MOTION OF FALLING OBJECTS WITH RESISTANCE

Transcription

1 DOING PHYSICS WIH MALAB MECHANICS MOION OF FALLING OBJECS WIH RESISANCE Ian Cooper School of Physics, Uniersity of Sydney DOWNLOAD DIRECORY FOR MALAB SCRIPS mec_fr_mg_b.m Computation of the displacement, elocity and acceleration for the motion of an object acted upon by a resistie force FR and its weight FG m g. he equation of motion is soled by analytical means (integration of the equation of motion) and by a finite difference numerical method. mec_fr_mg_b.m Computation of the displacement, elocity and acceleration for the motion of an object acted upon by a resistie force FR and its weight FG m g. he equation of motion is soled by analytical means (integration of the equation of motion) and by a finite difference numerical method. mec_stokes.m Computation of the displacement, elocity and acceleration for the motion of an object acted upon by a resistie force F 6 R (Stokes Law), the buoyance force F buoy and its weight F G numerical method. R F m g. he equation of motion is soled by a finite difference mec_drag.m Computation of the displacement, elocity and acceleration for the motion of an object 1 acted upon by a resistie force FR CD6AF and its weight FG m g. he equation of motion is soled by a finite difference numerical method. mec_tt_ball.m Experimental data for a table tennis ball falling from rest. Used to plot the actual measurements of displacement and time from a ideo recording with the predicted alues of displacement using a finite difference approach. 1

2 INRODUCION We will consider the ertical motion of objects through fluids near the Earth s surface where the acceleration due to graity is assumed to be constant g = 9.8 m.s -. he motion of falling objects is usually described with constant acceleration. his is only approximately true. For example, in introductory physics textbooks, two objects of different mass when dropped simultaneously from rest will hit the ground at the same time. his is an idealized situation and ignores the effects of the air resisting the motion of the falling objects. Air resistance, a friction which increases with increasing speed, acts against graity, so the speed of falling objects tends toward a limit called terminal elocity (terminal speed). Motion through any real fluid (liquid or gas) gies rise to forces resisting the motion. o a reasonable approximation, fluid resistance tends to depend on either the first power of the speed (a linear resistance) or the second power (a quadratic resistance). Our two models for the resistie force F R are Model (1) Model () FR FR low speeds high speeds where and are constants of proportionality. Model (1) for linear resistance is often applicable when the object is moing with low speeds. In the motion through a fluid, the resistie force FR is often called the iscous drag and it arises from the cohesie forces between the layers of the fluid. he S.I. units for the constant are N.m -1.s -1 or kg.s -1. Model () for quadratic resistance is more applicable for higher speeds. In the motion through fluids, the resistie force FR is usually called the drag and is related to the momentum transfer between the moing object and the fluid it traels through. he S.I. units for the constant are N.m -.s - or kg.m -1. Many problems in the mathematical analysis of particles moing under the influence of resistie forces, you start with the equation of motion. o find elocities and displacements as functions of time you must integrate the equation of motion.

3 he equation of motion for an object can be deried from Newton s Second Law 1 a Fi Newton s Second Law m i For the ertical motion of an object through a fluid, the forces acting on the object are the graitational force F G (weight) and the resistie force F R. In our frame of reference, we will take down as the positie direction. herefore, the ertical acceleration a of an object through a fluid is a g / m Model (1) / / a g m Model () In Model If the object is moing down then > / / / 1 / a g m g m g m If the object is moing up then < / / / 1 / a g m g m g m 3

4 MODEL 1 F ma m g R he mscript used for the plots is mec_fr_mg_b.m Analytical Approach F ma m g R For the ertical motion of an object through a fluid, the forces acting on the object are the graitational force F G (weight) and the resistie force F R. In our frame of reference, we will take down as the positie direction. he equation of motion of the object is determined from Newton s Second Law. d ma m FR mg a g dt m where a is the acceleration of the object at any instance. he initial conditions are / t x a m When a =, the elocity is constant = where is the terminal elocity m g mg terminal elocity We start with the equation of motion then integrate this equation where the limits of the integration are determined by the initial conditions (t = and = ) and final conditions (t and ) d m g a g dt m m m g du u du d dt m u mg mg t u du dt t logeu mg loge m u u m mg mg / m t e mg m g m g e / m t mg 4

