Dynamics ( 동역학 ) Ch.3 Kinetic of Particles (3.1)

Transcription

1 Dynamics ( 동역학 ) Ch.3 Kinetic of Particles (3.1)

2 Introduction This chapter exclusively deals with the Newton s second Law Newton s second law: - A particle will have an acceleration proportional to the magnitude of the resultant force acting on it in the direction of the resultant force. - This law is a Law of Nature experimentally proven and not the results of an analytical proof. - Newton s second law cannot be used when the particle s speed approaches the speed of light, or if the size of the particle is extremely small (~ size of an atom) - Acceleration must be evaluated with respect to a Newtonian frame of reference, i.e., one that is not accelerating or rotating. 3-

3 3.1: Cartesian Coordinates Newton's second law (for a constant mass) n i1 F i ma This can be called the force balance law where many individual forces maybeactingonthemassm. We might be given the force and asked to determine the acceleration: F a m 3. Or we might be given the acceleration and asked to evaluate the force needed to produce that acceleration: F ma

4 3.1: Cartesian Coordinates In Cartesian coordinates, motions in the x, y and z directions can be viewed as individual rectilinear problems. Applying (3.1) to a Cartesian coordinate system gives F yj 3.4 m xi If we break up the applied force into two components, one in the i direction and one in the j direction, (F =F x i+f y j), we have i : Fx mx j : Fy my We can easily extend this coordinate system to three-dimensional motion just add a unit vector k as shown in the figure. k : Fz mz

5 Equations of Motion Newton s second law provides F ma Solution for particle motion is facilitated by resolving vector equation into scalar component equations, e.g., for rectangular components, F i F j F k ma i a j a k F F x x x ma y mx x z F F y y ma my y x y F F z z z ma mz z

6 Dynamic Equilibrium Alternate expression of Newton s second law, F ma 0 ma inertial vector With the inclusion of the inertial vector, the system of forces acting on the particle is equivalent to zero. The particle is in dynamic equilibrium

7 Procedure for obtaining Equation Of Motion (EOM) 1. Select a convenient inertial coordinate system. Rectangular, normal/tangential, or cylindrical coordinates may be used.. Draw a free-body diagram showing all external forces applied to the particle. Resolve forces into their appropriate coordinate components. 3. Draw the kinetic diagram, showing the particle s inertial force, ma. Resolve this vector into its appropriate components. 4. Apply the equations of motion in their scalar component form and solve these equations for the unknowns. 5. It may be necessary to apply the proper kinematic relations to generate additional equations. 3-7

8 Example 3.1: Analysis of a Spaceship The main engine of a space craft is firing, causing the craft to accelerate in a straight line The craft is moving along the i direction. The acceleration magnitude recorded by the ship's navicomputer is 0. m/s. The mass of the spacecraft is 1 x 10 6 kg. Determine the magnitude of the force being exerted by the engine on the spacecraft this is called the engine's thrust. Neglect any change in mass due to fuel being burned. GOAL Find the magnitude T of the engine's thrust. GIVEN Mass and acceleration of the spaceship. 3-8

9 DRAW The simplified system is shown in Figure 3.3. The only force acting on the spacecraft is the thrust ASSUME No additional assumptions are needed. FORMULATE EQUATIONS Equation tells us that T mx SOLVE There's not much to do here except plug in the given values: = (1.0 x10 6 kg)(0. m/s ) =.0 x 10 5 N 3-9

10 Example 3.: Forces Acting on an Airplane The plane's acceleration vector is currently (1.0i+1.1j+0.050k) m/s, where i and j indicate east and north, respectively, and k points up, away from the earth's surface. Knowing that the mass of the jetliner is 1.5 x 10 5 kg, calculate the individual components of the force acting on it. GOAL Find the force components acting on the plane. GIVEN Mass and acceleration of the airplane. DRAW The simplified system is shown in the figure. The acceleration vectors are along with the i, j, k unit vectors. The plane has acceleration components in the same (orthogonal) directions, 3-10

