Free Fall and Terminal Velocity

6 Key excerpts on "Free Fall and Terminal Velocity"

• 

  eBook - PDF

  Physics at a Glance

  Full Physics Content of the New GCSE

  • Tim Mills(Author)
  • 2008(Publication Date)
  • CRC Press

    (Publisher)

  H o w e v e r , t h i s i s a l s o t h e d e f i n i t i o n o f g r a v i t a t i o n a l f i e l d s t r e n g t h . 13 FORCES AND MOTION Terminal Velocity Driving force remains constant but air resistance increases as vehicle speeds up Terminal velocity occurs when the accelerating and resistive force on an object are balanced. Velocity Terminal velocity Air resistance Driving force Air resistance Time Driving force Driving force > air resistance Vehicle accelerates Velocity Terminal velocity (parachute open) Resistive forces Acceleration Terminal velocity Deceleration Speed decreases New, lower terminal velocity Reaction of ground Air resistance Air resistance increases as speed increases Weight Air resistance Weight Air resistance Large surface area greatly increases air resistance Weight Air resistance Slowing down reduces air resistance Weight Wear on moving parts Lubrication Friction Nuisance Drag on vehicles Streamlined shapes Use wheels rather than sliding Reduced fuel economy Useful Grip of tyres/shoes on ground Parachute Shuttlecock Brakes Time Weight Decelerating Accelerating Key ideas: • Drag/resistive forces on objects increase with increasing speed for objects moving through a fluid, e.g. air or water. • When accelerating and resistive forces are balanced, Newton’s First Law says that the object will continue to travel at constant velocity. Questions 1. What happens to the size of the drag force experienced by an object moving through a fluid (e.g. air or water) as it speeds up? 2. What force attracts all objects towards the centre of the Earth? 3. Why does a car need to keep its engine running to travel at constant velocity? 4. A hot air balloon of weight 6000 N is released from its mooring ropes. a. The upward force from the hot air rising is 6330 N. Show the initial acceleration is about 0.5 m/s 2 . b. This acceleration gradually decreases as the balloon rises until it is travelling at a constant velocity.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  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 .

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  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.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Workshop Physics Activity Guide Module 1

  Mechanics I

  • Priscilla W. Laws, David P. Jackson, Brett J. Pearson(Authors)
  • 2023(Publication Date)
  • Wiley

    (Publisher)

  In other words, all objects will fall with the same acceleration in the absence of air resistance. We say that objects near the surface of Earth have a free-fall acceler- ation of 9.8 m/s 2 . (If you’re being careful with units, you should have found that a y = −9.8 N/kg, but because we are calculating an acceleration, it is appropri- ate to convert the units from N/kg to m/s 2 .) The minus sign in our answer is a result of our choice of coordinate system and tells us that the acceleration points in the negative y-direction (down). This result is only strictly true when there’s no air resistance; for many objects, this will not be the case, as the following activity makes clear. 6.5.3. Activity: Air Drag a. Using the same motion-sensor setup as in Activity 6.3.2, track the motion of a dropped (uncrumpled) coffee filter. Because the motion of the coffee filter is strongly affected by air resistance, it may float to the left or right after being dropped. Thus, you’ll probably need to try the experiment a few times before you are able to track its UNIT 6: GRAVITY AND PROJECTILE MOTION 179 motion. Once you have clean data, make a sketch of the velocity-time graph (you can ignore any small bumps in your actual data; just try to sketch the basic shape of the graph). b. Notice that the coffee filter reaches a constant speed after a short period of time. We call this the object’s terminal speed. What is the (approx- imate) terminal speed of the coffee filter? What is the coffee filter’s acceleration once it reaches its terminal speed? c. Below, draw a free-body diagram of the coffee filter (mass m) as it is falling to the ground; include a standard coordinate system next to your diagram. For this situation, in addition to the gravitational force, there is a drag force from the air that cannot be ignored. This drag force points in the opposite direction of the motion and its magnitude changes with time (the magnitude actually depends on the object’s speed).

  Sign up to read

  Learn more about book
• 

  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.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Fundamentals of Physics, Extended

  • David Halliday, Robert Resnick, Jearl Walker(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  138 CHAPTER 6 FORCE AND MOTION—II 6.2 THE DRAG FORCE AND TERMINAL SPEED Learning Objectives After reading this module, you should be able to . . . 6.2.1 Apply the relationship between the drag force on an object moving through air and the speed of the object. 6.2.2 Determine the terminal speed of an object falling through air. Key Ideas ● When there is relative motion between air (or some other fluid) and a body, the body experiences a drag force D → that opposes the relative motion and points in the direction in which the fluid flows relative to the body. The magnitude of D → is related to the relative speed v by an experimentally determined drag coefficient C according to D = 1 _ 2 CρAv 2 , where  is the fluid density (mass per unit volume) and A is the effective cross-sectional area of the body (the area of a cross section taken perpendicular to the rela- tive velocity v → ). ● When a blunt object has fallen far enough through air, the magnitudes of the drag force D → and the gravi- tational force F → g on the body become equal. The body then falls at a constant terminal speed v t given by v t = √ _____ 2F g _____ CρA . The Drag Force and Terminal Speed A fluid is anything that can flow—generally either a gas or a liquid. When there is a relative velocity between a fluid and a body (either because the body moves through the fluid or because the fluid moves past the body), the body experiences a drag force D → that opposes the relative motion and points in the direction in which the fluid flows relative to the body. Here we examine only cases in which air is the fluid, the body is blunt (like a baseball) rather than slender (like a javelin), and the relative motion is fast enough so that the air becomes turbulent (breaks up into swirls) behind the body.

  Sign up to read

  Learn more about book