Tutorial Test 5 D weling robot Phys 70: Planar rigi boy ynamics The problem statement is appene at the en of the reference solution. June 19, 015 Begin: 10:00 am En: 11:30 am Duration: 90 min Solution. (a) The equations of motion. The kinematic variables are the angular position θ an the raial coorinate r of the weling tip on the arm. These variables have the following time erivatives of interest, t θ = ω, θ = α, t t r = v, t r = a. The time erivatives of r carry a prime, to enote that they are taken wrt the reference frame rotating with the robotic arm. We now set up the equations of motion for the weling robot. To robot consists of two pieces, the weling tip an the arm, which are rigi boies by themselves but o not form a rigi boy jointly. Thus, we have to set up the equations of motion for these two rigi boies separately. Here is a free boy iagram: F F M M ~ (on arm) -F (on tip) (on tip) (on arm) The moment M exerte by motor 1 an the force F exerte by motor are riving the motion. The irection of M is out of the page an the irection of F is parallel to the arm. The moment M causes the force F an the couple moment M on the tip as follows. As a consequence of M, the arm will unergo angular acceleration. But since the weling tip is confine to the arm, it must accelerate with it. That is, the tip will have a component of linear acceleration perpenicular to the arm, an the force F, exerte by the arm on the tip, will generate it. Then, by Newton s thir law, the tip will react with a force F on the arm. This force generates a moment which reuces M an thus reuces the acceleration. There is one more thing to take into account. If the arm accelerates with an angular acceleration α, then the weling tip accelerates with the same α, since it is not a mass point an it is mounte on the arm. A couple moment M is require to generate this angular acceleration α of the weling tip, an the weling tip will react back to the arm with a couple moment M, reucing the original moment M further. Now, the force EoM for the weling tip is F + F = m a, where we use m to enote both the mass of the tip an the bar, since they are equal. The acceleration a of the tip is not irectly one of the kinematic variables, an we use the relative motion formula for rotating axes to relate it to 1
the linear acceleration a in the rotating reference frame. Using the fact that the acceleration of the pinpoint A is a A = 0, this gives F + F = m ( α r ω r + ω v + a ). (1) We now ecompose the vector-value Eq. (1) into its scalar components, beginning with the component perpenicular to the arm. To extract it, we take the cross prouct r Eq. (1). We obtain The moment-eom for the arm reas where l is the length of the arm. The moment-eom for the weling tip reas r F = m ( r α + rv ω ) () M r F M = ml α, (3) 3 M = mk G α, (4) where k G is the raius of gyration for the weling tip. In the last three equations, the force F an the couple moment M are unknown. However, by aing up the Eqns. () - (4), we can eliminate them. We obtain l M = m 3 + r + kg α + rv ω. Since α, ω ˆk, the above is a scalar-value equation (posing as a vector-value equation). We multiply with ˆk an obtain M = m l 3 + r + kg α + rv ω. (5) This is our first equation of motion. Note that the factor multiplying α on the rhs is the combine mass moment of inertia of arm an tip. We now return to Eq. (1) an observe that we have not yet extracte the component parallel to the arm. To obtain it, we multiply Eq. (1) with e r, the unit vector along r. We obtain F = m ( ω r + a ), (6) which is our secon equation of motion. Consistency with angular momentum. The moment M measures the rate of change of the angular momentum L, M = t L. The angular momentum L = I total ω for the given situation is l L = m 3 + r + kg ω. Taking the time erivative of this equation, an noting that the raial position r of the tip is timeepenent (/t r = v ), we obtain l M = m 3 + r + kg α + rv ω,