11 ASSUME No additional assumptions are needed. FORMULATE EQUATIONS This problem, simply requires an application of F = ma A, into each components. Using (3.1) gives us SOLVE F 1 i+ F j + F 3 k = (1.5 x 10 5 kg)[(1.0i +1.1j k)m/s ] F 1 i+ F j + F 3 k = (1.5 x 10 5 kg)[(1.0i +1.1j k)m/s ] = (1.5 x 10 5 i +1.7 x 10 5 j x 10 3 k) N Matching coefficients tells us that i: F 1 = 1.5 x 10 5 N j: F = 1.7 x 10 5 N k: F 3 = 7.5 x 10 3 N 3-11

12 Example 3.3: Response of an Underwater Probe The rear thruster of the probe provides a propulsive force T, while a drag force F d : F d a bv where v is the probe's speed, a = 500 N, and b = 10 kg/m. The thrust T is equal to a constant 10,000 N, the probe has a mass of 10,000 kg, and it is initially moving to the right (positive x direction) at 10 m/s. How fast will it be moving 4 seconds later? GOAL Find the probe's speed at t = 4 s. GIVEN Mass of the probe and form of forces acting on it. 3-1

13 DRAW The free-body diagram and the associated inertial response are shown in figure. ASSUME No additional assumptions are needed. FORMULATE EQUATIONS Applying (3.5) gives us m x T SOLVE Dividing by m and using the given drag expression yields 1 x T a bv m F d To determine the change in speed, we'll use the MATLAB function ode45. The initial conditions are x(0) = 0, x (0) = 10. We'll need to define a column vector y such that y(1,1) = x and y(,1) = x. Our initial condition y0 is therefore given by y0 = [0; 10] ;

14 The time duration of the integration is 4 seconds, and thus our integration vector tspan is specified by tspan= [0 4] ; Typing [t,y] =ode45( probe, tspan, y0) ; will produce the desired position and speed data as the columns of the y matrix after which typing plot (t,y(:, ) ) ; produces the plot shown in Figure 3.8. From the calculated data, the speed at 4 seconds is given by v(4) = 13.5 m/s 3-14

15 We can easily solve for this by setting to zero in : T a T Fd T a bv / b 0 v 30.8 m s If we let the time integration last a good bit longer than 4 seconds, 00 to be precise, we will get the plot shown in Figure

16 Example 3.4: Sliding Ming Bowl A bowl of mass m is suddenly released. Treating it as a particle, find its acceleration at the time of release. GOAL Find the bowl's acceleration. GIVEN Physical description of sloped surface. DRAW Figure shows a schematic of our system. ASSUME The bowl remains in contact with the sloped surface. FORMULATE EQUATIONS Because the motion will be along the slope (in the b 1 direction), it makes sense to express our equation of motion in the b 1, b coordinate set. Applying gives us m rb 1 mg j N1b1 N b 3-16

17 Two sets of unit vectors are shown: i, j for right-left/up-down motion and b 1, b for motion along the slope and orthogonal to it. The coordinate r is introduced to track motion along the slope. The gravity force (mg) is oriented vertically, but the two reaction forces created by the bowl/slope interface (N 1 and N ) are aligned with the b 1 and b directions. b 1 b i cos θ -sin θ j sin θ cos θ m rb m rb 1 mg j N1b1 N b b cos b N b N mg b sin b1 : mr N1 mg sin b : 0 N mg cos 3-17

18 b1 : mr N1 mg sin SOLVE The bowl remains in contact with the slope. This means no motion in the b direction; The second equation of motion has zero on the left-hand side. There are two equations with three unknowns:, N 1, and N. r The conclusion must therefore be that either (1) we need a third equation or () one of the unknowns is, in fact, known. Two reaction forces were included (for full generality) N 1 and N but we were told that the bowl/slope interface is frictionless, N 1 = 0. b : 0 N mg cos 3-18

