Terminal Velocity

5 Key excerpts on "Terminal Velocity"

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  Available until 9 Apr |Learn more

  A First Course in Differential Equations, Modeling, and Simulation

  • Carlos A. Smith, Scott W. Campbell(Authors)
  • 2016(Publication Date)
  • CRC Press

    (Publisher)

  From Equations 2.11, 2.5a, and 2.13b dy v dt g r e d t y rt t t y = = --∫ ∫ ∫ [ ] 1 0 0 30 26 A First Course in Differential Equations, Modeling, and Simulation After integration and some simple rearrangement (again, practice), we get y g r t g r e rt = -+ --30 1 2 [ ] (2.14) At this point we stress once again that the starting point was a physical law, Newton’s second law, followed by a couple of equations describing the phenomena due to the forces of gravity and air resistance. As mentioned in Section 2.4, the model of the drag force due to air resistance ( F d = – Pv y ) is rather crude. Experiments suggest that a better model may be for the drag force to vary with the square of the velocity. Think about the solution using this new model; it is the topic of Problem 2.14 at the end of the chapter. Before concluding the section, let’s discuss the meaning of Terminal Velocity , and explain it considering Equation 2.12a, m dv dt mg Pv y y = --(2.12a) As discussed, the first term on the right-hand side is the force due to gravity, which is constant and down. The second term is air drag force and opposes the motion of the object; thus, it is up. When the object is released, the velocity is slow, but as it continues falling, the velocity increases and therefore the drag force also increases. If enough dis-tance is available (that means, before getting to the ground), the up drag force becomes equal to the down gravity force and the velocity reaches a constant value, dv y / dt = 0. This constant velocity is the Terminal Velocity, which can easily be calculated from Equation 2.12a, v m P g y terminal = -Substituting v y terminal into Equation 2.13b it can be seen, mathematically, that the time to reach Terminal Velocity is infinite. One might ask though, technically speaking, how long it would take to reach it. In Chapter 5 we develop an engineering rule of thumb that answers this question.

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  eBook - ePub

  Doing Physics with Scientific Notebook

  A Problem Solving Approach

  • Joseph Gallant(Author)
  • 2012(Publication Date)
  • Wiley

    (Publisher)

  Eq. (4.13b) becomes

  (4.19)

  where v is the (positive) magnitude of the (vertical) velocity vector. As the object moves faster, its acceleration decreases. When the object reaches Terminal Velocity its acceleration is zero.

  (4.20)

  When the object’s speed is vT , its acceleration is zero and it continues to fall at that constant velocity.

  Figure 4.4 shows the vertical speed as a function of time for an object dropped from rest (vox = 0) with and without air resistance.

  Figure 4.4

  Terminal Velocity

  Figure 4.5

  A tale of two trajectories

  With no air resistance, the speed increases linearly with time. With air resistance, the falling speed approaches Terminal Velocity.

  The plot uses k = 1/5s (which is about the baseball value) so the Terminal Velocity is approximately vT = 49 m/s.

  Trajectory

  When we ignored air resistance, the calculation to find the trajectory was relatively easy and straightforward. Now the equations are more complicated, but the basic ideas are the same. Use Solve Exact and Eq. (4.17a) for the x-coordinate to find the time.

  Now Substitute (with Evaluate ) the time into Eq. (4.17b) for the y-coordinate, and then Simplify in-place the second term and Expand in-place the argument of the natural logarithm.

  We can edit this result by-hand and put the trajectory in a more standard form. Here is the trajectory in terms of the magnitude and direction of the initial velocity.

  (4.21)

  This is not the equation of a parabola!

  The expression analogous to Eq. (4.8) for the range of the projectile under the influence of linear air resistance cannot be expressed in terms of elementary functions. If you try to find the range with Solve Exact and Eq. (4.21) , SNB cannot find an algebraic solution.

  Even so, we can find the range for particular problems using Eq. (4.21) and Solve Numeric

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  Guide to Mechanics

  • Philip Dyke, Roger Whitworth(Authors)
  • 2017(Publication Date)
  • Red Globe Press

    (Publisher)