which coincies with Eq. (5). Test OK! (b) The situation now simplifies because the weling tip stays at its outer position at all times, r l, v, a = 0. The remaining equation of motion, Eq. (5) therefore simplifies to 4 M = m 3 l + kg α. }{{} I The optimal strategy uner these circumstances is to first accelerate as har as possible, an then ecelerate as har as possible, subject to the two constraints that 1. The arm accumulates a rotation angle of θ = π uring that motion, an. It arrives at the final position with angular velocity zero. Accelerating phase. The epenence of M on ω is M = M 0 (1 ω/ω 0 ), such that we obtain the following linear ifferential equation for ω. t ω = M ) 0 (1 ωω0. I The solution of this LDE, with bounary conition ω(t = 0) = 0, is where the characteristic time scale τ is Integrating Eq. (7) in time, we obtain ω(t) = ω 0 (1 e t/τ ), (7) θ(t) = ω 0 τ τ = Iω 0 M 0. ( ) e t/τ 1 + t/τ. (8) The accelerating phase ens at a time T 1, whose numerical value remains to be etermine. Decelerating phase. The epenence of M on ω now is M = M 0 ( 1 ω/ω 0 ), such that we obtain the following linear ifferential equation for ω. t ω = M ) 0 ( 1 ωω0. I The solution of this LDE, with bounary conition ω(t = T 1 ) = ω(t 1 ) = ω 0 ( 1 e T 1 /τ ), is Integrating Eq. (9) in time, we obtain (( ω(t + T 1 ) = ω 0 1 + ω(t ) ) 1) e t/τ 1. (9) ω 0 θ(t + T 1 ) = θ(t 1 ) + τ(ω 0 + ω(t 1 )) ( 1 e t/τ ) ω 0 t. (10) The ecelerating phase lasts for a time T, whose numerical value remains to be etermine. θ(t 1 ) on the rhs of the last expression is given by θ(t 1 ) = ω 0 τ ( e T 1/τ 1 + T 1 /τ ), cf Eq. (8). 3
Numerical solution. We now solve Eqs. (9) an (10) for T 1, T, such that ω(t 1 + T ) = 0 an θ(t 1 + T ) = π. We fin T 1 =.158 s, T = 0.580 s. The total uration of the motion thus is T 1 + T =.737 s. The angle covere in the accelerating phase is.685 ra = 154, an the ecelerating phase is 0.456 ra = 6. (c) Optimal strategy in the general case. A goo strategy is to get the weling tip as quickly as possible to the centre A of the arm, such that the mass moment of inertia T total is minimize, allowing for rapi acceleration an eceleration of the arm. Depening on the relative strength of motors 1 an, however, if motor is weak it may happen that the weling tip oes not reach the central position A in time, an must turn aroun sooner. If motor is strong enough for the weling tip to reach position A, the optimal strategy is: Motor 1: 1. Accelerate har for a time T 1.. Decelerate har for a time T Motor : 1. Accelerate har inwars for a time T 3.. Har ecelerate inwar motion for a time T 4 (arrive at A with v = 0). 3. Stay at the centre A for a time T 5. 4. Accelerate har outwar for a time T 6. 5. Har ecelerate outwar motion for a time T 7. The motion is thus escribe by the seven parameters T 1,..., T 7. Given these 7 parameters, the motion follows from (numerically) integrating the couple equations of motion Eqs. (5), (6). We have the following constraints: 1. ω(t 1 + T ) = 0.. θ(t 1 + T ) = π. 3. v (T 3 + T 4 ) = 0. 4. r(t 3 + T 4 ) = 0. 5. v (T 3 + T 4 + T 5 + T 6 + T 7 ) = 0. 6. r(t 3 + T 4 + T 5 + T 6 + T 7 ) = l. 7. T 1 + T = T 3 + T 4 + T 5 + T 6 + T 7. We thus have 7 constraints for 7 parameters. If the values of the given quantities are right (motor sufficiently strong) we can expect a solution. Marking: (a) 5 points, (b) 5 points, (c) points. 4
M M 0 F 0 F v 0 ω0 ω v Motor 1 Motor Figure 1: Characteristics of the two motors. Left: Moment M exerte on the arm by Motor 1, as a function of the angular velocity ω of the arm. Right: Force F exerte on the weling tip by Motor, as a function of the velocity v of the weling tip with respect to the arm. Problem 1. We iscuss the simplifie esign of a weling robot in two imensions; see the figure below. The weling robot consists of an arm (= thin ro) capable of rotating about the en point A, an a weling tip, which can freely move along the ro, from the outermost point of the arm all the way to the center point A. The esign invokes two electric motors, one to turn the arm, an one to move the weling tip along the