Dynamics Applying Newton s Laws Introducing Energy

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1 Dynamics Applying Newton s Laws Introducing Energy Lana Sheridan De Anza College Oct 23, 2017

2 Last time introduced resistive forces model 1: Stokes drag

3 Overview finish resistive forces energy work

4 Model 1: Stokes Drag Find an expression for how the velocity changes with time. (calling down positive) F net = ma mg bv = ma m dv dt ( ) m dv mg bv dt = mg bv = 1 Integrating with respect to t, we can set v = 0 at t = 0: v ( ) m t mg bv dv = 1 dt 0 0 v(t) = mg b (1 e bt/m )

5 Model 1: Stokes Drag Exercise: Check that the solution v(t) = mg b (1 e bt/m ) satisfies the equation m dv = mg bv dt

6 maximum (or terminal) speed v T. s of the object and v is the velocity of the object relative to e sign indicates that S R is in the opposite direction to Model 1: Stokes Drag S v. e of mass m released from rest in a liquid as in Figure 6.13a. acting on the sphere are the resistive force S R 52b S Exercise: Check that the solution v(t) = mg v and b (1 e bt/m ) satisfies, let us describe its motion. 1 We model the sphere as a par- the equation m dv v 0 a g = mg bv The sphere dt approaches a v v T v v T a v T b c t The time constant t is the time at which the sphere reaches a speed of 0.632v T. Time constant, τ = m b. The solution can also be written v(t) = mg b (1 e t/τ ). t

7 Model 1: Stokes Drag, Question Time constant, τ = m b. Suppose two falling objects had different masses, but similar shapes and fall through the same fluid, and so have the same constant b. Which one reaches 99% of its terminal velocity first? (A) the one with more mass (B) the one with less mass

8 Model 1: Stokes Drag, Question Time constant, τ = m b. Suppose two falling objects had different masses, but similar shapes and fall through the same fluid, and so have the same constant b. Which one reaches 99% of its terminal velocity first? (A) the one with more mass (B) the one with less mass

9 Resistive Forces There are two main models for how this happens. Either the resistive force R is proportional to v, or is proportional to v 2

10 Resistive Forces There are two main models for how this happens. Either the resistive force R is proportional to v, or is proportional to v 2

11 Model 2: The Drag Equation For this model R v 2. Appropriate for high Reynolds Number fluids and/or objects moving at high speed. This is the one commonly used for airplanes, skydivers, cars, etc.

12 Model 2: The Drag Equation For this model R v 2. Appropriate for high Reynolds Number fluids and/or objects moving at high speed. This is the one commonly used for airplanes, skydivers, cars, etc. There is some turbulence in these cases.

13 Model 2: The Drag Equation Altogether, the conditions for use are: fast moving objects low viscousity fluids turbulent flow in the fluid

14 Model 2: The Drag Equation The magnitude of the resistive force is: R = 1 2 DρAv 2 This is called the drag equation. D is the drag coefficent (depends on object s shape) ρ is the density of air A is the cross-sectional area of the object perpendicular to its motion

15 Model 2: The Drag Equation It is straightforward to find an expression for v T in this case also. Equilibrium F net = 0. F net = 1 2 DρAv 2 T mg = DρAv 2 T = mg v T = 2mg DρA

16 Model 2: The Drag Equation Finding the net force and using Newton s second law: F net = mg 1 2 DρAv 2

17 Model 2: The Drag Equation Finding the net force and using Newton s second law: F net = mg 1 2 DρAv 2 ma = mg 1 2 DρAv 2 1 g dv dt = 1 DρA 2mg v 2

18 Model 2: The Drag Equation Finding the net force and using Newton s second law: F net = mg 1 2 DρAv 2 ma = mg 1 2 DρAv 2 1 g dv dt = 1 DρA 2mg v 2 Let k = DρA 2mg. 1 g dv dt = 1 k 2 v 2

19 Model 2: The Drag Equation The integrating factor trick does not work cleanly here. But we can still separate variables and integrate with respect to t: 1 v g 0 1 g dv dt 1 1 k 2 v 2 dv dt dt = = 1 k 2 v 2 t 0 1 dt

