A four-bar linkage is shown in the figure (the ground "link"...

A four-bar linkage is shown in the figure (the ground "link" $O C$ is considered the fourth bar). If the drive link $O A$ has a counterclockwise angular velocity $\omega_{0}=10 \mathrm{rad} / \mathrm{s},$ determine the angular velocities of links $A B$ and $B C$.

Key Concepts

Instantaneous Center of Rotation

The instantaneous center of rotation (ICR) is a point in a rigid body or mechanism that, at a specific instant, has zero velocity relative to the ground. Identifying the ICR simplifies the analysis of rotational motion in planar mechanisms, as the angular velocity of a link can be determined based on the linear velocities of points on the link relative to this center.

Four-Bar Linkage

A four-bar linkage is a closed-chain mechanism consisting of four rigid members connected by pivot joints. It is a fundamental mechanism in kinematics used to convert one type of motion into another. The analysis of these linkages involves understanding the geometric relationships among the links and how the motion of one link affects the others.

Vector Loop Equations

Vector loop equations are used to represent the kinematic constraints of closed-chain mechanisms by summing the position vectors around the loop to zero. This method is essential for deriving relationships between the angular positions, velocities, and accelerations of different links within a mechanism.

Relative Velocity Analysis

Relative velocity analysis is a technique used to determine the velocities of various points or links in a mechanism by considering the motion of one component relative to another. In the context of linkages, it involves writing velocity equations based on known angular velocities and geometrical constraints to solve for the unknown angular velocities of other links.

Transcript

00:01 Okay, in this question, we have a few bars that are connected by four length points, o, a, b, and c.

00:09 We're told that the bar oa is rotating about point o, with an angular velocity omega -0 of 10 radiance per second.

00:20 We have the lengths for each of these bars, and we're given some angles to start with.

00:26 And what we'd like to do is find the angular velocities of links ab and bc.

00:34 So first we can find the velocities of some of these points.

00:39 So the velocity of point a should be pretty simple to find.

00:43 That'll just be the length of oa times omega of that link oa, which is just omega 0, 0 ,0 here.

00:56 Omega or oa is 0 .08, omega 0 is 10.

01:00 So this should be 0 .8 meters per second then.

01:09 Next, we can draw a velocity diagram, which i'll switch colors for here, for comparing velocities a and b, which will be the two important moving points here.

01:29 Well, velocity a is moving up into the left, velocity of b will be moving directly left.

01:40 And then we'll have the velocity of b with respect to a.

01:46 Velocity of b with respect to a, velocity of a, velocity of b.

01:52 And we'll have all of the angles.

01:56 If we look at this carefully, we end up with two 75 degree angles and one 30 degree angle, just based on the geometry of the situation.

02:17 And we know that vb has to be moving directly to the left because its bar rotates about point c at the bottom here.

02:29 Point a is moving up into the left and that's where we get the 30 degree angle from.

02:37 So we can use the law of signs then to help find vb.

02:42 We know that vb over the sign of 75 degrees should equal va over the sign of its opposite angle, which is 75 degrees, which then tells us that va equals vb.

03:02 So we know what both va and vb are now...

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