Motion of a Point. Figure 1 Dropped vehicle is rectilinear motion with constant acceleration. Figure 2 Time and distance to reach a speed of 6 m/sec

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1 Introduction Motion of a Point In this chapter, you begin the subject of kinematics (the study of the geometry of motion) by focusing on a single point or particle. You utilize different coordinate systems and manipulate the differential relations between acceleration, velocity and displacement. Example 13.1 Straight line motion with constant acceleration Here you are dropping an object (point center of mass) with constant vertical gravitational acceleration. The displacement, I, axis is taken vertically downward. The question is when and where does the object reach a speed of 6 m/s? Figure 1 shows the object and the rules to evaluate its motion. The rules have been formulated in terms of the basic concept that the area under the acceleration time curve (AU_at) is the change in the velocity. Likewise, the area under the resulting velocity time curve (AU_vt) is the change in velocity. Those rules are written in a similar form in the text. The top three rules are the same for all rectilinear motions, while the last three rules are problem dependent. There is a third general relation that the area under the acceleration displacement curve (AU_as) is equal to the change in the square of the speed. The results are given in Figure 2. Figure 1 Dropped vehicle is rectilinear motion with constant acceleration Figure 2 Time and distance to reach a speed of 6 m/sec Copyright J.E. Akin. All rights reserved. Page 1 of 25

2 Example 13.2 Two piecewise constant accelerations Here a cheetah starts from rest and reaches a maximum speed of 75 mi/hr by constant acceleration over 4 seconds. Then, the acceleration stops and it runs with constant speed for another 6 seconds. In other words, you have a motion described by two segments of constant horizontal acceleration. You seek the total distance travelled. You simply repeat the previous rules for each of the two independent periods of constant acceleration. The velocity history and expanded rules are in Figure 3, and the final distance is found from Figure 4. Figure 3 Two consecutive constant acceleration motions Copyright J.E. Akin. All rights reserved. Page 2 of 25

3 Figure 4 Results after 10 seconds for cheetah speed and position Example 13.3 Acceleration as a function of time When the acceleration is given as a function of time, you simply integrate that history (find the area under the a t curve) to get the change in velocity. Integrating again (get the area under the v t curve) gives you the change in position. The rules in Figure 5 give the kinematics for acceleration as a function of time raised to a power (times a units correcting constant). That is more general that the example train acceleration that is Copyright J.E. Akin. All rights reserved. Page 3 of 25

4 assumed linear in time (n = 1). Knowing the initial time and speed, you desire to find the change in position at a later time. The velocity and later position are in Figure 6. Figure 5 Kinematics for acceleration proportional to time raised to a power Figure 6 Later speed and position of train Example 13.4 Acceleration as a function of velocity When the acceleration depends on the velocity of the point, the calculus rules can become more complex. However, they are just based on the chain rule, as shown in the text example. The kinematic rules in Figure 7 assume a general power law relation. They include logic for two special cases, n = 1 and n = 2. The example Copyright J.E. Akin. All rights reserved. Page 4 of 25

5 plane drag parachute is taken to be proportional to the velocity squared (n = 2). The example goal is to find the time and distance to slow the speed to a specific value. The rules do not have to be arranged for that case. You just supply the know items and TK attempts to find all other variables (see Figure 7). Figure 7 Kinematics for acceleration as a power law function of velocity Figure 8 Time and distance for the parachute to slow the aircraft Copyright J.E. Akin. All rights reserved. Page 5 of 25

6 The displacement and velocity time histories of the slowing aircraft are seen in Figure 9, where the arrows indicate the sought position and time. Figure 9 Position and speed history of the aircraft Example 13.6 Motion in two Cartesian components This description of the center of mass of a helicopter assumes that both the vertical and horizontal components of acceleration are linear functions of time. The expected flight path is shown in Figure 10. Starting from rest you need to determine the actual flight path at any time. The kinematic rules in Figure 11 illustrate two component Cartesian motion. For both components of motion you utilize the area under each acceleration time curve to obtain the changes in the velocity components. A second pair of integrations of the velocity time curves give the two change in position components. Figure 10 Expected flight path Copyright J.E. Akin. All rights reserved. Page 6 of 25

7 Figure 11 Kinematics rules for x and y accelerations linear in time A List Solve yields the x and y coordinates for each listed time over the desired interval from 0 to 10 seconds. Those lists are processed by a Plot sheet to produce the actual flight path shown in Figure 12. The detail results for the final time are in Figure 13. Figure 12 The computed helicopter flight path over 10 seconds Copyright J.E. Akin. All rights reserved. Page 7 of 25

8 Figure 13 Helicopter flight results at t = 10 seconds Example 13.7 Skier projectile motion The skier in Figure 14 (left) leave the slope above a vertical drop at 10 m/s, and later lands on the lower inclined slope. You are to determine when and where the skier lands on the slope, the final speed, and its relative component tangent to the lower slope. The contact distance, d, along the slope is also unknown. The kinematics rules in Figure 15 assume constant acceleration components. Thus, they are more general that the given example. This actual example is obtained by setting the horizontal acceleration to zero, and the vertical (gravitational) acceleration to 9.81 m/s^2 in the input column of the Variable Sheet in Figure 16. Copyright J.E. Akin. All rights reserved. Page 8 of 25

