Dynamics ( 동역학 ) Ch.2 Motion of Translating Bodies (2.1 & 2.2)

Transcription

1 Dynamics ( 동역학 ) Ch. Motion of Translating Bodies (. &.)

2 Motion of Translating Bodies This chapter is usually referred to as Kinematics of Particles. Particles: In dynamics, a particle is a body without considering its size or dimension. Only the motion as an entire unit will be considered Any rotational motion will be neglected. Then what is rigid body? -

3 . Straight-line Motion Consider the system shown below, a dragster (a custom built racing car) being tested on a dragstrip. We are interested in the position of the dragster at any time after the race starts. Its position along the track is shown diagrammatically below O: The start of the race : the distance from the starting position to the dragster -3

4 Straight-line Motion From the speedometer, the speed is 00 mph and the acceleration is 0.8 g. (0.8 times the acceleration of graity). (a) Speed readout; (b) acceleration readout. This situation illustrate the key questions of this section: How do we determine where something is? How fast is it moing? What s its acceleration? -4

5 Straight-line Motion A mass particle P moes along this line and is currently 4 m from the origin. Defining the current time as t, (in seconds), At some time t in the future (t =t + t), the mass has moed m further from the origin. t m t 6m 4 Mathematically, and and the difference between the two positions is indicated by. -5

6 Straight-line Motion: Displacement Define position and displacement Position: the distance of an object from a reference point Displacement: the changes in position The location of the particle relatie to the origin can be presented as a ector because we need two pieces of information to pin it down: its distance from its origin and its orientation with respect to the origin (in front or behind). For the current situation the orientation is simply represented by the sign of the answer: + means to the right and - indicates to the left of the origin. The full-blown ector representation is r P/O where P/O stands for P with respect to O. -6

7 Straight-line Motion: elocity The position of the particle is defined as r P O To find the particle s aerage elocity from t to t ag. t Velocity is the rate of change in time of the displacement When t is 0 and t is 4 seconds, then m ag 0.5m / s 4s Velocity is also a ector quantity, need the direction and magnitude -7

8 Straight-line motion: Acceleration To find the elocity at time t (instantaneous elocity) All we hae to do is take the limit as t goes to zero: P t lim t0 t0 t t t t t lim t t t0 t t t d lim t Instantaneous elocity may be positie or negatie. elocity is referred to as particle speed.. Magnitude of The particle s acceleration is simply its rate of change of elocity, requires both a magnitude and direction (relatie to some coordinates). a P t t t t P P d lim lim t0 0 t t t.3-8

9 Straight-line motion The motion of a particle is known if the position coordinate for particle is known for eery alue of time t. Consider particle with motion gien by 3 6t t d t 3t a d d t 6 at t = 0, = 0, = 0, a = m/s at t = s, = 6 m, = ma = m/s, a = 0 at t = 4 s, = ma = 3 m, = 0, a = - m/s at t = 6 s, = 0, = -36 m/s, a = - 4 m/s -9

10 Acceleration Instantaneous acceleration may be: - positie: increasing positie elocity or decreasing negatie elocity - negatie: decreasing positie elocity or increasing negatie elocity. -0

11 Straight-line Motion Figures.5,.6, and.7 illustrate what these plots look like when the particles motion is gien by (for which sin t and cos t ) t cost -

12 Straight-line Motion A typical dynamic problem requires Requires to sole for the acceleration of a particle and then, Wants to determine its elocity and position oer time, which requires integration, whether analytically or with the computer. Mathematically, for any two alues of time t and t, we hae t t t.4 t t t t.5 t -

13 Eample.: Speed Determination ia Integration A piston P with an initial speed of 0 = 3 cm/s eperiences an acceleration defined by = (a + a t) for one second (a =cm/s and a = - cm/s 3 ). What is the piston s final speed? GOAL Determine the speed of the piston. GIVEN Piston s initial speed and acceleration profile DRAW Figure.0 shows our system ASSUME No additional assumptions are needed FORMULATE EQUATIONS SOLVE.4 a a t 0 a 0 at t cm / s cm / s 4cm / s -3

14 Eample. You re traeling at 48 km/h and hae only 6.7 m in which to stop. What constant deceleration is needed to aoid a collision? 6.7 m GOAL Determine the constant deceleration magnitude that will reduce your speed from 48 km/h to zero in 6.7 m. GIVEN Initial and final speed as well as distance traeled. DRAW Figure. shows our one-dimensional situation. gies the position of the bicyclist, who starts at (0) = 0 (t = 0) and comes to a complete stop at (t) = 6.7 m. let 0 (0) and 0 (0). -4

15 Eample. ASSUME Acceleration is constant. FORMULATE EQUATIONS We ll use to represent the aerage acceleration. t.4 t 0 a 0 at More generally, for any final time t, we hae t a.6 0 t Using (.6) in (.5) and ealuating from 0 to t gies us t t a t a 0-5

16 Eample. SOLVE We know that 0 = 0 and 0 = 48 km/h= 3.3 m/s. using this data along with the known distance to the mammoth in (.7) yields at 6.7m 3.3m / st.8 where t is the time at which the cyclist reaches =6.7m.This equation has two unknowns ( a and t ), and so we ll inoke (.6), ealuated at t = t : 0 3.3m / s at Using this alue for a a in (.8) gies us 3.3m / s t 3.3m / s t 6.7m 3.3m / st t. 00s Used in (.9), this yields a 3.3m / s. 37g.9-6

