Question

b) A planar manipulator has link lengths L1 2m and L2-1 m.Use the inverse kinematic equations to find the joint angles which will place the end point at the following positions (x ,y=1+ i) Write the forward kinematic equations for the end point. [2 marks] ii) Calculate the link L2 joint angle iii) Calculate the link L1 joint angle [5 marks] [5 marks] [Q1 Total: 20 Marks] Question 2 a) Explain the principle of Reed switch sensors and give examples of two common applications [4 marks] b) The standard binary and gray codes are given in the Table Q2. Answer the following Table Q2 Standard binary and gray code 2 4 Decimal number Standard Binary code 000001 000001 010 011 100 101 110 111 Gray code 011 010 110 111 101 100 i) Sketch a 3 bits absolute standard binary encoding disk and absolute gray encoding disk. Assume, the inner ring corresponds to bit 0, black sectors are "on" and white sectors are "off". Zero degree is on the right-hand side, with the angle increasing counter clockwise

please help with part B of question

Answer 1

The manipulator is planar, There is only one translation along X axis between the links apart from joint rotation.

cl -s10 11cl s パ 0 1s1

c2 -s2 0 12c2 1s2 c2 0 12s2 T:

2 CS01 0010 s2 e2 0 0 2200 CS 1 C 0 0 1 0010 1100 1100 T2

c12 -s12 0 11c1 12c12 s12 c12 0 1s112s12 0 0

$\Rightarrow x = l_1cos\theta_1+l_2cos(\theta_1+\theta_2);y = l_1sin\theta_1+l_2sin(\theta_1+\theta_2)$

Given:

$x = l_1cos\theta_1+l_2cos(\theta_1+\theta_2) = \sqrt(2)$

$y = l_1sin\theta_1+l_2sin(\theta_1+\theta_2) = 1+\sqrt(2)$

and

11 = 2: 12 = 1

which gives:

$x = 2cos\theta_1+cos(\theta_1+\theta_2) = \sqrt(2)$

$y = 2sin\theta_1+sin(\theta_1+\theta_2) = 1+\sqrt(2)$

square and add

$x^2+y^2 = (1+\sqrt(2))^2+2 = 4+1+4(cos(\theta_1-\theta_1-\theta_2)$

$\Rightarrow 5+4cos(\theta_2)=1+2\sqrt(2)+2+2$

$\Rightarrow 5+4cos(\theta_2)=5+2\sqrt(2)$

$\Rightarrow cos(\theta_2) = \frac{1}{\sqrt(2)}$

$\Rightarrow \theta_2 = 45^0$

Now consider:

$2cos\theta_1+cos(\theta_1+45) = \sqrt(2)$

2cosĄ + cos(θ1 ) cos45-sin ( θί ) sin 45-V (2)

$2 \sqrt(2)cos\theta_1+cos(\theta_1) -sin(\theta_1)= 2$

by observation we get $\theta_1= 45^0$

$\Rightarrow \theta_2 = 45^0$ and $\theta_1= 45^0$