- Use the planar rigid body velocity equation relating points A and B on link AB. In this equation, you know the velocity of A (speed v_0 in the negative i direction) and the direction of the velocity of B (along the j axis). Balance i and j components to give two equations in terms of two unknowns: v_B and omega.
- Use the planar rigid body acceleration equation relating points A and B on link AB. In this equation, you know the acceleration of A (zero since A is traveling with a constant speed along a straight line path) and the direction of the acceleration of B (along the j axis). Balance i and j components to give two equations in terms of two unknowns: a_B and alpha.
Discussion
The are other ways to work this problem using methods that you might consider to be easier than the vector approach above. For example, you could write x = L*cos(theta) and perform time differentiation of this expression to arrive at answers for theta_dot and theta_dot_dot. We encourage you to use the vector approach above, however, since this approach will work on all problems in this part of the course regardless of the complexity of the problem. For problems more complex than this one, the "simple" approach is not a very good one to pursue.
For more details on the vector approach applied to a problem very similar to this one, see the lecture solutions for Example on page 42 of the lecture notes for Divisions 1 and 3 (Problem 5/81), and Example: Simple Mechanism on page 8 of the lecture notes for Division 2.
As always, if you have questions related to this problem, please add a comment to this post.
5 comments:
I found the angular velocity and the angular acceleration using the cross product and two equations. But the book says they want the answer with respect to x. How and why would you solve with respect to x?
The author is asking us to find the angular velocity and angular acceleration as functions of the variable x. That is, we want general expressions for these in terms of x. At the end, if you are given the position of A (a numerical value for x), you can then find numerical values for omega and alpha.
x comes into the problem through the position vector. For example, if you write v_B = v_A + omega x r_B/A, you will need to use: r_B/A = -x*i + y*j. From the constraints on the motion of the bar you know that y = sqrt(L^2 - x^2). Therefore, this position is strictly a function of x. And when you use it in the velocity equation, everything is a function of x.
The same idea holds for the acceleration equation.
Let us know if this does not answer your question.
Yes thank you, that was very helpful. I was solving the problem with respect to the hypotenuse, so all I needed to do was substitute in x. Thanks again.
Two questions: Instead of saying y = sqrt(L^2 - x^2), can i say y = xtan(theta)? And for finding alpha is it necessary to know the precise a_b direction, j or -j... does it affect the answer at all? Thanks.
It would not be incorrect to write y in terms of x and theta. However, since the problem asked for the answer in terms of x, you really need to have y written solely in terms of x, not x and theta. Writing the answer in terms of x and theta might imply that x and theta are independent when they are dependent ( x = L*cos(theta)).
Just make an assumption at the beginning as to whether a_B is in the -j or +j direction. The math will take care of the signs as you solve. If your answer for a_B is negative, it will mean that your original assumption was incorrect.
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