  These values are g = k for R / speed and p g = D † for R / (speed) 2 . Each of these values is called the Terminal Velocity , for obvious reasons. As this velocity is approached, the acceleration gets smaller and smaller. Ultimately, accelera-tion becomes negligible, and so, according to Newton's first law, the net external force on the particle will also be negligible. The Terminal Velocity must, therefore, be given by putting a ˆ 0 in equation (5.20); thus: R ˆ mg With R ˆ mk j v j , this implies j v j ˆ g = k , and with R ˆ mDv 2 , this implies v 2 ˆ g = D , which is consistent with previous results. Also, if initially the speed of the particle exceeds the terminal speed, R mg > 0, whereas if initially the speed of the particle is less than the terminal speed (the case considered above), R mg < 0. In each case the sign of a , the acceleration, ensures that the speed approaches the terminal speed for large t . Figure 5.3 summarises these findings. Note that the arguments leading to the results in Figure 5.3 are independent of the form of R . This completes what we want to say about vertical motion under gravity. Here is an extended example which may be considered a case study. Motion Under Gravity 111 Speed (> Terminal speed) t v = Speed v > v v < v v Speed (< Terminal speed) Figure 5.3 Speed plotted against time to illustrate terminal speed Example 5.9 A helicopter is stationary at a height of 1000 m above the ground when a parachutist jumps out. When the parachute is not open, air resistance can be assumed to be negligible. After the parachute opens, resistance is assumed to be proportional to the square of the speed through the formula R ˆ mv 2 = 100, where m is the mass of the parachutist and his parachute. (a) If all is well, the parachutist will count 5 s, then pull the rip-cord. Assuming that the parachute opens instantaneously, calculate the parachutist's velocity as a function of time and draw a graph of v against t .

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  An Introduction to Physical Science

  • James Shipman, Jerry Wilson, Charles Higgins, Bo Lou, James Shipman(Authors)
  • 2020(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 40 Chapter 2 ● Motion Air resistance is the result of a moving object colliding with air molecules. Therefore, air resistance (considered a type of friction) depends on an object’s size and shape, as well as on its speed. The larger the object (or the more downward area exposed) and the faster it moves, the more collisions and the more air resistance there will be. As a skydiver accelerates downward, his or her speed increases, as does the air resistance. At some point, the upward air resistance balances the downward weight of the diver and the acceleration goes to zero. The skydiver then falls with a constant velocity, which is called the Terminal Velocity, v t (●●Fig. 2.11a). Wanting to maximize the time of fall, skydivers assume a “spread-eagle” position to provide greater surface area and maximize the air resistance (Fig. 2.11b). The air resistance then builds up faster and Terminal Velocity is reached sooner, giving the sky- diver more fall time. This position is putting air resistance to use. The magnitude of a skydiver’s Terminal Velocity during a fall is reached at about 200 km/h (125 mi/h). Acceleration is used in a variety of practical applications. You probably use an acceler- ometer. See Physical Science Today 2.1: Rotating Tablet Screens. v 5 v o v 5 0 v o v g v v v g g Figure 2.10 Up and Down An object projected straight upward slows down because the accelera- tion due to gravity is in the opposite direction of the velocity, and the object stops (v 5 0) for an instant at its maximum height.

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  Thermodynamics, Kinetics, and Microphysics of Clouds

  • Vitaly I. Khvorostyanov, Judith A. Curry(Authors)
  • 2014(Publication Date)
  • Cambridge University Press

    (Publisher)

  The Terminal Velocity V t of a falling body in the steady state dV/dt = 0, as described by (Eqn. 12.2.5a), follows from (Eqn. 12.2.5a) by projection to the vertical direction. The force F gr is directed downward, and the other three forces are upward. Thus, taking into account the signs, V t is obtained by equating the drag force F D to the difference of the gravitational force F gr and the sum of the buoyancy force F b and electrical force F el , v ρ ρ ρ - - = - - = = mg F F g qE F C V A ( ) (1/2) , b el b F b D D F t 2 (12.2.5b) Solving for V t , we obtain v v ρ ρ ρ ρ ρ ρ ρ = - -         = - -         = -         - V mg F F AC g qE AC g AC K 2 | | | 2 ( ) 2 | 2 | | |1 | . t b el F D b b F F D b b F F D el 1/ 2 1/2 1/2 1/2 (12.2.6a) K el is the ratio of the electrical force to the difference of the gravitation and buoyancy force, ρ ρ = - K qE g | | v . el b F b (12.2.6b) The notation |…| refers to the absolute value, since here we consider the positive differences. Equation (12.2.6a) shows that the effect of the electrical forces can be accounted for by multiplying by the factor |1 − K el | 1/2 , which is equal to 1 in the absence of the electrical charge. Hereafter, we omit this factor for the sake of simplicity, keeping in mind that it can be accounted for in the final equation for V t . Then the expression for V t becomes v ρ ρ ρ = -         V g AC 2 | | . t b b F F D 1/2 (12.2.6c) The sign |…| of absolute value means that we consider falling bodies if they are denser than the fluid (r b > r F ) and rising bodies in the opposite case r b < r F . For hydrometeors in the air, r b >> r F , and the term with r F under the absolute value sign can be neglected, however, if r b /r F ∼ 1, as, for example, for sand particles in the ocean or for dropsondes or radiosondes in the air or for falling objects in laboratory devices with dense liquids, then r F should be retained.

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