20 Model 2: The Drag Equation The integrating factor trick does not work cleanly here. But we can still separate variables and integrate with respect to t: 1 v g 0 1 g 1 g 1 dv 1 k 2 v 2 v 0 dv dt = 1 k 2 v 2 t dt dt = 1 1 k 2 v 2 dv = t 0 1 dt

21 Model 2: The Drag Equation The integrating factor trick does not work cleanly here. But we can still separate variables and integrate with respect to t: 1 v g 0 1 g 1 g 1 dv 1 k 2 v 2 v We need to find 1 1 k 2 v 2 dv. Use partial fractions: 0 dv dt = 1 k 2 v 2 t dt dt = 1 1 k 2 v 2 dv = t 1 1 k 2 v 2 = 1 (1 + kv)(1 kv) = 1 2 ( 0 1 dt ) 1 (1 + kv) + 1 (1 kv)

22 Model 2: The Drag Equation After integrating and some algebra: ( ) 1 e 2kgt v(t) = k 1 + e 2kgt We can write this a little more conveniently as a hyperbolic function. v(t) = ( ) 2mg DρAg DρA tanh 2m t

23 Comparing the two models The two types of resistive forces predict different curves, but depending on the constants, in each model the curve might approach the terminal velocity faster or slower. [ (1 e x ) vs. tanh x] 1 Figure from Wikipedia.

24 Figure 6.16b, in which the resistive force is plotted as a function of the square of the terminal speed. This graph indicates that the resis- Determining tive force is proportional to the square regime: of the speed as which suggested by model? Equation 6.7. Finalize Here is a good opportunity for you to take some actual data at home on real coffee filters and see if you can reproduce the results shown Experimental in Figure If you data: have shampoo and a marble as mentioned in Example 6.8, take data on that system too and see if the resistive force is appropriately modeled as being proportional to the speed a All values of v T are approximate. Resistive force (N) a The data points do not lie along a straight line, but instead suggest a curve Terminal speed (m/s) 3 4 Resistive force (N) b Terminal speed squared (m/s) 2 The fit of the straight line to the data points indicates that the resistive force is proportional to the terminal speed squared. Figure 6.16 (Example 6.10) (a) Relationship between the resistive force acting on falling coffee filters and their terminal speed. (b) Graph relating the resistive force to the square of the terminal speed. The resistive force R = mg at the terminal velocity. Plot R against v T. Example 6.11 Resistive Force Exerted on a Baseball A pitcher hurls a kg baseball past a batter at 40.2 m/s (5 90 mi/h). Find the resistive force acting on the ball at this speed. 1 Figures from Serway and Jewett. AM

25 Section 6.4 Motion in the Presence of Resistive Forces Drag Equation Example ar N ng n- as e- 26. Review. (a) Estimate the terminal speed of a wooden sphere (density g/cm 3 ) falling through air, taking its radius as 8.00 cm and its drag coefficient as (b) From what height would a freely falling object reach this speed in the absence of air resistance? Page 171, # The mass of a sports car is kg. The shape of the body is such that the aerodynamic drag coefficient is and the frontal area is 2.20 m 2. Ignoring all other sources of friction, calculate the initial acceleration the car has if it has been traveling at 100 km/h and is now shifted into neutral and allowed to coast. 28. A skydiver of mass 80.0 kg jumps from a slow-moving (You aircraft can use ρand air = reaches 1.20 kg ma 3 terminal.) speed of 50.0 m/s. (a) What is her acceleration when her speed is 30.0 m/s? What is the drag force on the skydiver when her speed is (b) 50.0 m/s and (c) 30.0 m/s? 29. Calculate the force required to pull a copper ball of

26 Drag Equation Example Page 171, #27 (You can use ρ air = 1.20 kg m 3.)

27 Drag Equation Example Page 171, #27 (You can use ρ air = 1.20 kg m 3.) Answer: a = m/s 2 (negative sign indicates opposite direction to car s motion)

28 Drag Equation, One more point What if the object is not dropped from rest? If an object starts out with a large velocity, then begins to experience an unbalanced resistive force it will have an acceleration opposite to its direction of motion. Consider what happens during a skydive.

