The velocity of AB
The radial acceleration of AB is given by
Since A is fixed, the tangential component is zero. Also, the radial acceleration is parallel to the link AB.
1. Draw velocity vector of Ab parallel to link AB on a suitable scale.
2. draw the line at an angle perpendicular to BC from B on the velocity diagram.
3. Similarly, draw an extension line from D on the velocity diagram at an angle perpendicular to link CD. Both extension lines intersect at C to give the velocity diagram.
Measure the velocities we get
Acceleration Diagram :
We Know that
arAB = VAB2 / AB = 2.852 /0.06 = 135.38 m/s2
arBC = VBC2 /BC = 2.352 /0.2 = 27.61 m/s2
arCD = VCD2 /CD = 2.712 /0.1 = 73.44 m/s2
1. Since AB have only the radial acceleration draw the radial component parallel to link AB
2.Draw the radial component of BC from B and draw tangential component perpendicular to radial component
3Draw the radial component of CD from D and draw tangential component perpendicular to radial component
From acceleration diagram we get the acceleration of eac link
aB = 135.38 m/s2
aBC = 37.78 m/s2
aC = 100.37 m/s2
To find the acceleration due to masses we use similarity ratios
since the mass m1 is at the orgin the acceleration due to the mass is zero
B2- distance between Band mass m2
(Since the center of mass is at center)
Plotting the point in the acceleration diagram and measure it to get true acceleration due to the mass. The acceleration due to the mass is drawn in green color.
From the drawing we get
Then the inertia force is given by
F = -ma
The negative sign indicates the forces are acting in opposition to the direction of acceleration.
Also, the forces are displaced from the center of mass. Let the distance be h.
Then we get
F*h = I
I - Moment of inertia
h - Perpendicular distance between the force and center of gravity.
- Angular acceleration.
= at / Length of link
2 = 25.78/0.1 = 257.8 rad/s
3 = 90.65 /0.2 = 453.25 rad/s