arm. weling tip A robotic arm (initial position) final position In the initial position (Pos. 1), the arm is at rest (ω = 0), an the weling tip is at rest with respect to the arm (v x y = 0). In the final position (Pos. ), the arm is again at rest (ω = 0), an the weling tip is again at rest with respect to the arm (v x y = 0). Furthermore, the arm in its final position is at a 180 egree angle from the initial position. Information about the motors. The robotic arm is riven by two electric motors, one rotating the arm (motor 1), an one moving the weling tip on the arm inwars or outwars (motor ). The moment which motor 1 generates to rotate the robotic arm epens linearly on the angular frequency of the arm. It is M 0 = 30 Nm at ω = 0, an zero at ω 0 = ra/s. The force which motor generates to move the weling tip in an out on the robotic arm epens linearly on the velocity v of the weling tip with respect to the arm. It is F = 0 N at v = 0 an F = 0 at v 0 = m/s. These characteristics are plotte in Fig. 1. We have to consier what happens when the voltage across Motor 1 an/or Motor is reverse. The effect is the same, an so we isplay it below for Motor 1 only. Namely, what happens is that M 0 M 0 an ω 0 ω 0. For Motor 1, the epenency of M on ω for the reverse voltage (= maximum negative voltage) is plotte in Fig., along with the same epenence for the original (= maximum positive) voltage. Analogously, for Motor, F 0 F 0 an v 0 v 0 uner reversal of the voltage. There is one point worth of note about these characteristics: The motor can be use both as an accelerating evice an as a brake. An, it always brakes harer than it accelerates. To see this, consier the vertical ashe line in Fig.. At that point near ω 0, Motor 1 can only eliver a weak accelerating moment, but, with the maximum negative voltage applie, it elivers a large ecelerating moment. Dimensions, masses an constraints. The robotic arm may be consiere as a thin ro. It has a length of l = 1 m, an a mass of 10 kg. The weling tip also has a mass of 10 kg, an a raius of gyration of 10 cm. The weling tip is free to move on the arm for its whole length l, i.e., the centre of mass of the weling tip may move between the pin point A an the outermost point of the arm. The weling tip must be steere such that it is never slamme into its extreme inner an out position on the arm, but rather arrives at these points with zero relative velocity. 5
M M 0 maximum positive voltage -ω 0 ω 0 ω -M 0 maximum negative voltage Figure : M vs. ω characteristics (Motor 1) for maximum positive an maximum negative voltage. Questions. (a) [5 points] Derive an state the equations of motion for this weling robot. Hint & consistency check: Consier the situation where Motor 1 elivers zero moment, an the weling tip is moving inwars. What oes conservation of angular momentum say is going to happen? Is this reflecte in your equations of motion? (b) [5 points] Consier the simple case where Motor is not use at all, i.e., the weling tip stays in its outer position at all times. The motion of the arm begins at the angular position θ = 0 with angular velocity ω = 0, an ens at the angular position θ = 180 with angular velocity ω = 0. Devise an optimal (or goo) strategy for moving the arm. Calculate the time neee to move the arm from the initial to the final position. Hint: Analytical arguments go a long way, but in the en you will nee to recur to numerics. (c) [5 points] Now using both motors, evise an optimal (or goo) strategy for moving the arm from its initial position at θ = 0 to its final position at θ = 180. The weling tip starts an ens in the outermost position on the arm, with velocity zero. Explain the working principles behin your strategy. How many free parameters oes your strategy have? How many bounary conitions nee to be matche? Can they be matche by your strategy? ( ) [5 extra points] For the strategy evise in part (c), compute the time neee to move the robotic arm from the initial to the final position. Hint: Rely on numerics. 6