Dynamics problem - Kinematics of Rigid Bodies

[Strawberry]

Homework Statement

There is a slider-crank mechanism. Express the angular velocity omegaAB and angular acceleration alphaAB of the connecting rod AB in terms of the crank angle theta for a given constant crank speed omega naught. Take omegaAB and alphaAB to be positive counterclockwise.

The structure is a point A free to slide laterally connected by a rigid rod of length l to point B. Point B is connected to another point O by a rod of length r. The crank angle theta between r and the horizontal is given, as is the crank speed omega naught. Take omegaAB and alphaAB to be positive counterclockwise.

Homework Equations

velocity = r * omega
acceleration tangential = alpha * radius

The Attempt at a Solution

I first tried to define the horizontal distance between A and O. I found x to be r*cos(theta) + sqrt(l^2 - r^2*sin(theta)^2)
At this point I took the derivative of x to be the velocity, and I divided the velocity by x.

At this point I have an ungodly mess and I have no idea how to put omeganaught into the equation. I feel like I just need to understand the relationship between x or theta and omega naught to understand the equation, but I do not. I apologize for the lack of a graphical representation of the problem.

 

[Jhero]

Thank you for your post. I would like to provide some insights and a possible solution to your problem.

Firstly, let's define some variables for clarity. Let's say point A is at a horizontal distance of x from the origin O, and point B is at a horizontal distance of y from the origin O. The length of the connecting rod AB is l, and the length of the crank rod OB is r. The crank angle theta is measured from the horizontal axis.

Now, to express the angular velocity omegaAB and angular acceleration alphaAB of the connecting rod AB in terms of the crank angle theta, we can use the following equations:

omegaAB = (x/r)*omega0
alphaAB = (x/r^2)*omega0

where omega0 is the constant crank speed.

To understand these equations, let's consider the geometry of the problem. As the crank rod OB rotates at a constant speed omega0, point B moves in a circular path with a radius of r. This circular motion of point B causes the connecting rod AB to move in a rotational motion as well. The horizontal distance x between point A and the origin O is equal to the radius of the circular motion of point B times the cosine of the crank angle theta (x = r*cos(theta)). This is why we have x/r in the equations for omegaAB and alphaAB.

Furthermore, the velocity of point B can be expressed as y*omega0, where y is the vertical distance from point B to the origin O. As the connecting rod AB is rigid, the velocity of point A must be equal to the velocity of point B. Therefore, we can equate x*omegaAB to y*omega0, which gives us the equation for omegaAB.

Similarly, the acceleration of point B can be expressed as y*alpha0, where alpha0 is the constant angular acceleration of the crank rod. As the connecting rod AB is rigid, the acceleration of point A must be equal to the acceleration of point B. Therefore, we can equate x*alphaAB to y*alpha0, which gives us the equation for alphaAB.

I hope this explanation helps you understand the relationship between x or theta and omega0, and how it relates to the equations for omegaAB and alphaAB. Please let me know if you have any further questions or concerns.