29 Now for something a little bit different.

30 Energy Energy is almost impossible to clearly define, yet everyone has a good intuitive notion of what it is. Energy is a property of physical systems. It tells us something about the states or configurations the system can be in. In fact, it is possible to find the dynamics of a system purely from understanding the distribution of energy in the system.

Energy Energy is almost impossible to clearly define, yet everyone has a good intuitive notion of what it is. Energy is a property of physical systems.

31 Energy Energy is almost impossible to clearly define, yet everyone has a good intuitive notion of what it is. Energy is a property of physical systems. It tells us something about the states or configurations the system can be in. In fact, it is possible to find the dynamics of a system purely from understanding the distribution of energy in the system. Importantly, it can neither be created or destroyed, but it can be transferred between systems, and take different forms.

32 Types of Energy motional energy (kinetic) energy as a result of object s configurations or stored energy (potential) heat, light, sound - can carry away energy from a mechanical system

33 Systems and Environments To make some predictions about physical objects, you must identify a system of interest. Some object or collection of objects. System

34 Systems and Environments To make some predictions about physical objects, you must identify a system of interest. Some object or collection of objects. System They are modeled. The model may not be exactly accurate, just good enough to make the predictions you want.

35 Systems and Environments To make some predictions about physical objects, you must identify a system of interest. Some object or collection of objects. System They are modeled. The model may not be exactly accurate, just good enough to make the predictions you want. Outside influences (eg. forces or energies) can be included in the description, but the source of these effects is not described. Everything outside the system is the system s environment.

36 does not move rse, Work we apply a gest that when, we must cone of the force. nt represents a N through the rtant. Moving ving it 2 cm if with the system. After on, insert the system. For example, the work done by the hammer on the nail Let s consider how identifies the environment the nail as can the system, effect the system by and the force from the hammer exchanging energy with it. represents the influence from the Take the systemenvironment. to be a block. An external force (from the environment) acts on it. system) undert force of magt. r orce on the itude Dr of Figure 7.2 An object undergoes This force affects a the displacement block: it can D S r under accelerate the it. u, where u is action of a constant force S F. It can also change the energy of the block. u S F S

e of the force. nt Work represents a N through the rtant.

represents the influence from the environment.

2 An object undergoes For a constant applied force Work is defined as:

37 e of the force. nt Work represents a N through the rtant. Moving ving it 2 cm if system) undert force of magt. represents the influence from the environment. r orce on the itude Dr of Figure 7.2 An object undergoes For a constant applied force Work is defined as: u, where u is u S F S a displacement D r S under the action of a constant force F S. (7.1) fined in terms xplore how to Work done W = by F a r constant force

38 Vectors W 5 F Dr Properties cos u5 S F? D and Operations S r (7.3) Multiplication by a vector: S r is a shorthand notation for F Dr cos u. with our discussion of work, let us investigate some properties igure 7.6 The shows two Dot vectors Product S A and S B and the angle u between inition of the dot product. In Figure 7.6, B cos u is the projecerefore, Let Equation A = 7.2 A x means i + Athat y j S A? S B is the product of the the projection of B = S B onto B x i + S A. B 1 y j, nd side of Equation 7.2, we also see that the scalar product is is, S S S S A? B 5 B? A A B = A x B x + A y B y The output of this operation is a scalar. to stating that A S? B S equals the product of the magnitude of B S and the projection of A S Equivalently, nother way of combining vectors that proves useful in physics and is not commutative. A B = AB cos θ u B cos u S B S A. = AB cos u B S S A Figure 7.6 The scalar product S S S A? B equals the magnitude of A multiplied by B cos u, which is the projection of S B onto S A.

39 Vectors W 5 F Dr Properties cos u5 S F? D and Operations S r (7.3) Multiplication by a vector: S r is a shorthand notation for F Dr cos u. with our discussion of work, let us investigate some properties igure 7.6 The shows two Dot vectors Product S A and S B and the angle u between inition of the dot product. In Figure 7.6, B cos u is the projecerefore, Let Equation A = 7.2 A x means i + Athat y j S A? S B is the product of the the projection of B = S B onto B x i + S A. B 1 y j, nd side of Equation 7.2, we also see that the scalar product is is, S S S S A? B 5 B? A A B = A x B x + A y B y The output of this operation is a scalar. to stating that A S? B S equals the product of the magnitude of B S and the projection of A S Equivalently, nother way of combining vectors that proves useful in physics and is not commutative. Properties A B = AB cos θ The dot product is commutative: A B = B A If A B, A B = AB. If A B, A B = 0. u B cos u S B S A. = AB cos u B S S A Figure 7.6 The scalar product S S S A? B equals the magnitude of A multiplied by B cos u, which is the projection of S B onto S A.

