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The mass \(m\) of a raindrop increases as it picks up moisture during its vertical descent through still air. If the air resistance to motion of the drop is \(R\) and its downward velocity is \(v,\) write the equation of motion for the drop and show that the relation \(\Sigma F=d(m v) / d t\) is obeyed as a special case of the variable-mass equation.

Short Answer

Expert verified
The equation of motion is \( mg - R = m \frac{dv}{dt} + v \frac{dm}{dt} \). This shows that \( \Sigma F = \frac{d(mv)}{dt} \) holds for variable mass.

Step by step solution

01

Understand the Exercise

We are asked to derive the equation of motion for a raindrop gaining mass as it falls. The raindrop has mass \(m\), velocity \(v\), and is subjected to air resistance \(R\). We will show that \(\Sigma F = \frac{d(mv)}{dt}\) holds.
02

Identify Forces Acting on the Raindrop

The two forces acting on the raindrop are gravity pulling it down and air resistance acting upward. If \(g\) is the acceleration due to gravity, the gravitational force is \(mg\), and the air resistance is \(R\), opposing the motion.
03

Write the General Variable Mass Equation

The general equation of motion for a body with changing mass is \( \Sigma F = \frac{d(mv)}{dt} = m \frac{dv}{dt} + v \frac{dm}{dt} \), where \( \Sigma F \) is the net external force.
04

Substitute Forces into the Equation

The net force \( \Sigma F \) is the gravitational force minus the air resistance: \( \Sigma F = mg - R \). Substitute this into the equation to get \( mg - R = m \frac{dv}{dt} + v \frac{dm}{dt} \).
05

Simplify and Analyze

The equation \( mg - R = m \frac{dv}{dt} + v \frac{dm}{dt} \) shows the relationship between the forces and the changing mass and velocity. It confirms \( \Sigma F = \frac{d(mv)}{dt} \) as a special case for this system when the mass is variable.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Variable Mass Systems
In variable mass systems, the mass of an object is not constant, but changes over time. This can occur in various scenarios like rockets burning fuel or, as in our exercise, a raindrop gaining mass while falling. Normally, in classical mechanics, we treat systems with a fixed mass. However, when mass is variable, it introduces complexity to the calculation of motion.

To tackle this, we use the concept of variable mass systems which is guided by a fundamental equation of motion:
  • Launch vehicles losing mass as they burn fuel.
  • Rain collecting during descent adds to its mass.

The general formula for such systems is: \[ \Sigma F = \frac{d(mv)}{dt} = m \frac{dv}{dt} + v \frac{dm}{dt} \]
This equation helps us consider both the change in velocity and the change in mass over time, ensuring that all forces on the system are accounted for accurately.
Dynamics
Dynamics is a branch of physics that deals with the study of forces and motions. In simple terms, it explains why things move the way they do. When we analyze dynamics, we often look at the forces causing motion and how they influence the object's speed and trajectory.

In our exercise, understanding dynamics involves recognizing the two primary forces: gravity pulling the raindrop downward and air resistance acting upward. These forces not only influence the velocity but also the way mass is acquired by the raindrop.

Under the influence of these forces:
  • Gravity, a constant force, acts to accelerate the drop.
  • Air resistance, which increases with speed, opposes this motion.
Both these forces are vital in crafting the raindrop's equation of motion, showing the complex but fascinating interaction between force, mass, and velocity in variable mass systems.
Newton's Second Law
Newton's Second Law is our primary tool for understanding motion. It states that the force acting on an object is equal to the mass of the object times its acceleration: \( \Sigma F = ma \). In the context of a variable mass system, like our raindrop, this law needs an adjustment.

When mass isn't constant, Newton's equation adapts to express the change in momentum over time. This adjustment is reflected in: \[ \Sigma F = \frac{d(mv)}{dt} = m \frac{dv}{dt} + v \frac{dm}{dt} \]
This altered formula allows us to succeed in predicting how the object accelerates even as its mass changes. It highlights the sum of forces acting on the raindrop, incorporating both how fast it picks up speed, and how rapidly it gathers moisture, maintaining the integrity of Newton's formidable principles in a variable scenario.

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Most popular questions from this chapter

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Air is pumped through the stationary duct \(A\) with a velocity of 50 ft/sec and exhausted through an experimental nozzle section \(B C\). The average static pressure across section \(B\) is 150 lb/in. \(^{2}\) gage, and the specific weight of air at this pressure and at the temperature prevailing is \(0.840 \mathrm{lb} / \mathrm{ft}^{3} .\) The average static pressure across the exit section \(C\) is measured to be \(2 \mathrm{lb} / \mathrm{in} .^{2}\) gage, and the corresponding specific weight of air is \(0.0760 \mathrm{lb} / \mathrm{ft}^{3}\). Calculate the force \(T\) exerted on the nozzle flange at \(B\) by the bolts and the gasket to hold the nozzle in place.

The upper end of the open-link chain of length \(L\) and mass \(\rho\) per unit length is released from rest with the lower end just touching the platform of the scale. Determine the expression for the force \(F\) read on the scale as a function of the distance \(x\) through which the upper end has fallen. (Comment: The chain acquires a free-fall velocity of \(\sqrt{2 g x}\) because the links on the scale exert no force on those above, which are still falling freely. Work the problem in two ways: first, by evaluating the time rate of change of momentum for the entire chain and second, by considering the force \(F\) to be composed of the weight of the links at rest on the scale plus the force necessary to divert an equivalent stream of fluid.)

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