19 m r mg sin r g sin where the minus sign tells us that the bowl accelerates downslope. The second equation gives us something we were not asked to find the normal force exerted by the slope on the vase: 0 N mg cos N mg cos CHECK Our acceleration is proportional to g, which has the units of acceleration, and the normal force N is proportional to mg, which has the units of force. So the units seem fine. The solution, r g sin, tells that the acceleration is less than g, it makes sense that the acceleration is less than what a freely falling object would experience. The magnitude of the acceleration is proportional to sin θ. The same logic supports the solution for the normal force. If the slope were horizontal, we would expect to get N =mg,whichis exactly the solution of the equation (mg cos0 = mg). 3-19

20 Example 3.5: Particle in an Enclosure A particle m is suspended in a rigid container by two massless, inextensible strings, AB and BC. A force is applied to the container so that it accelerates to the right at x. Under this constant acceleration, the tension in string is four times greater than that of string AB. Determine both the acceleration x and the tension in AB. GOAL Find the acceleration of the suspended mass m at B and find the tension in the strings. GIVEN The orientation of the supporting strings, the acceleration of the enclosure, and the relative tension in the strings. DRAW Figure 3.13 shows a free-body and inertial response diagram for the suspended mass. ASSUME We are given that the strings are massless and inextensible. BC 3-0

21 FORMULATE EQUATIONS We will need to apply (3.1), and realize that for this case we have a sum of three forces: two tension forces and a gravity-induced body force. T b T c mg j mx i To solve this, we will have to create the coordinate transformation arrays that let us express b 1 and c 1 in terms of i and j: i j b 1 cos θ sin θ b -sin θ cos θ SOLVE T cos 4T cos Let T = 4T 1 = 4T, decompose (3.1) into the i and j directions: mx 3.13 T sin 4T sin mg Use (3.14) to solve for T: mg T sin Substitute for in to find the acceleration 3g x 5 tan 3-1

22 Chapter 3.: Polar Coordinates In polar coordinates, the force will be broken down F r e r F e θ Fr ma r F ma or m m r r F r 3.16 r r F 3.17 Unlike the Cartesian case, the polar representation is strongly coupled. A force component in the e r direction affects both r and, A force component in the e θ direction affects r,, and. 3-

23 With the Cartesian representation, you could change a particle's acceleration in the x direction by applying a force in that direction: Fx x m There's no direct way to alter only r or by applying a force in a particular direction, however. The value of r or depends on the force as well as on the particle's positions and velocities. In the direction, for example, we have F r r m r There's not much else to say, and so let's move on to some examples. 3-3

24 Example 3.6 Forces Acting On A Payload A robotic arm that operates in a horizontal plane is shown in Figure The programming for the arm is such that the 4 kg payload with mass follows a path given by r t a a t a 1 3t where a 1 =0.7m,a =0.1 m/s, a 3 = 0.05 m/s, and Θ(t) = b 1 t+b t where b 1 = 0.8 rad/s and b = 0.03 rad/s. What are the forces acting on the payload at t = 1.1 s? Ignore gravity. GOAL Determine the forces acting on the payload at a particular time. GIVEN Description in time of the motion variables. DRAW Figure 3.15a shows a schematic of the system. 3-4

25 Figure 3.14 Mass in a robotic arm. 3-5

26 Figure 3.15 Schematic and FBD=IRD of mass. All that matters to us is the payload. The arm doesn't come into play except as the means by which the payload is moved. Thus our diagram includes only the payload itself. Because the motion is given in terms of r and θ, we should use a polar coordinate representation, which means our unit vectors are e r and e θ. Figure 3.15b shows that all the applied forces are represented by the two net forces F r and F θ. Keep in mind that these are the net forces. It's true, for instance, that a contact force developed between each gripper and the payload, and for some problems we might want to know what these individual forces are equal to. In this problem, however, all we care about is the overall force exerted on the payload. 3-6