40 Vectors Properties and Operations Multiplication by a vector: The Dot Product Try it! Find A B when A is a vector of magnitude 6 N directed at 60 above the x-axis and B is a vector of magnitude 2 m pointed along the x-axis.

41 Vectors Properties and Operations Multiplication by a vector: The Dot Product Try it! Find A B when A is a vector of magnitude 6 N directed at 60 above the x-axis and B is a vector of magnitude 2 m pointed along the x-axis. (1 J = 1 Nm) A B = (6 N)(2 m) cos(60 ) = 6 J

42 Vectors Properties and Operations Multiplication by a vector: The Dot Product Try it! Find A B when A is a vector of magnitude 6 N directed at 60 above the x-axis and B is a vector of magnitude 2 m pointed along the x-axis. (1 J = 1 Nm) Now find A B when: A B = (6 N)(2 m) cos(60 ) = 6 J A = 1 i + 2 j ; B = 1 i 4 j

43 Vectors Properties and Operations Multiplication by a vector: The Dot Product Try it! Find A B when A is a vector of magnitude 6 N directed at 60 above the x-axis and B is a vector of magnitude 2 m pointed along the x-axis. (1 J = 1 Nm) Now find A B when: A B = (6 N)(2 m) cos(60 ) = 6 J A = 1 i + 2 j ; B = 1 i 4 j A B = (1)( 1) + (2)( 4) = 9

44 Work Again, W = F r Recall that the dot product, A B = AB cos θ, so we can also write work as W = F r cos θ

45 Work Again, W = F r Recall that the dot product, A B = AB cos θ, so we can also write work as W = F r cos θ If there are several forces acting on a system, each one can have an associated work. In other words, we can ask what is the work done on the system by each force separately. The net work is the work done by the net force (if the system can be modeled as a particle). It is also the sum of all the individual works.

46 Work done by 180 individual forces Chapter 7 Energy of a System S F is the only force that does work on the block in this situation. S n mg S u r S S F Figure 7.3 An object is dis- have done a co tion, however, the chair, but does not move the situation d Also notice is zero when th application. Th Figure 7.3, the the gravitation dicular to the tion of D r S. The sign of done by the ap is in the same

47 Units of Work The unit of work is the Joule, J. 1 J = 1 Nm Work can be positive or negative! For work done on a system: Positive energy is transferred to the system. Negative energy is transferred from the system.

48 Question Quick Quiz The gravitational force exerted by the Sun on the Earth holds the Earth in an orbit around the Sun. Let us assume that the orbit is perfectly circular. The work done by this gravitational force during a short time interval in which the Earth moves through a displacement in its orbital path is (A) zero (B) positive (C) negative (D) impossible to determine 1 Serway and Jewett, page 180.

49 Question Quick Quiz The gravitational force exerted by the Sun on the Earth holds the Earth in an orbit around the Sun. Let us assume that the orbit is perfectly circular. The work done by this gravitational force during a short time interval in which the Earth moves through a displacement in its orbital path is (A) zero (B) positive (C) negative (D) impossible to determine 1 Serway and Jewett, page 180.

50 Summary Solution to the Drag equation resistive force mode Energy, systems, environments Work Collected Homework! due Fri, Oct 27. (Uncollected) Homework Serway & Jewett, Read ahead in Chapter 7 about work, springs, and kinetic energy. Ch 7, onward from page 204. Questions: Section Qs 3, 5, 7, 11

Dynamics Applying Newton s Laws The Drag Equation

Dynamics Applying Newton s Laws The Drag Equation Lana Sheridan De Anza College Feb 8, 2019 Last time introduced resistive forces model 1: Stokes drag Overview model 2: the Drag Equation finish resistive