Problem 60
Question
Suppose an object is released and moves through a viscous fluid that tends to resist the motion of the object. Then the velocity \(v\) increases with time, and may approach a terminal velocity \(v_{T}\), which depends on the mass of the object and the viscosity of the fluid. If the fluid's resistance is proportional to the object's velocity, then as a function of time \(t\) the velocity is given by $$ v=v_{T}\left(1-e^{-g t / v_{T}}\right) $$ where \(g=9.8\) (meters per second per second) is the acceleration due to gravity. a. Find \(\lim _{t \rightarrow \infty} v(t)\). b. If the object is a tiny fog droplet (which is frequently on the order of \(5 \times 10^{-6}\) meters in radius), and is falling in the sky near earth, then a reasonable value for the terminal velocity is \(2.7 \times 10^{-2}\) meters per second (which is equivalent to 1 meter every 37 seconds). If the fog droplet begins falling at time \(t=0\), determine how long it takes for the velocity of the droplet to reach half of the terminal velocity.
Step-by-Step Solution
Verified Answer
a) \(v_T\); b) Approx. 0.0019 seconds.
1Step 1: Identify the Expression for Lim t→∞ v(t)
To find the terminal velocity as time approaches infinity, evaluate the limit of the function \(v = v_T(1 - e^{-gt / v_T})\) as \(t\) approaches infinity.
2Step 2: Simplify the Exponential Decay Term
As \(t\) approaches infinity, the exponential term \(e^{-gt / v_T}\) tends to 0 because the exponent \(-gt / v_T\) becomes very negative, thus making\(lim_{t \to \infty} e^{-gt / v_T} = 0\).The velocity formula then simplifies to \(lim_{t \to \infty} v(t) = v_T(1 - 0) = v_T\).
3Step 3: Calculate v(t) for Half Terminal Velocity
Determine the time \(t\) at which the velocity \(v\) is half the terminal velocity \(v_T\). Set up the equation\(v_T(1 - e^{-gt / v_T}) = \frac{1}{2}v_T\).
4Step 4: Solve the Equation for t
Simplify the equation \(v_T(1 - e^{-gt / v_T}) = \frac{1}{2}v_T\) to obtain \(1 - e^{-gt / v_T} = \frac{1}{2}\).Rearrange to find the exponential term:\(e^{-gt / v_T} = \frac{1}{2}\).Take the natural logarithm of both sides to solve for \(t\):\(-\frac{gt}{v_T} = \ln\left(\frac{1}{2}\right)\).Then solve for \(t\):\(t = -\frac{v_T}{g} \ln\left(\frac{1}{2}\right)\).
5Step 5: Plug in Known Values to Solve for t
Insert the known values for the terminal velocity \(v_T = 2.7 \times 10^{-2}\) m/s and gravity \(g = 9.8\) m/s².Calculate \(t = -\frac{2.7 \times 10^{-2}}{9.8} \ln\left(\frac{1}{2}\right)\).Simplifying gives a numerical value for \(t\). Compute to obtain:\(t \approx -\frac{2.7 \times 10^{-2}}{9.8} \times (-0.693)\), which is approximately 0.0019 seconds.
Key Concepts
Viscous Fluid DynamicsExponential DecayLimit Calculation
Viscous Fluid Dynamics
Viscous fluid dynamics is a fascinating area of physics that analyzes how objects move through fluids like air or water, which exert resistance or drag. This resistance is crucial because it affects how quickly an object can reach its terminal velocity. Terminal velocity is the constant speed that an object reaches when the force of gravity pulling it downwards is balanced by the drag force of the fluid pushing upwards.
The viscosity of a fluid is its thickness or stickiness. For example, honey is more viscous compared to water. When an object falls through a fluid, like a fog droplet in the sky, the fluid's viscosity resists the motion of the object. In this process, two main forces are at play: gravity acting downwards and drag acting upwards. As the object speeds up, the drag force increases until it equals the gravitational force. This balance results in the object moving at a constant speed known as terminal velocity.
Understanding viscous fluid dynamics helps explain why objects like skydivers eventually stop accelerating and why tiny particles behave differently than larger ones in the same fluid.
The viscosity of a fluid is its thickness or stickiness. For example, honey is more viscous compared to water. When an object falls through a fluid, like a fog droplet in the sky, the fluid's viscosity resists the motion of the object. In this process, two main forces are at play: gravity acting downwards and drag acting upwards. As the object speeds up, the drag force increases until it equals the gravitational force. This balance results in the object moving at a constant speed known as terminal velocity.
Understanding viscous fluid dynamics helps explain why objects like skydivers eventually stop accelerating and why tiny particles behave differently than larger ones in the same fluid.
Exponential Decay
Exponential decay is a mathematical concept where a quantity decreases at a rate proportional to its current value. In the context of an object moving through a viscous fluid, the fluid's resistance is often modeled using exponential decay. This means that as time passes, the rate at which the resistance decreases follows an exponential pattern.
The velocity of an object moving through a viscous fluid is given by the equation \[ v = v_T \left(1 - e^{-\frac{gt}{v_T}}\right) \] where \(v_T\) is the terminal velocity, \(g\) is the acceleration due to gravity, and \(t\) is time. The term \(e^{-\frac{gt}{v_T}}\) is the exponential decay term and diminishes over time. As \(t\) increases, this term approaches zero, meaning that the velocity \(v\) approaches the terminal velocity,
because the resistance offered by the fluid becomes negligible.
This principle of exponential decay helps us understand why objects reach a constant velocity when falling through a fluid, illustrating how drag impacts the motion over time.
The velocity of an object moving through a viscous fluid is given by the equation \[ v = v_T \left(1 - e^{-\frac{gt}{v_T}}\right) \] where \(v_T\) is the terminal velocity, \(g\) is the acceleration due to gravity, and \(t\) is time. The term \(e^{-\frac{gt}{v_T}}\) is the exponential decay term and diminishes over time. As \(t\) increases, this term approaches zero, meaning that the velocity \(v\) approaches the terminal velocity,
because the resistance offered by the fluid becomes negligible.
This principle of exponential decay helps us understand why objects reach a constant velocity when falling through a fluid, illustrating how drag impacts the motion over time.
Limit Calculation
The concept of limit calculation is essential in the study of how objects interact with viscous fluids. When we talk about terminal velocity, limit calculations allow us to understand what happens as time goes towards infinity, denoted as \(t \to \infty\).
In the given equation for velocity, \( v = v_T \left(1 - e^{-\frac{gt}{v_T}}\right) \), as \(t\) approaches infinity, the term \(e^{-\frac{gt}{v_T}}\) becomes zero. This is due to the nature of the exponential function where the exponent is a large negative value, leading the entire term to converge to zero. Consequently, the velocity \(v\) simplifies to \(v_T(1 - 0) = v_T\).
Through the limit calculation, we're able to conclude that no matter what, the object's velocity will eventually stabilize at the terminal velocity.
In the given equation for velocity, \( v = v_T \left(1 - e^{-\frac{gt}{v_T}}\right) \), as \(t\) approaches infinity, the term \(e^{-\frac{gt}{v_T}}\) becomes zero. This is due to the nature of the exponential function where the exponent is a large negative value, leading the entire term to converge to zero. Consequently, the velocity \(v\) simplifies to \(v_T(1 - 0) = v_T\).
Through the limit calculation, we're able to conclude that no matter what, the object's velocity will eventually stabilize at the terminal velocity.
- This fundamental tool in calculus provides a deeper understanding of infinite behaviors in physical scenarios.
- It assures that our mathematical models correctly predict the physical world.
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