9 Figure 14 Assumed and calculated skier projectile path (neglecting air resistance) There are two approaches to this problem. As stated, you are to determine the time and distance where the snow shape intersects the center of mass projectile path. That solution yields the landing time (1.6 seconds), distance, and velocity vector. The last three lines of the rule set give that single solution. Another solution of interest (not requested in the text) is to computer the skier path down to the landing point. To accomplish that general time history solution you would have to comment out the last three lines requiring a single common point on the projectile path. Knowing the landing time, you could fill the time (t) list and solve for the x and y positions for each time in the list. Plotting those two coordinates against each other yields the full skier path shown on the right in Figure 14. Example 13.8 Angular motion of a jet turbine In this example the acceleration as a function of velocity kinematics covered in Example 13.4 are simply converted to angular motions. That is, angular acceleration yields angular velocity which yields angular displacement (rotation). The stated problem is that engine shut down experiences an angular acceleration that decreases in direct proportion to the angular velocity. Like Example, 13.4 a more general set of kinematic rules are provided in the rules of Figure 17. Copyright J.E. Akin. All rights reserved. Page 9 of 25

10 Figure 15 Kinematic rules for constant x and y accelerations Copyright J.E. Akin. All rights reserved. Page 10 of 25

11 Figure 16 Single point results where skier path intersects snow shape Copyright J.E. Akin. All rights reserved. Page 11 of 25

12 Figure 17 Angular kinematic rules for acceleration related to velocity raised to a power Note that the rules in Figure 17 include logic for the special cases that require logarithmic integrals. Given the starting and ending rotational speeds, the time and number of revolutions to reach the final state are in Figure 18. Copyright J.E. Akin. All rights reserved. Page 12 of 25

13 Figure 18 Rotational kinematics for a slowing turbine Example 13.9 Tangential and normal components in circular motion Figure 19 shows a circular path (left) upon which a motor cyclist starts from rest and accelerates, tangent to the circle, with a magnitude linearly proportional to time. You wish to find the arc distance travelled and the acceleration as a function of time. The simple kinematic rules for circular motion are in Figure 20. Figure 19 Path and acceleration for a motor cyclist on a circular track Copyright J.E. Akin. All rights reserved. Page 13 of 25

14 Figure 20 Kinematics for circular motion with linear tangential acceleration in time Figure 21 Circular motion results at t = 10 seconds Copyright J.E. Akin. All rights reserved. Page 14 of 25

15 Figure 22 Arc distance of the cyclist as a function of time Example Relating circular motion components to Cartesian components This example is a continuation of the helicopter flight path in Example It differs from the above motor cycle example in that instead of having a constant radius, the flight path has an instantaneous (changing) radius of curvature at all times. Basically, given the previously computed path you are to find the instantaneous curvature and the components parallel and normal to its direction (Figure 23). Figure 23 Changing radius of curvature of a general path Copyright J.E. Akin. All rights reserved. Page 15 of 25

16 The extension of the linear acceleration, in time, rules of Example 13.6 are in Figure 24. The additional results at t = 4 seconds are in Figure 25. Figure 24 Kinematics converting between either Cartesian or normal and tangential components Copyright J.E. Akin. All rights reserved. Page 16 of 25

17 Figure 25 Helicopter curvature and tangential and normal acceleration at t=4 seconds Copyright J.E. Akin. All rights reserved. Page 17 of 25

18 Example Polar and Cartesian components The robot in Figure 26 has its radial and angular position programmed as trigonometric functions of time so as to form a closed path. Given those time histories you are to compute the path (Figure 27), its velocities and accelerations (omitted in the text, but not in Problem13.172). The longer set of governing rules for polar coordinate motion, and its conversion to Cartesian components are in Figure 28. Figure 26 A robot programmed with radial and angular positions Figure 27 Prescribed robot hand path Copyright J.E. Akin. All rights reserved. Page 18 of 25

19 Figure 27 Kinematic rules for planar polar coordinate motion, and its conversion Copyright J.E. Akin. All rights reserved. Page 19 of 25

20 Figure 28 Robot path results at 0.8 seconds (of a 1 second cycle) Copyright J.E. Akin. All rights reserved. Page 20 of 25

21 Example Radial position as a function of angle The cam follower system has its radial position defined as cos, in meters. A slightly more general definition is used by introducing three constants to be user supplied. The calculus of the more general form is given in Figure 29, along with a typical shape. The kinematics of this problem can be broken into two parts. First, the specific calculus for the generalized shape, and its use with the chain rule to obtain time derivatives. Secondly one has the general relations between polar components and Cartesian components. While the text example only addresses velocity components, the rules given here also include consideration of acceleration. The specific geometry rules are in Figure 30, while the more general rules are in Figure 31. The specific cam shape is graphed in Figure 32. Figure 29 A typical cam follower system Figure 30 Kinematic rules for the specific example cam shape Copyright J.E. Akin. All rights reserved. Page 21 of 25

22 Figure 31 General kinematics for polar and Cartesian components Figure 32 Specific given cam shape Copyright J.E. Akin. All rights reserved. Page 22 of 25

23 For the given conditions at 45 degrees, the results are listed in Figure 33. By completing a List Solve over the full angular range, the position (Figure 32), velocity components (Figure 34) and acceleration components (Figure 35) were generated. In addition, the Coriolis and centripetal contributions to the acceleration were extracted and are in Figures 36 and 37, respectively. Figure 33 Cam follower results at the 45 degree position Copyright J.E. Akin. All rights reserved. Page 23 of 25

24 Figure 34 Velocity as a function of angular position of the follower Figure 35 Acceleration as a function of angular position of the follower Copyright J.E. Akin. All rights reserved. Page 24 of 25

25 Figure 36 Coriolis acceleration contribution as a function of angular position of the follower Figure 36 Centripetal acceleration contribution as a function of angular position of the follower Copyright J.E. Akin. All rights reserved. Page 25 of 25