00:01
Hi, in this question we have the input torque is given by it is ij2 into alpha2 where ij2 is the so moment of inertia of the link 2 and alpha2 is the angular acceleration.
00:16
So, we have alpha2 is nothing but it is d omega2 by dt.
00:20
So, we have so alpha2 is omega2 is constant therefore, alpha2 will be 0 therefore, t2 will be equal to 0.
00:28
Then we have so for link 2 so we have to consider for link 2 we have so x and y components of g2 are given by so ag2 x is given by it is r2 into alpha2 and ag2 y is given by r2 into alpha2.
00:56
So, this is equal to it is 2 into 0 and this is also 2 into 0 which is equal to 0.
01:03
So, this is also 0 so here velocity remains constant.
01:07
Then we have for link 3 so again taking the x and y component we have ag3 x is equal to minus rs into alpha3 and ag3 y is equal to 0.
01:22
So, now we have this is equal to minus 4 into 52 .17 which is equal to minus 208 .68 inch per second square...