5 herefore, we can express the elocity as e / m t elocity 1 e / increases to decreases to In eery case, the elocity tends towards the limiting alue. m t Plots of the elocity as functions of time t m =. kg = 5. kg.s -1 g = 9.8 m.s - = 3.9 m.s -1 Initial alues for elocity [m.s -1 ] blue: 1 red: magenta: cyan: 5

6 he acceleration a as a function of time t is e / m t d d a e dt dt / a e m m t / m t a e m a g e / m t a / m t a and decreases to a and a increases to t a Initial alues for elocity [m.s -1 ] blue: 1 red: magenta: cyan: 6

7 We can now calculate the displacement x as a function of elocity t e dx dx dt dt / m t x t / m t dx e dt m x t e / m t m m x t e / m t m x t 1 e t / m t t x m x t e / m t 1 Initial alues for elocity [m.s -1 ] blue: 1 red: magenta: cyan: So far we hae only considered the case where the initial elocity was either zero or a positie quantity ( ), i.e., the object was released from rest or projected downward. We will now consider the case where the object was project ertically upward ( < ). Note: in our frame of reference, the origin is taken as x =, the position of the object at time t = ; down is the positie direction and up is the negatie direction. When the object is launched upward at time t =, the initial elocity has a negatie alue. Let u be the magnitude of the initial elocity u u herefore, the equation for the elocity as a function of time t can be expressed as e u e / m t / m t 7

8 We can now find the time t up it takes for the object to rise to its maximum height x up aboe the origin (remember: up is negatie). At the highest point =, therefore, u e t up / m t up m u loge 1 he maximum height x up reached by the object in time t = t up is m x t 1 e m xup tup u e For the parameters / m t / m tup 1 m =. kg = 5. kg.s -1 g = 9.8 m.s - u = 1 m.s -1 = 3.9 m.s -1 he time to reach maximum height is t up =.57 s he max height h up reached is h up =.13 m x up = -.13 m 8

9 We can find the displacement x as a function of elocity d d a g / m dx dx g / m 1 d d dx mg / / mdx / m mg / We can integrate this equation by a substitution method or an algebraic manipulation method. Substitution Method u du d u d du u u u u / mdx du u x u u u / mdx du 1 du u u u u u / m x u loge u u u log u e u / m x log e m x loge b Algebraic manipulation d / m dx d 1 1 x u 1 1 / m dx d / m x log / m x log e m x loge b e QED 9

10 For the object projected up with an initial elocity = - u where u >, the maximum height reached x up occurs when = m x loge b m u xup u loge b m u xup loge 1 u b Note: up is negatie and down is positie in our frame of reference. 1

11 Numerical Approach F ma m g R We can also find the elocity and displacement of the object by soling Newton s Second Law of motion using a finite difference method. We start with d a g / m dt mg In the finite difference method we calculate the elocity and displacement x at N discrete times t k at fixed time interals t t 1, t,, t k,, t N t =t - t 1 t 1 = t k = (k-1) t k = 1,, 3,, N he acceleration a is approximated by the difference formula d t t t dt t herefore, the elocity t t k k1 k k tk at time t k + is / m t k1 k k / k1 tk 1 1 k t k k g t t t g t m t t t t g Hence, to calculate the elocity tk we need to know the elocity at the two preious time steps tk 1 and t k. We know t 1 = and (t 1 ) = () =. We estimate the elocity at the second time step t ( t ) ( t ) a( t ) t ( t ) / m t t where we hae assumed a constant acceleration in the first time step. We can improe our estimate of t ( ) by using an aerage alue of the acceleration in the first time step a( t1) a( t) ( t1) t ( t) 1 ( t1) / m t1 t t ( t) We can now calculate the elocity (t) at all times from t = t 1 to t = t N. 11