17 Eample. CHECK We can work backwards and ask how long it takes to reach 3.3 m/s if we accelerate at 3.3m/s. Clearly, the answer is one second, matching the result just found. In one second, starting from rest and assuming a constant acceleration of 3.3m/s, we would trael a distance 3.3m / s s 6. 7m -7

18 Determination of the Motion of a Particle Recall, motion of a particle is known if the position is known for all time t. Typically, conditions of motion are specified by the type of acceleration eperienced by the particle. Determination of elocity and position requires two successie integrations. Three classes of motion may be defined for: - Case : acceleration gien as a function of time, a = f(t) - Case : acceleration gien as a function of position, a = f() - Case 3: acceleration gien as a function of elocity, a = f() Recall that Kinematics is a math. Don t get confused or don t spend too much time here. All you need to know is the mathematical relationship between acceleration, elocity, and position. -8

19 Case: When Acceleration is constant (or gien as a function of a time) Consider when acceleration is constant first. Then recall in the preious slides and just integrate. t t t t a t a t at t.0 Letting t start at zero, denoting (0) as 0 and letting t be replaced by the general t gies us where. t Integrating once again will yield the particle s position as a function of time: where 0 indicates the position at t =0. these results are alid ONLY if the acceleration is CONSTANT! t t 0 at. a t t t

20 Case: When Acceleration is constant (or gien as a function of a time) If the acceleration isn t constant but depends eplicitly on t, youcan still integrate by hand to find elocity and position (as long as the time dependence is relatiely simple). But if you hae something annoyingly comple such as g.3 then you need to use a computer Recall that the acceleration is just the time rate of change of speed: d a.4 Net, consider what elocity is, namely, the time rate of change of position, d.5-0

21 Case: When Acceleration is constant (or gien as a function of a time) Soling both equations for gies d a d Equate these two epressions for and you will get d d a Which, when rearranged, becomes dad Time doesn t show up eplicitly in (.6). That s one of the interesting things about this relationship. It shows another way of looking at position, speed, and acceleration and how they relate to one another. d a d.6 -

22 Case: When Acceleration is constant (or gien as a function of a time) Now let s see what we can get by letting the acceleration be constant. Because both the left- and right-hand sides of (.6) are eact differentials, meaning that we can immediately integrate them: d a d Because ais constant, carrying through the integration gies us which can be epressed more simply as a a.7 -

23 Case: When Acceleration Depends on Position This situation occurs, for instance, when the body is attached to a spring. We can rewrite (.6) to show this dependency: d a d (.9 ) What we hae now are two eact differentials, which means we can integrate both sides without haing to find any sort of integrating factor: d a d ad.0 a d. -3

24 Case 3: When Acceleration Depends on Speed This is the case, for eample, when a particle is eperiencing a drag force that depends on speed, as most do. Again we need to use our acceleration/speed relationships. and a d d d a Taken together, these relationships imply that d d d a a d d d a..3-4

25 EXAMPLE.3 Constant Acceleration / Speed / Distance Relation In one time trial, you see that the car, starting from rest, has moed eactly 00 m at the point where it hits 00 km/hr. You did not time the trial. You need to get the (aerage) acceleration information. GOAL GIVEN DRAW Find the magnitude of the car s aerage acceleration. Speed and distance traeled. ASSUME Acceleration is constant -5

26 FORMULATE EQUATIONS First, conert km/hr to m/s: 000m / km 3600s / h 00km / h 7.7m / s Because you re looking for the aerage acceleration, all you need to find is the magnitude of the constant acceleration that, oer 00 m, brings the car to 00 km/hr. Equation (.7) is ideally suited for the task. Using (.7) you will hae a SOLVE 7.7m / s 0 a00m 0 7.7m / s a 3.86m / s 0. 39g 00m -6

27 . Cartesian Coordinates The motions of a particle is a ector quantity. They must be described in both direction and magnitude. Therefore, they must be defined in relation to a coordinate. Coordinate system: A reference frame that describes an eact position of an object to define where an object is in space now and how to get to where it moes net. The first set of coordinates is the simplest form Cartesian coordinates and is also known as rectangular coordinates. -7

28 Therefore we can epress P's position ector as r P O i yj (.7) The position ector is written r P/O rather than just r P to make it clear that it's the position P of with respect to a specific point, namely, the origin O. There are two different types of reference frame: a fied reference frame and a moing reference frame. -8

29 We need to relate these two sets of unit ectors. -9

30 We hae and b cos i sin j b sin i cos j.8.9 These equations can easily be represented in an array called a coordinate transformation array: i j b b cos sin i cos θ sin θ b -sin θ cos θ b sin cos j In addition, the inerse transformation is also possible i j sin cos cos b sin b -30

31 The elocity of the particle in Cartesian system inoles differentiating its position ector with respect to time. P d rp d d i O i yj from (.7).3 di dy j y dj ( product rule ) The unit ectors are not time dependent. So their time deriaties are zero. Then, P i yj.3 i yj.33 a P In 3 dimensional space, r a P O P P i i i yj zk yj zk yj zk -3