27 ASSUME Because r and θ are given as explicit functions of time, they and their derivatives are not unknowns. Thus we have two equations with two unknowns (F r and F θ ). FORMULATE EQUATIONS Applying (3.16) and (3.17) gives us SOLVE m r r Fr r r F m Differentiating our expressions for r and θ leads to r t a a t a t r r 1 t a t a3 t b1t t b1 t b a b b 3 t t t 3 3-7

28 and evaluating at t =1.1s,wefind for a final result of r r r m m / s m / s rad rad / s rad s / Using these values in our equations of motion yields (4 kg) [ 0.1 m/s ( m)(0.734 rad/s) ] = F (4 kg)[(0.871 m)( rad/s ) + (0.10 m/s)(0.734 rad/s)] F r = N, F θ = 1.0 N 3-8

29 Example 3.9 No-Slip In A Rotating Arm Figure 3.19 shows an arm that rotates in a horizontal plane about O. Attached to the end of the arm is a small enclosure that holds a particle P. The surface farthest from O is flat and angled at 80, as shown. What is the minimum coefficient of static friction µ s that would permit P to remain fixed with respect to the angled surface? Assume that and is constant. GOAL Calculate the minimum value of µ s such that P won't slide relative to the spinning enclosure. GIVEN Orientation of the enclosure and distance of P from center of rotation. DRAW Figure 3.0 shows the system under consideration along with two sets of unit vectors, both fixed to the rotating arm, while Figure 3.1 shows the relevant free-body/inertial response diagram. This problem is an example of when it's convenient to set up the unit vectors in a way that's not so convenient when it comes time to produce the transformation array. 3-9

30 Figure 3.19 Rotating arm with enclosed mass. 3-30

31 Figure 3.0 Unit vectors for rotating arm. 3-31

32 Figure 3.1 FBD=IRD for constrained mass. 3-3

33 Logically, it makes sense to align the b 1, b vectors as shown in Figure 3.0, with b 1 directed along the surface that P contacts and b normal to that surface. We know that there will be a normal force between the constraining surface and P and a sliding force developed along the surface (shown as N and S, respectively, in Figure 3.1). Although nice from the standpoint of physics, we derived the coordinate transformation arrays for angles that are small and for which it was then easy to say cosφ 1 andsinφ Φ. So what do we do? Just allow the 80 angle of our problem to become much smaller and rotate the unit vectors b 1 and b along with it. Doing so will give us Figure 3.. It's now straightforward to create our transformation. b 1 and e r are essentially opposed; hence they are related by cosφ. The same holds between b and e θ. Getting to the tip of b 1 means going in the opposite direction to e r andthenjustalittle bit in the e θ direction. 3-33

34 Figure 3. Unit vectors aligned to simplify construction of coordinate transformation array. 3-34

35 Hence b1 cos er sin e.similarly,gettingtothetipofb means going in the opposite direction to e θ andthenjustabitalong the negative direction: b cose r sin e r. The entire transformation array is thus given by b 1 b e r -cos Φ -sin Φ e θ sin Φ -cos Φ where Φ, as shown in Figure 3-, should be set equal to 80 to match the example's conditions. FORMULATE EQUATIONS A force balance gives us r r e r r e Nb Sb1 m r 3-35

36 ASSUME Assume that the mass does not move relative to the enclosure. SOLVE Using the constraints to simplify our set of unknowns, we will need to apply the given system constraints r L, r r 0, 0 in our equation of motion. Expressing everything in terms of and gives ml er Nb Sb 1 ml cos80b 1 ml sin 80b Nb Sb 1 Equating coefficients leads to: b1 : S ml cos80 b : N ml sin

37 When S is a maximum, S=µ s N. Using (3.8) and (3.9) gives us L cos80 L sin 80 s s cot CHECK Why is the value we just solved for equal to the minimum value of µ? It all has to do with our assumptions. We approached the problem by assuming that the mass was stationary with respect to the enclosure. For this to be the case, the coefficient of static friction has to be large enough, that is, sufficiently large so that the forces trying to get the mass to slide aren't able to overcome the static friction. 3-37