12 he acceleration at each time step is a tk g t m he displacement at each time step is dx x dt t x t t k1 k x t k t x t x t t t k k k1 k 1

13 EXAMPLE F ma m g R he mscript mec_fr_mg_b.m can be used for simulations for the motion of an object acted upon a resistie force of the form Input parameters FR (Model 1). m = kg = 1 kg.s -1 = 1 m.s -1 t = 1x1-4 s t max =. s Outputs N numerical approach A analytical approach terminal elocity = 1.96 m.s -1 acceleration t a elocity t 13

14 displacement t x For the input parameters used in this simulation, there is excellent agreement between the alues calculated using the numerical and analytical approaches. Howeer, you always need to be careful in using numerical approaches to sole problems. In this instance, you need to check the conergence of results by progressiely making the time step t smaller. When the time step is t = 5x1 - s the numerical and analytical results do not agree. he time step is too large for accurate results using the numerical approach. 14

15 MODEL F ma m g R he mscript used for the plots is mec_fr_mg_b.m Analytical Approach F ma m g R For the ertical motion of an object through a fluid, the forces acting on the object are the graitational force F G (weight) and the resistie force F R. In our frame of reference, down is the positie direction. he equation of motion of the object is determined from Newton s Second Law. d dt / / m a m FG FR mg a g where a is the acceleration of the object at any instance. he initial conditions are m t x a g / m When a =, the elocity is constant = where is the terminal elocity mg m g mg We start with the equation of motion then integrate this equation where the limits of the integration are determined by the initial conditions (t = and = ) and final conditions (t and ). Since the acceleration depends upon its a more difficult problem then for the linear resistie force example. We hae to do separate analytical calculations for the motion when the object is falling or rising. Velocity of the object is always positie (falling object) Equation of motion a g m 15

16 d a g dt m d d m g dt mg g m m d dt d m mg 1 1 dt d m 4 g 1 1 dt m 4 g m t dt d g t log m log e e d 4 g t log m log e e 4 g t loge loge m 4 g t loge m 4 g t m 4 g 4 g 4 g g e m mg t e K K e 1 K 1 K 1 K 1 K 1 1 g t g t g t e g t e e e g g t alid only if 16

17 We can now calculate the displacement x as a function of elocity d d a g / m dt dx d / mm g / m g / dx m d V d dx g x dx g x loge g d x loge g alid only if We can now inestigate what happens as time t g t t e t In falling, the object will finally reach a constant elocity (a = ) which is known as the terminal elocity. 1 t V x loge g In falling, as time t increases the objects displacement x just gets larger and larger. 17

18 Example Small rock dropped from rest: m =.1 kg = kg.m -1 = m.s -1 = 31.3 m.s

19 Example Small rock thrown ertically downward ( < ) m =.1 kg = kg.m -1 = +5. m.s -1 = 31.3 m.s

20 Example Small rock thrown ertically downward ( > ) m =.1 kg = kg.m -1 = +4. m.s -1 = 31.3 m.s -1 For problems in which the object is projected ertically upward, you hae to diide the problem into two parts. (1) Calculate the time to reach its maximum height and calculate the maximum height reached for the upward motion. () Reset the initial conditions to the position at maximum height where the initial elocity becomes = and do the calculations for the downward moement of the object.

