00:01
Calculate the angular velocities of the bar bd and ed and the velocity at point d.
00:14
Then calculate the angular acceleration of bar bd and ed.
00:26
From the instantaneous method here we have the link.
00:33
2, 3, 4 and the fixed one is 1.
00:42
The instantaneous centers are like i -12, i -2, i -2, i -3 -i -34 and i -14 or i -1 -3, where link 1 is equals to a -e, which is the fixed link, link 2, is a, b, d, 3 is bd and 4 is d .e.
01:25
The instantaneous i -24 is the line joining the extension of 2 -3 and 3 4 with i -1 -2 and i -14.
01:35
I -1 -3 is the instantaneous center with the line joining the extension of i -14 and with i -12 and i -3.
01:47
According to the kennedy theorem, the theorem, link 2 and 3 omega 2 by omega 3 is equals to i23 .i -13 divided by i23 .i12.
02:08
Substituting the values, where omega 2 is 3 writing per second divided by omega 3 equals 0 .3 plus 0 .12 by 0 .12 from here we get the value for omega 3 is equal to 0 .857 radian per second.
02:29
Omega 3 is the angular velocity of bar bd.
02:35
Now for link 3 and 4 using the kennedy theorem we can write omega 4 by omega 3 is equal to i34 .i13 divided by i34 .i1 .4.
02:55
Substituting the values, we have omega 4 by 0 .857 equals 1.
03:06
Again, the omega 4 or the omega of bar de is equals to 0 .857 radiance per second...