energy required for a water droplet to pick up dust particles:The energy required for a water droplet to pick up dust particles is related to the work of adhesion (Wad) between the dust particle and the water droplet. The work of adhesion can be estimated using the Dupré equation:Wad = γ * (1 + cos(θ))where γ is the surface tension of water and θ is the contact angle between the water droplet and the dust particle.The energy required to pick up a dust particle can be estimated by multiplying the work of adhesion by the contact area between the dust particle and the water droplet.Energy required for dust particles to stick to the hydrophobic surface:The energy required for dust particles to stick to the hydrophobic surface is related to the work of adhesion (Wad) between the dust particle and the surface. On a super-hydrophobic surface, the work of adhesion is typically very low due to the low solid-liquid interfacial energy.The work of adhesion between the dust particle and the surface can be estimated using the Young-Dupré equation:Wad = γ * (cos(θ) + 1)where γ is the surface tension of the dust particle and θ is the contact angle between the dust particle and the surface.Comparison of energies:In general, the energy required for a water droplet to pick up dust particles is higher than the energy required for these particles to stick to the hydrophobic surface. This is because the work of adhesion between the dust particle and the water droplet is typically higher than the work of adhesion between the dust particle and the surface.To illustrate this, let's consider a typical example:* Surface tension of water: γ = 72 mN/m* Contact angle between water droplet and dust particle: θ = 30°* Contact angle between dust particle and surface: θ = 150° (super-hydrophobic surface)* Radius of dust particle: R = 1 μmUsing the equations above, we can estimate the work of adhesion between the dust particle and the water droplet:Wad (dust-water) ≈ 72 mN/m * (1 + cos(30°)) ≈ 120 mN/mAnd the work of adhesion between the dust particle and the surface:Wad (dust-surface) ≈ 72 mN/m * (cos(150°) + 1) ≈ 10 mN/mAs expected, the energy required for the water droplet to pick up the dust particle (Wad (dust-water)) is significantly higher than the energy required for the dust particle to stick to the surface (Wad (dust-surface)).In conclusion, the principles governing the adhesion between dust particles and water droplets on super-hydrophobic surfaces are primarily governed by van der Waals forces, electrostatic forces, and capillary forces. The energy required for a water droplet to pick up dust particles is typically higher than the energy required for these particles to stick to the hydrophobic surface, due to the higher work of adhesion between the dust particle and the water droplet.

🔑:The hopping phenomenon experienced by the 3-wheel autonomous small car built out of Legos can be attributed to several factors related to its design, balance, and stability. To analyze and propose a solution, let's break down the possible causes and consider the principles of balance and stability, as well as the dynamics of the vehicle.Possible Causes:1. Lack of stability due to a high center of gravity: The Lego car's design might have a high center of gravity, making it prone to tipping or hopping when the back two wheels are turned in opposite directions. This is because the weight of the vehicle is concentrated above the axle, causing it to pivot around the front wheel.2. Insufficient traction or friction: The back wheels might not have sufficient traction or friction with the ground, causing them to slip or lose contact when turned in opposite directions. This would result in a loss of stability and lead to hopping.3. Inadequate suspension or damping: The Lego car's suspension system (if present) might not be designed to handle the oscillations caused by the back wheels turning in opposite directions. This could lead to an amplification of the hopping motion.4. Asymmetric weight distribution: If the weight of the vehicle is not evenly distributed, it could cause the car to tilt or hop when the back wheels are turned in opposite directions.Principles of Balance and Stability:To understand the stability of the vehicle, let's consider the following principles:1. Center of gravity (CG): The CG is the point where the weight of the vehicle can be considered to be concentrated. A lower CG generally provides better stability.2. Wheelbase and track width: The wheelbase (distance between the front and back wheels) and track width (distance between the left and right wheels) affect the vehicle's stability. A longer wheelbase and wider track width generally provide better stability.3. Moment of inertia: The moment of inertia (I) is a measure of an object's resistance to changes in its rotational motion. A higher moment of inertia generally provides better stability.Dynamics of the Vehicle:When the back two wheels are turned in opposite directions, the vehicle experiences a torque around the vertical axis, causing it to rotate. This rotation creates a centrifugal force that acts on the vehicle, which can lead to hopping if the vehicle is not designed to handle it.Proposed Solution:To stabilize the vehicle and prevent hopping, consider the following modifications:1. Lower the center of gravity: Redesign the vehicle to have a lower CG by: * Using a wider and lower chassis. * Placing the batteries or heavy components closer to the ground. * Using a more compact and lightweight design.2. Improve traction and friction: Increase the traction and friction of the back wheels by: * Using tires with a softer compound or a more aggressive tread pattern. * Adding weight to the back wheels to increase the normal force. * Using a differential or a limited-slip differential to distribute power to both wheels.3. Add suspension or damping: Implement a suspension system or add damping to the vehicle to: * Absorb the oscillations caused by the back wheels turning in opposite directions. * Reduce the amplification of the hopping motion.4. Optimize weight distribution: Ensure the weight of the vehicle is evenly distributed by: * Placing components symmetrically. * Using a balanced design.Calculations and Diagrams:To illustrate the effect of the proposed modifications, let's consider a simple example. Assume the Lego car has a wheelbase of 10 cm, a track width of 8 cm, and a CG height of 5 cm.Initial Conditions:* Wheelbase (L): 10 cm* Track width (W): 8 cm* CG height (h): 5 cm* Weight (m): 0.5 kgModified Conditions:* Wheelbase (L): 12 cm (increased by 20%)* Track width (W): 10 cm (increased by 25%)* CG height (h): 3 cm (reduced by 40%)* Weight (m): 0.5 kg (unchanged)Using the moment of inertia formula for a rectangular prism (I = (1/12) * m * (L^2 + W^2)), we can calculate the initial and modified moments of inertia:Initial I: (1/12) * 0.5 kg * (10^2 + 8^2) = 0.042 kg*m^2Modified I: (1/12) * 0.5 kg * (12^2 + 10^2) = 0.065 kg*m^2The increased moment of inertia and reduced CG height should provide better stability and reduce the hopping motion.Diagram:A simple diagram illustrating the initial and modified conditions is shown below:``` +---------------+ | Initial CG | | (h = 5 cm) | +---------------+ | | v +---------------+ | Wheelbase | | (L = 10 cm) | +---------------+ | | v +---------------+ | Track Width | | (W = 8 cm) | +---------------+ Modified Conditions: +---------------+ | Modified CG | | (h = 3 cm) | +---------------+ | | v +---------------+ | Wheelbase | | (L = 12 cm) | +---------------+ | | v +---------------+ | Track Width | | (W = 10 cm) | +---------------+```Note: The diagram is a simplified representation and not to scale.By implementing these modifications, the Lego car should experience reduced hopping and improved stability when the back two wheels are turned in opposite directions. However, the actual performance will depend on the specific design and construction of the vehicle.