Robotics 2 E Part 15 pdf

410 Solutions to the Exercises or For a DC motor with a characteristic T= T^-To^, Expression (a) becomes A solution using MATHEMATICA language follows. The numerical data given in the problem are substituted by symbols (here, for reasons of convenience, we denote <f> = v and0= w). fl=.01 v"[t]+.5 (Cos[.51])*2 v"[t]- .125(Sin[t])v'[t] l jl=NDSolvel{fl==0,vIO]==0,v'[0]==0},{v[t]},{t,0,2}l bl=Plot[Evaluate[v[t]/.jl],{t,0,2}^xesLabel->{"t","v"}] £2=.01 w'ltl+.S (Cos[.5 t])A2 w'W- .125(Sin[t])w[t] l j2=NDSolve[{f2==0,w[0]==0},{w[t]},{t,0,2}] b2=Plot[Evaluate[w[t]/.j2] ,{t,0,2},AxesLabel->{"t","w"}] FIGURE 3E-6.1 a) Rotation angle v; b) Speed w of the column versus time. 13 Solution to Exercise 3E-6a) The rotation angle 0 of the disc depends upon the torques acting on the mecha- nism. The driving torque Tmust be equalized by inertia torques, in keeping with expres- sion (3.165): where / is the common moment of inertia of the disc / 0 and the moving mass I r . Obviously: Solutions to the Exercises 411 Therefore, For a DC motor with a characteristic T=T^- r 0 0, the Expression (a) becomes For the same data as in solution 3E-6, and where 7\ = 0.1 Nm and T 0 = 0.025 Nm/sec and RQ = 0.5 m, the following solution using MATHEMATICA is given. The equation is 0.010+0.5 (cos 0.51] 2 0-0.125 sin 10-0.1+0.025 0 = 0. f01=.01 v"[t]+.5 (Cos[.5 t]) A 2 v"[t] 125 (Sin[t]) v'[t] 1 +0.025 v'[t] j01=NDSolve[{f01==0,v[0]==oy[0]==0},{v[t]},{t,0,2}] b01=Plot[Evaluate[v[t]/.j01],{t,0,2}^VxesLabel->{"t","v"}] f02=.01 w1t]+.5 (Cos[.5 t])A2 w'[t] 125 (Sin[tl) w[t] 1 +0.025 w[t] j02=NDSolve[{f02==0,w[0]==0,w'[0]==0},{w[t]},{t,0,2}] b02=Plot[Evaluate[w[t] /.J02] ,{t,0,2},AxesLabel->{"t","w"}] FIGURE 3E-6a).l a) Rotation angle v; b) Speed w of the column versus time. 14 Solution to Exercise 3E-7 To answer the questions we use Formula (3.39). Thus, from this formula, it follows that when the number of winds W s = 2 is doubled, we have the following expressions for the response time: 412 Solutions to the Exercises From these formulas, it follows that when the voltage is doubled, u s = 2, we have the following expressions for the response time: From these formulas it follows that when the mass of the armature is doubled, m s = 2, we have the following expressions for the response time: 15 Solution to Exercise 4E-1 Case a) From geometrical considerations, the motion function n(jc) becomes Differentiating (a), we obtain Thus, Substituting the given data into (c), we obtain for y By differentiating (b), we obtain the following dependence from Expression (4.3) [the case where x = 0]: Solutions to the Exercises 413 From (c) and the Relationship (4.4) we obtain Substituting the numerical data into (c) and (d), we obtain Case b) From the geometry of the given mechanism, we have AD = CE. Then, the motion function Yl(x) is defined as follows: y = n(</0 = AOsin0 = 0.2sm30° = 0.1m. Thus, y = n'(0)0 = 0.2cos30° 5 = 0.866 m / sec, and y = n"(0)0 2 =-0.2sin30° 5 2 = -2.5m/sec 2 . 16 Solution to Exercise 4E-2 From the Formula (4.24) and its derivatives, we have: Here, it follows from the description of the problem that and therefore Thus, from (a) we obtain 414 Solutions to the Exercises To find the angle 0 corresponding to the maximum pressure angle a max , we differ- entiate (b): From (c), it follows that 2-0.08 cos (4-0)+fecos(4-0)-Mcos 2 (4-0)+sin 2 (4-0)] = 0 or On the other hand, from (b), we have Substituting tana = tan 20° = 0.324 into (e), and from (d), we obtain Solving Equation (f) by any method (for instance, graphically, by the method of Newton, or by computer) we obtain which from (d) gives for h The solution in MATHEMATICA language is al=(2 Sin[8 f] 364 (Cos[4 f]+(Cos[4 f])A2))/(l-Cos[4 f]) 364 bl=FindRoot[al==0,{f,.5}] {f-> 0.369625} 17 Solution to Exercise 6E-1 Condition (6.17) states that horizontal component .A/, of the acceleration takes the form Solutions to the Exercises 415 where A is the vibrational amplitude. Assuming that the vibrations S have the form: Then the accelerations are Thus, Case a) Condition (a) can then be rewritten in a form that takes into account that ft> = 2flrf=2*:50 = 3141/sec: From (a) follows or Case b) From the condition (a) and taking into account (b) of the previous case, it follows also for this case that Now we obtain or and 416 Solutions to the Exercises 18 Solution to Exercise 7E-1 Here we use Equation System (7.1). Since the two levers press the strip from both sides (upper and lower), the mechanism must develop a friction force P = F/2 at every contact point. Thus, the equations for forces and torques with respect to point O become and Here, R x and R y are the reaction forces in hinge 0; Nis