Question

Find K

r" = kt

r = 1 t = 0

r = 2 t =3

Imagine that you are an engineer working in à mahufaclug created a machine that is used to adhere one part to another. Figure 2 shows Part A being adhered to Part B. You will assume that you have constrained the manipulator such that 0 for all time. Additionally, you will control the acceleration of the end-point of the manipulator, and this expression is given by f kt where k is a constant. Part B Part A Figure 2: The two-link robotic manipulator at the position . Part A must be adhered to part B. The machine will operate during three second time intervals. Your job as a robot designer is to determine a value of k such that over a three second time interval the end-point will have an ending position equal tor- 2 m. You may assume the manipulator starts at r 1 m when tO s. Assume that the manipulator starts from rest and remains at 0 0 throughout the motion. The derivation for your solution should be shown in r technical report and should be in variable form. That is, you should only substitute numbers once you have an expression for k in variable form. Please create plots of the position, velocity, and acceleration of the end-point of the machine as time varies from t 0 seconds to t 3 seconds with your value of k. Include this plot and the mathematical analysis in your technical report. The plot must be computer generated and include appropriate labels. You should plot the position, velocity, and acceleration ont your technical report would also be appropriate. Please note that the plot should be a continuous time plot. he same plot. Including a figure of the manipulator at θ 0° in

Answer 1

Black - r(t) vs t Purple - v(t) vs t Orange - a(t) vs t given r^�� = kt implies d^2 r/dt^2 = kt integral d^2 r/dt^2 = integral kt dt dr/dt = kt^2/2 + c_1 again integrating with respect to r(t) = kt^3/2 times 3 + c_1 t + c_2 where c_1 & c_2 are the constants of integration given r(t - 0 sec) = 1 m and r(t = 3 sec) = 2m r(0 sec) = c_2 implies c_2 = 1m r(3 sec) = 3^3 k/3 times 2 + 3c_1 + 1m \r\n r(t) = kt^3/6 + c_1 t + c_2 at t = 0 sec implies c_2 = 1m at t = 3 sec implies 2m = 27k/6 + 3c_1 + c_2 (2m - c_2) = 27k/6 + 3c_1 (2m - c_2) - 27k/6 = 3c_1 2m - c_2/3sec - 27k/3 times 6 = c_1 (1/3)m/sec - 27k s^3/6 times 3 sec = c_1 c_1

Black - r(t) vs t

Purple - v(t) vs t

Orange - a(t) vs t

28 09:05 5 1.915, 3.34 0,1.833 1.16, 1.16 0.562, 0 1.915,0 3.562, 0 -1 4 es

Here x stands for time and Y is for postion.

in this case

Y is for velocity and x is for time.