In the acceleration equation, the k direction was ignored because it would be zero in the cross product. However, shouldn’t it be considered in the part of the equation without a cross product?
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Fall 2008 -- Purdue University -- West Lafayette, IN
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9 comments:
In the solution video, I actually determined the acceleration of point B on the arm (see figure provided in the post). Since P and B have the same acceleration, the answer is correct.
If I had found the acceleration of P directly, I would need to use the following form of the acceleration equation: a_P = a_O + alpha x r_P/O + omega x (omega x r_P/O). In this case, the r_P/O vector would include a z-component. As you have stated, this z-component would not have made a contribution since k x k = 0.
In short, the form of the acceleration equation a_P = a_O + alpha x r_P/O - omega^2 r_P/O is valid only when O and P are in the same plane. This will be true for virtually all of the problems we will do in this course (unfortunately, it is not true is this particular example...). I will talk about this in class on Friday.
Thanks for pointing out this. You have made a good observation about the video solution provided. Let me know if you have further questions on this.
In this equation, a_P = a_O + alpha x r_P/O + omega x (omega x r_P/O), is there a cross product between omega and r_P/O ???
Well, if it is not the cross product then I don't understand why it wont contribute to the acceleration, at least as far as the equation goes.
I agree that you have found the acceleration of B and that P and B will have the same value.
But why will finding the acceleration of P give the wrong value????
in the omega^2 * (r_P/O) part of the acceleration, the z-component does contribute......thats why I don't understand the mathematics behind it,......even though it gives the right answer....
The one equation:
a_P = a_O + alpha x r_P/O + omega x (omega x r_P/O)
is valid for any two points O and P on a body. The other equation
a_P = a_O + alpha x r_P/O - omega^ * r_P/O
is valid for any two points lying in a plane perpendicular to the axis of rotation (the case for just about all problems that we are doing this semester...except Problem 5/12).
Put in different words, the first equation reduces to the second equation when O and P lie in the x-y plane. The second equation is NOT valid for a point P that has a different z-coordinate than O.
The second equation is easier to use.
Does this help? If not, let me know.
Alright, I have understood the equations, but just to reconfirm the earlier one which you wrote involves a cross product while the latter is a special cease of the former.....??
special case*
Yes, the latter (w/o cross product) is a special case of the former (w/ cross product).
um just want to make sure of this but we did the exact same problem in class just with different # values and in finding the acceleration, the professor only used a_p = a_0 + alpha X r_p....and that was it; none of the omega stuff that is added on at the end. Is there something I'm missing or was it a mistake in class that went unnoticed?
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