21 Velocity of the object negatie and moing up and Equation of motion Note: up is the positie direction a g m alid only if and d a g dt m d d m g dt mg g m m d dt m t dt m Standard Integral d dx 1 x atan C a x a a 1 m m g t atan m g g t atan atan atan g g he time tup to reach maximum height occurs when = t up g g atan atan atan he elocity as a function of time t is atan g atan t g t -1 tan atan atan tan and 1

22 he displacement x as a function of elocity is d d a g / m dt dx d / mm g / m g / dx m d d dx g x dx d g and loge x g x loge g he maximum height x up reached by the object occurs when = x up loge g

23 Example Small rock thrown ertically upward ( < = - u u > ) m =.1 kg = kg.m -1 = - 1. m.s -1 = 31.3 m.s -1 he terminal elocity is m g / -1 m g / / 1 m.s m.s

24 When = the object reaches its maximum height x up (up is negatie) x x up up loge g m he time t up to reach the maximum height t t up up atan g s he calculations agree with the alues for t up and x up determined from the graphs. From the graphs: x = time t =.75 s elocity = 11.1 m.s -1 ime to fall from max height to origin t down = ( ) s = 1.6 s takes slight longer to fall then rise to and from origin to max height Launch speed = 1. m.s -1 slightly greater than return speed = 11.1 m.s -1 In the absence of any resistie forces a g at a s At maximum height t 1.45 s x m up up 4

25 Numerical Approach F ma m g R We can also find the elocity and displacement of the object by soling Newton s Second Law of motion using a finite difference method. We start with d a g / m dt (4) In the finite difference method we calculate the elocity and displacement x at N discrete times t k at fixed time interals t t 1, t,, t k,, t N t =t - t 1 t 1 = t k = (k-1) t k = 1,, 3,, N he acceleration a is approximated by the difference formula d t t t dt t k1 k k herefore, the elocity t t k t k tk at time t k + is g / m t / k1 t t t g / m t / k k k1 d a t g m t t dt 3 / / k k k Hence, to calculate the elocity tk 3 3 we need to know the elocity at the two preious time steps tk 1 and t k. We know t 1 = and (t 1 ) = () =. We estimate the elocity at the second time step t (13) ( t) ( t1) a( t1) t where we hae assumed a constant acceleration in the first time step. We can improe our estimate of t ( ) by using an aerage alue of the acceleration in the first time step a( t1) a( t) ( t1) t ( t) We can now calculate the elocity (t) at all times from t = t 1 to t = t N. 5

26 he displacement at each time step is dx x dt t x t t k1 x t k t (15) xt xt tt k k k1 k Proided the time step is small enough, there is excellent agreement between the numerical alues and analytical alues for the acceleration, elocity and displacement. 6

27 REYNOLDS NUMBER and MOION HROUGH A FLUID o characterize the motion of an object moing through a fluid, it is useful to define a dimensionless quantity called the Reynolds number N R N R R F F where the fluid is characterized by its iscosity F and density F. he effectie radius of the object is R and its elocity is. Different alues of the Reynolds number N R determine the different regimes of flow in which different laws of the resistie force are alid. Reynolds number N R Resistie force F 6 R to 1 Stokes law 1 to 3 ransition region 3 to 3 constant drag coefficient C D F ~ 1 C A drag drag D F > 3 Some ariation in drag coefficient C D with F drag only roughly proportional to F Low Reynolds Number ( < N R < 1) For a small object moing through a fluid with a low elocity, the flow around the object is essentially laminar where the fluid flows in layers and there is no turbulence. In this situation, it is found experimentally that the iscous drag force F acting on the object by the fluid is directly proportional to the elocity of the object. For a small sphere of radius R, the drag force F drag is also directly proportional to the radius R. drag For a small sphere of radius R, the iscous drag force iscous drag force F 6 where the drag force drag R SOKES LAW F Fdrag is gien by Stokes Law F drag is always in the opposite direction to the elocity. he quantity F is called the iscosity of the fluid [Pa.s kg.m -1.s -1 ]. Viscosity is a measure of the frictional force acting on a fluid flowing oer a solid surface. 7

