226 226 Introduction to Robotics T, 12, 12 02 Yı m2 T B 21 yo 11,11 PC e mi хо Figure 6.6 A 2-DOF robot arm. Therefore,

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PLEASE show ONLY how Vd squared final equation was found . I tried subbing in the equations given in the solution above but did not get the simplified ones they got.
226 226 Introduction to Robotics T, 12, 12 02 Yı m2 T B 21 yo 11,11 PC e mi хо Figure 6.6 A 2-DOF robot arm. Therefore, the total velocity of the center of mass of link 2 is: vý = x +jß = 0;(( +0.251; +hıl,C2) + 02 (0.2513) +0102 (0.512 +212C2) + + + (6.7)
Example 6.4 Using the Lagrangian method, derive the equations of motion for the 2- shown in Figure 6.6. The center of mass for each link is at the center of the link. The moments of inertia are I and I2. Solution: The solution of this example robot arm is in fact similar to the solution of Example 6.3. However, in addition to a change in the coordinate frames, the two links have distributed masses, requiring the use of moments of inertia in the calculation of the kinetic energy. We follow the same steps as before. First we calculate the velocity of the center of mass of link 2 by differentiating its position: *D = l1 C1 +0.512C12 → ÅD=-119101 -0.5l2S12 (01 + 02) YD=44S1 +0.512S12 → İd=11C 02 +0.512C12 (01 + 2) +