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The terminal velocity of a raindrop that is \(4.0\) \(\mathrm{mm}\) in diameter is approximately \(8.5 \mathrm{~m} / \mathrm{s}\) under controlled, windless conditions. The density of water is \(1000 \mathrm{~kg} / \mathrm{m}^{3}\). Recall that the density of an object is its mass divided by its volume. (a) If we model the air drag as being proportional to the square of the speed, \(F_{\mathrm{drag}}=-b v^{2}\), what is the value of \(b\) ? (b) Under the same conditions as above, what would be the terminal velocity of a raindrop that is \(8.0 \mathrm{~mm}\) in diameter? Try to use your answer from part (a) to solve the problem by proportional reasoning instead of just doing the same calculation over again.

Short Answer

Expert verified
The value of the constant \(b\) is approximately \(10.22 \, kg/m\). The terminal velocity for a raindrop of \(8.0 \, mm\) diameter is approximately \(17.0 \, m/s\).

Step by step solution

01

Calculate the mass of the raindrop

First, obtain the mass of the raindrop by utilizing the formula for the volume of a sphere and the given density of water. The volume \(v\) of a sphere with radius \(r\) (in this case, half of the raindrop's diameter) is given by \(v = \frac{4}{3} \pi r^3\). Substituting the radius \(4.0 mm = 4.0 \times 10^{-3} m = 2.0 \times 10^{-3} m\) and the density \(\rho = 1000 kg/m^3\), the mass \(m\) is then calculated using the relation \(m = \rho v\).
02

Equate the forces to find b

At terminal velocity, the force due to gravity \(F_g\) equals the drag force \(F_{drag} = -bv^2\). The gravitational force acting on the raindrop is given by \(F_g = mg\), where \(m\) is the mass of the raindrop and \(g=9.8m/s^2\) is the acceleration due to gravity. Setting the forces equal gives \(mg = bv^2\). Solve this equation for \(b\).
03

Calculate the terminal velocity of a raindrop with diameter 8.0 mm

Using proportional reasoning, note that the diameter of the bigger raindrop is doubled, meaning its volume (and therefore also its mass, assuming the density remains constant) is increased by a multiplier of 8. Since the terminal velocity is proportional to the square root of the mass (inferred from the equation in Step 2), we can multiply the original terminal velocity by the square root of the multiplier, which gives the terminal velocity of the larger raindrop.

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Key Concepts

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

Air Drag
Air drag is a crucial factor when considering the motion of objects through a fluid such as air. As an object moves, it interacts with air molecules. This interaction results in a force opposite to the direction of motion, known as air drag or air resistance. The magnitude of this drag force can vary based on several parameters like the speed of the object, its surface area, and air density.

For a small raindrop falling under gravity without wind interference, air drag can be mathematically modeled as proportional to the square of its velocity, indicated by the formula:
  • \( F_{\text{drag}} = -b v^2 \)
Here, \(b\) is a constant that incorporates factors like the shape of the object and the density of the air. The negative sign signifies that the drag force opposes the direction of velocity.

Understanding air drag is essential for determining when an object achieves terminal velocity, the point where the drag force balances gravitational force. Beyond this point, the object no longer accelerates and continues to move at a constant speed downward.
Density of Water
Density is a fundamental property of matter defined as mass per unit volume. For water, this density is relatively high at \(1000 \text{ kg/m}^3\), meaning each cubic meter of water weighs about 1000 kilograms.

In the context of a raindrop, knowing the density of water helps calculate its mass when its volume is known. For a sphere (which raindrops approximate), the formula for volume is:
  • \( v = \frac{4}{3} \pi r^3 \)
Substituting the radius derived from the raindrop's diameter into the equation gives the volume. Multiplying by the density of water then provides the mass of the raindrop.

With this mass, other forces like gravity can be calculated, allowing us to determine the corresponding drag force necessary for terminal velocity. Understanding the density of water was thus critical in deriving forces acting on a falling raindrop in the problem scenario.
Proportional Reasoning
Proportional reasoning is a powerful mathematical tool that helps draw conclusions about the relationships between quantities without extensive calculations. It involves understanding how one quantity changes in relation to another.

In the raindrop example, instead of recalculating the terminal velocity by repeating all the steps with a new diameter, we use the principle of proportional reasoning. Here, if the diameter of the raindrop is doubled, its volume—and thus its mass—changes by the cube of the size change multiplier (since volume is a cubic measure). The mass increases by a factor of 8 when the diameter doubles from 4 mm to 8 mm.

Since terminal velocity is based on balancing drag with gravitational force (
  • \( mg = bv^2 \)
), it shows a relationship where terminal velocity is proportional to the square root of mass. Therefore, when the mass increases eight-fold, terminal velocity increases by a factor of \(\sqrt{8}\). This method provides an efficient calculation shortcut and illustrates a broader understanding of how physical properties influence motion.

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