Problem 66 - Physics for Scientists and Engineers with Modern Physics

Question

(II) The terminal velocity of a \(3 \times 10^{-5} \mathrm{~kg}\) raindrop is about \(9 \mathrm{~m} / \mathrm{s}\). Assuming a drag force \(F_{\mathrm{D}}=-b v\), determine \((a)\) the value of the constant \(b\) and \((b)\) the time required for such a drop, starting from rest, to reach \(63 \%\) of terminal velocity.

Step-by-Step Solution

Answer

(a) \(b \approx 3.267 \times 10^{-5} \; \text{kg/s}\), (b) \(t \approx 0.653 \; \text{s}\).

Step 1: Define the Forces

The forces acting on the raindrop are gravity and the drag force. Gravity is given by the equation \( F_g = mg \), where \( m = 3 \times 10^{-5} \; \text{kg} \) and \( g = 9.8 \; \text{m/s}^2 \). Thus, \( F_g = 3 \times 10^{-5} \times 9.8 \approx 2.94 \times 10^{-4} \; \text{N} \). The drag force is \( F_D = -bv \). At terminal velocity (\( v_t = 9 \; \text{m/s} \)), \( F_g = F_D \).

Step 2: Calculate the Constant b

At terminal velocity, the drag force equals the gravitational force: \( b v_t = mg \). Solving for \( b \), \( b = \frac{mg}{v_t} = \frac{2.94 \times 10^{-4}}{9} \approx 3.267 \times 10^{-5} \; \text{kg/s} \).

Step 3: Use the Exponential Velocity Equation

The velocity of the raindrop as a function of time is given by \( v(t) = v_t (1 - e^{-(b/m)t}) \). We need to find \( t \) when \( v(t) = 0.63 v_t \).

Step 4: Set Up the Equation for Time

Substitute \( v(t) = 0.63 v_t \) into the expression: \( 0.63 v_t = v_t (1 - e^{-(b/m)t}) \). Simplify to get: \( 0.63 = 1 - e^{-(b/m)t} \).

Step 5: Solve for Time t

Rearrange the equation: \( e^{-(b/m)t} = 1 - 0.63 = 0.37 \). Take the natural logarithm on both sides: \( -(b/m)t = \ln(0.37) \). Thus, \( t = -\frac{ ext{ln}(0.37) \cdot m}{b} \approx \frac{(\ln(0.37) \cdot 3 \times 10^{-5})}{-3.267 \times 10^{-5}} \approx 0.653 \; \text{s} \).

Key Concepts

Drag Force

Drag force is a crucial concept in understanding how objects move through fluids such as air or water. It is the resistive force that opposes the motion of an object. When an object falls through the air, like a raindrop descending from the sky, it experiences a drag force that depends on its velocity. This force can be mathematically expressed as \( F_{\text{D}} = -b v \), where \( b \) is a constant representing the drag coefficient, and \( v \) is the velocity of the object.

The negative sign indicates that the drag force acts in the opposite direction to the object's motion. As the velocity of the object increases, the drag force grows stronger. This relationship means that as an object accelerates, the drag force increases until it eventually balances the downward force of gravity.

• The drag coefficient \( b \) is determined by factors such as the shape and size of the object, as well as properties of the fluid through which it moves.
• In our example, when the raindrop reaches its terminal velocity, the drag force equals the gravitational force, leading to a net acceleration of zero.

Exponential Velocity Equation

The exponential velocity equation is used to describe how an object's velocity changes over time as it approaches terminal velocity. When an object falls freely under the influence of gravity and experiences a drag force, its velocity as a function of time \( v(t) \) can be expressed with the equation:
\[v(t) = v_t \left(1 - e^{-\left(\frac{b}{m}\right)t}\right)\]This equation reveals how quickly the object accelerates to its terminal velocity \( v_t \). Here, \( t \) is the elapsed time since the object started falling, \( m \) is the mass of the object, and \( e \) is the base of the natural logarithm.

• As time progresses, the term \( e^{-\left(\frac{b}{m}\right)t} \) decreases, making the velocity \( v(t) \) inch closer to \( v_t \).
• Early in the fall, when \( t \) is small, the velocity is still increasing rapidly.
• As the object approaches terminal velocity, changes in velocity become minimal.

Time to Reach Terminal Velocity

Understanding the time it takes for a falling object to reach a certain percentage of its terminal velocity is important for predicting its motion. In this context, we calculate the time needed for a raindrop to reach 63% of its terminal velocity. This specific percentage is significant as it often represents a practical benchmark in many calculations due to mathematical properties of exponential growth.

The solution involves setting \( v(t) = 0.63v_t \) in the exponential velocity equation and solving for \( t \). The simplified equation becomes:

• Equation: \( 0.63 = 1 - e^{-\frac{b}{m}t} \)
• Rearranging gives \( e^{-\frac{b}{m}t} = 0.37 \)
• Taking the natural logarithm: \( -\frac{b}{m}t = \ln(0.37) \)
• Solving for \( t \): \( t = \frac{\ln(0.37) \cdot m}{-b} \)

Using the given values of the raindrop mass and the drag coefficient, the time is calculated to be approximately 0.653 seconds. This means the raindrop, starting from rest, quickly reaches 63% of its ultimate velocity in under a second.