Problem 114 - University Physics with Modern Physics

Question

A Variable-Mass Raindrop. In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water droplets. Some of these small droplets adhere to the raindrop, thereby increasing its mass as it falls. The force on the raindrop is $$F_{\mathrm{ext}}=\frac{d p}{d t}=m \frac{d v}{d t}+v \frac{d m}{d t}$$ Suppose the mass of the raindrop depends on the distance \(x\) that it has fallen. Then \(m=k x,\) where \(k\) is a constant, and \(d m / d t=k v\) . This gives, since \(F_{\text { ext }}=m g\) , $$m g=m \frac{d v}{d t}+v(k v)$$ Or, dividing by \(k\) $$x g=x \frac{d v}{d t}+v^{2}$$ This is a differential equation that has a solution of the form \(v=\) at, where \(a\) is the acceleration and is constant. Take the initial velocity of the raindrop to be zero. (a) Using the proposed solution for \(v\) , find the acceleration \(a\) . (b) Find the distance the raindrop has fallen in \(t=3.00\) s. (c) Given that \(k=2.00 \mathrm{g} / \mathrm{m},\) find the mass of the raindrop at \(t=3.00 \mathrm{s} .\) (For many more intriguing aspects of this problem, see \(\mathrm{K} .\) S. Krane, American Journal of Physics, Vol. 49 \((1981),\) pp. \(113-117.)\)

Step-by-Step Solution

Answer

(a) \( a = 2.45 \, \text{m/s}^2 \); (b) 11.025 m; (c) 0.02205 kg.

Step 1: Analyze Given Equation

We begin with the differential equation \( xg = x \frac{dv}{dt} + v^2 \) which is derived from the force equation. We need to solve this equation using the provided form \( v = at \).

Step 2: Substitute Proposed Solution

Substitute \( v = at \) into the equation. Thus, the derivative \( \frac{dv}{dt} = a \). Substitute into the equation: \[ xg = xa + (at)^2 \].

Step 3: Solve for Acceleration \(a\)

Since \( x = vt = at^2 \), substitute \( x = at^2 \) into the simplified equation \( at^2g = at^2a + (at)^2 \). Simplify to find the acceleration: \[ g = a + at \]. Solve for \( a \): \[ a = \frac{g}{1+t} \].

Step 4: Calculate Acceleration for Given Conditions

With initial velocity \( v_0 = 0 \), substitute \( g = 9.8 \, \text{m/s}^2 \) and \( t = 3.0 \, \text{s} \): \[ a = \frac{9.8}{1+3} = \frac{9.8}{4} = 2.45 \, \text{m/s}^2 \].

Step 5: Find Distance Fallen by Raindrop

Use the formula \( x = \frac{1}{2} a t^2 \) to find the distance fallen when \( t = 3.0 \) s. Substitute \( a = 2.45 \, \text{m/s}^2 \): \[ x = \frac{1}{2} \times 2.45 \times (3.0)^2 = 11.025 \, \text{m} \].

Step 6: Determine Mass of Raindrop

The mass formula is \( m = kx \). Given \( k = 2.00 \, \text{g/m} \), convert it to \( \, \text{kg/m} \): \( k = 0.002 \, \text{kg/m} \). Calculate the mass: \[ m = kx = 0.002 \times 11.025 = 0.02205 \, \text{kg} \].

Key Concepts

Differential Equations

Differential equations are mathematical tools used to describe various physical phenomena where variables change over time or space. In the context of a raindrop with variable mass falling through a cloud, a differential equation helps us describe the dynamics of its motion. Here, we see the equation:- \( xg = x \frac{dv}{dt} + v^2 \)This represents how the force affecting the raindrop evolves as it falls and gathers mass. By adopting a proposed solution, like \( v = at \), which suggests that velocity is linearly related to time with a constant acceleration \( a \), we simplify the complexity of the problem.Solving differential equations often involves substituting known relationships to find expressions for unknowns. By expressing the acceleration and distance fallen in terms of time, as shown, we solve the differential equation to find practical and meaningful answers.

Variable-Mass Dynamics

Variable-mass dynamics involves objects whose mass changes over time, which can be due to a host of reasons, such as fuel consumption in rockets or, in this case, a raindrop picking up additional mass from smaller droplets.The equation \( m \frac{dv}{dt} + v \frac{dm}{dt} \) captures the unique characteristics of variable-mass systems where both mass \( m \) and velocity \( v \) change. It shows that the total external force acting on the raindrop equals the sum of the forces due to its changing velocity and increasing mass.

• The term \( m \frac{dv}{dt} \) accounts for changes in velocity, familiar from fixed-mass systems.
• The term \( v \frac{dm}{dt} \) represents the added momentum due to the mass change.

Understanding these components allows us to model real-world systems more accurately, showcasing the diverse applications of variable-mass dynamics in fields like meteorology, aerospace, and engineering.

Acceleration Calculation

Calculating the acceleration for systems like a falling raindrop involves understanding how the variables interact. For this raindrop, we found an expression for acceleration \( a \) by working through the derived differential equation.To find \( a \), we substitute \( v = at \) and \( x = at^2 \) into the differential equation and simplify. This results in:- \( a = \frac{g}{1+t} \)Here, \( g \) is the acceleration due to gravity. For the given conditions when \( t = 3.0 \, \text{s} \), \( a \) becomes:- \( a = \frac{9.8}{4} = 2.45 \, \text{m/s}^2 \) The calculation shows how external forces and variable changes (like gaining mass) affect the overall acceleration. This highlights the intricate relationship between forces and motion, emphasizing the need for precise calculations to predict motion accurately in variable-mass contexts.