28 For an object falling through a fluid, we make the assumption that there are three distinct forces acting on the object weight FG m g iscous drag force F 6 drag 4 3 buoyancy force buoy 3 R SOKES LAW F R g where in our frame of reference, down is taken as the positie direction. Newton s Second Law applied to the falling object gies he acceleration a is ma m g R R g 6 F F 4 3 F 3 a g 6 R / m R g / m he terminal elocity (a = ) of the object is F F F 4 3 m g 3 R F g 6 R F he motion of tiny droplets of water falling through the air, tiny particles of rock settling under the sea and the sedimentation of red blood cells in blood plasma, the rise of oil drops in water are examples of the motion of objects moing through a fluid with low Reynolds numbers. We can use the finite difference method to numerically find the elocity and displacement x of the object as functions of time. a t k1 d t t t dt t t t k1 k k 3 k1 3 g 6 R / m t R g / m k k 4 t 4 3 k k k1 3 t t t g 6 R / m t R g / m 8

29 Example: A small oil drop released in water will rise to the surface mscript mec_stokes.m Input parameters radius of oil drop R = 5x1-3 m density of oil drop = 9 kg.m -3 density of water F = 1 kg.m -3 iscosity of water F =.1 Pa.s max time t Max =.4 s No. time steps N = 1 Output parameters mass of oil drop m = 4.71x1-4 kg terminal elocity =.54 m.s -1 time to reach t( ) ~.3 s Reynolds number is in the range from to 1, hence Stokes Law F R is alid. he main reason the oil drop rises is because of the buoyancy force, the iscous drag force plays only a minor role. 9

30 3

31 Example: How do small water droplets fall? mscript mec_stokes.m Input parameters radius of water drop R = 4x1-5 m density of water drop = 1 kg.m -3 density of air F = 1. kg.m -3 iscosity of air F = 1.8x1-5 Pa.s max time t Max = 1. s No. time steps N = 1 Output parameters mass of water drop m =.7x1-1 kg terminal elocity =. m.s -1 time to reach t( ) ~.1 s Reynolds number is in the range from to 1, hence Stokes Law F R is alid. In the upper atmosphere, water apour condenses on hygroscopic nuclei (~ 1 1 / m 3 radii ~ 1-7 m) forming water droplets whose radii range from about 4x1-6 m to 4x1-5 m. For R < 4x1-5 m, the water droplets ery quickly each their ery low alues of terminal speed <. m.s -1. Because of the ery small alues of the terminal speeds of the water droplets, they can be kept afloat by ertical air currents. he aggregation of suspended water droplets forms clouds or near ground leel water droplets forms fogs. 31

32 In this example, the buoyance force acting on water droplets is insignificant. Stokes law is not alid for raindrops. A raindrop of only 1 mm in radius has a terminal elocity that is ery large ~ 1 m.s -1 and N R ~

33 Intermediate Reynolds Number (3 < N R < 3 ) For large fast moing objects such as cars, aeroplanes, cricket balls, baseballs, hailstones, and skydiers, the drag force to a good approximation aries with the square of its speed. he drag force F can be written as 1 F C A drag D F drag A effectie cross-sectional area of object [m ] density of fluid [kg.m -3 ] F C D drag coefficient [dimensionless] For objects of a define shape such as spheres, the drag coefficient is nearly constant oer a wide range of speeds and sizes. For larger Reynold numbers, the turbulence associated with the motion of the object through the fluid is mainly responsible for the frictional retarding force on the object as it moes through the fluid. his is the reason why the resistie force depends upon and it is much greater than the resistie force for laminar flow where the resistie force depends upon. For an object falling through a fluid, we make the assumption that there are two distinct forces acting on the object. he buoyant force can be assumed to be negligible. weight FG m g F C A 1 drag force drag D F where in our frame of reference, down is taken as the positie direction. Newton s Second Law applied to the falling object gies 1 ma m g CD A F he acceleration a is 1 a g CD A F / m 33