the normal force at the contact point between the strip and the lever. From (a), we express the normal force Nas From the Equation (b), we express the force Q developed by the spring as Reactions R x and Ry are, respectively, 19 Solution to Exercise 7E-la) Here we use equation system (7.6). Since the two rollers press the strip from both sides (upper and lower), the mechanism must develop a friction force F b = QI2 at every point of contact with the strip. Thus, the equations for forces and torques with respect to point O become Solutions to the Exercises 417 and From the Equation (a), it follows that From the Equation (b), it follows that or From the Equation (c) and the given mechanism it follows that and finally 20 Solution to Exercise 7E-lb We continue to use Equation System (7.1). Since the two levers press the strip from both sides (right and left), the mechanism must develop a friction force F= Q/2 at every contact point. Thus, the equations for forces and torques with respect to point O become and 418 Solutions to the Exercises 21 Solution to Exercise 7E-2 The angular frequency co of the oscillations of the bowl is CQ = 2nf=2x5Q = 3l4 I/sec. The motion S of the bowl is: S = 0.0001 sin 314 t m. The acceleration S of the bowl obviously is S = -0<w 2 sina£ = -0.0001-314 2 sin314tm/sec 2 . The maximal value of the acceleration S max is Smax = aa> 2 = 0.0001-314 2 = 10 ml sec 2 . The angle/3 = y-oc = 30° - 2° = 28°. From Expressions (7.33-7.34), we calculate the values of critical accelerations for the half-periods of both positive and negative oscil- lations. Thus, and The latter expression means that during the second half-period of oscillations slide conditions practically do not occur for the body on the tray . By applying Expression (7.35), we check whether rebound conditions exist on the tray, a situation that occurs when the acceleration exceeds the value S r . Thus, At any point of movement, no point of the bowl reaches this acceleration value. There- fore, there is no rebound in the discussed case. We can now proceed to calculate the displacement of the items. From the curves in Figure 7.25 it follows that the time t v at which the slide begins (section EM) and the groove lags behind the item, is defined as Solutions to the Exercises 419 At this moment in time, the speed V 0 of the item (and the bowl) is defined as The slide begins with this speed and is under the influence of the friction force F = -fj.m(g + y) acting backwards. For our engineering purposes, we simplify this def- inition to the form F= //mg. This force causes deceleration: W L = -ju = -0.6 • 9.8 = 5.88 m / sec 2 . This assumption gives a lower estimation of the displacement. The following gives the upper estimation: This condition exists during time t 2 , which is defined as The displacement S 1 is then or or 0.000053m < 81 < 0.000083m. It is interesting to observe the influence of the friction coefficient ju on the values of the critical accelerations for both oscillation directions. We show here the compu- tation in MATHEMATICA language. Results are given in Figure 7E-2. (For convenience in MATHEMATICA we use m for the friction coefficient.) gl=Plot[9.8 (Sin[2 Degree]+m Cos[2 Degree])/ (m Sin [30 Degree]+Cos [30 Degree]), {m,.2,l},AxesLabel->{"m","s""}] g2=Plot[9.8 (Sin[2 Degree]-m Cos[2 Degree])/ (m Sin[30 Degree]-Cos [30 Degree]), {m,.2,l},AxesLabel->{"m","s""}] Show[gl,g2] [...]... 0.00 015 m for the same vibrofeeder as in the previous Exercise 7E -2, obviously, we change the dynamics of the device However, its characteristics remain the same Therefore, we have the same values of the critical accelerations Scr and S'cr as before The changes in the dynamic behavior of the device take place because of the fact that the maximal value of the acceleration of the bowl Smax becomes higher,... 420 Solutions to the Exercises FIGURE 7E -2. 1 Dependence "critical acceleration s" versus friction coefficient "m" for the specific design of the vibrofeeder as in this and next exercises This displacement takes place 50 times every second Therefore, the total displacement H during one second is 0.0 026 5m . previous Exercise 7E -2, obviously, we change the dynamics of the device. However, its characteristics remain the same. Therefore, we have the same values of the critical accelerations . the specific design of the vibrofeeder as in this and next exercises. This displacement takes place 50 times every second. Therefore, the total dis- placement H during one second . inductance R electrical resistance Afl increment of electrical resistance s, x linear displacements or distance T temperature t time V speed v voltage Ay increment of voltage W energy w