34 he terminal elocity (a = ) of the object is mg C A D F We can use the finite difference method to numerically find the elocity and displacement x of the object as functions of time. a t k1 d t t t dt t k1 k k tk t k 1 t k1 g CD A F / m t t k1 1 tk tk t g CD A F / m t t k1 k1 Example mscript Falling raindrops mec_drag.m Input parameters radius of raindrop R = 5.x1-3 m (big raindrop) density of water drop = 1 kg.m -3 drag coefficient C D =.5 density of air F = 1. kg.m -3 iscosity of air F = 1.8x1-5 Pa.s max time t Max = 6. s No. time steps N = 1 Output parameters mass of raindrop m = 5.4x1-4 kg cross-sectional area A = 7.85x1-5 m terminal elocity = 14.8 m.s -1 time to reach t( ) ~ 4 s 34

35 Our model in assuming Fdrag aries from zero to about 1. should be OK since the Reynolds number A raindrop could fall a distance ~ 5 m. With no frictional forces acting the raindrops would reach the ground with speeds ~ 3 m.s -1 (> 1 km.s -1 )!!! his does not happen. Raindrops quickly each their terminal speed within seconds and reach the ground at speeds < m.s

36 Example A falling table tennis ball mscript mec_drag.m mec_tt_ball.m We can model a table tennis ball falling from rest using the numerical approach where the resistie force or drag force is 1 F C A drag D F Input parameters radius of table tennis ball mass of table tennis ball R =.x1-3 m m = 1.7x1-3 kg mass entered and not calculated from density drag coefficient C D =.445 density of air F = 1.5 kg.m -3 iscosity of air F = 1.789x1-5 Pa.s max time t Max = 3. s No. time steps N = 1 Output parameters cross-sectional area A = 1.6x1-3 m terminal elocity = 8.79 m.s -1 time to reach t( ) ~.5 s distance to reach x( ) ~ 17 m 36

37 Our model in assuming Fdrag mostly in the range from 3 to 3. should be OK since the Reynolds number is 37

38 But, how good is our model? We can test our model against experimental data. he data for the fall of a table tennis ball was taken from a paper by French. he data for time and displacement of the falling table tennis ball was stored in the mscript mec_tt_ball.m and this mscript was used to compare the ideo measurements with our displacements / time predictions using the finite difference approximation to sole the equation of motion. he following plot shows the French data and the theoretical alues for the displacement as functions of time. he agreement between our model s predictions and the actual measured alues is ery good. 38

39 Example mscript A falling steel ball mec_drag.m We can model a steel ball with a radius identical to that of a table tennis ball falling from rest using the numerical approach where the resistie force or drag force is 1 F C A drag D F We can then compare the fall of the steel ball with that of the table tennis ball. Input parameters radius of steel ball R =.x1-3 m density of steel ball = 78 kg.m -3 mass of table tennis ball m = 1.7x1-3 kg drag coefficient C D =.445 density of air F = 1.5 kg.m -3 iscosity of air F = 1.789x1-5 Pa.s max time t Max = 3. s No. time steps N = 1 Output parameters table tennis ball steel ball free fall F drag = terminal elocity [m.s -1 ] t = 3. s a [m.s - ] t = 3. s x [m] t = 3. s [m.s -1 ] t = 3. s Reynolds number N R.4x x1 4 It is true that heaier objects do fall faster than lighter ones. table tennis ball steel ball 39

40 Example mscript A falling skydier mec_drag.m We can model a falling skydier falling from rest using the numerical approach where the resistie force or drag force is 1 F C A drag D F Input parameters radius (estimate) mass R =.5 m m = 7 kg mass entered and not calculated from density drag coefficient C D =.5 density of air F = 1.5 kg.m -3 iscosity of air F = 1.789x1-5 Pa.s max time t Max =. s No. time steps N = 1 Output parameters cross-sectional area A =.7854 m terminal elocity = 54 m.s -1 = 19 km.h -1 time to reach t( ) ~ 18 s distance to reach x( ) ~ 76 m You can inestigate how the skydier can control their speed of fall by arying their effectie cross-sectional area. 4