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Sunday, September 9, 2007

Homework Hint: Problem 7/10


Suggestions:
  • The angular velocity of the panel has two components: Omega about the fixed Z-axis and omega about the moving y-axis (see figure to the left).
  • Write down the omega vector for the panel using the two components above.
  • To find the angular acceleration vector alpha, differentiate the omega vector above with respect to time. When doing so, note that the Z-axis is fixed; therefore, dK/dt = 0. Also, recall that dj/dt = omega x j.

10 comments:

Andrew Crandall said...

I'm having trouble putting together the acceleration of point P. How do we find a_rel? On the exam it was just v_rel^2/r, but in this problem r has a j and a k component so I don't think we can just divide them and then assign the it the direction like on the exam.

CMK said...

My suggestion on this problem is to put an observer on the panel (as indicated in the figure posted with this hint).

Now envision the motion that the observer sees of point P. This observer moving with the panel sees no motion of P since both the observer and point P are on the same rigid body. Therefore, (v_P/O)_rel = (a_P/O)_rel = 0.

This result is different from the exam problem no. 2 that you mention in this comment. Why is it different? In the exam problem, P is NOT attached to the rigid body to which the observer is attached. In that case, the observer sees a circular motion of P with the observed speed being a constant. Therefore in that case (a_P/O)_rel = -(u^2/r)*i (a centripetal component).

Very good question. Let me know if this explanation helps. If not, I will try to get you a better answer.

Andrew Crandall said...

I see my error now. I was confusing point O in the problem's drawing to the position of the observer relative to P. I see now that P and O are on the same rigid body.

jonbrueck said...

is there any way you could post the answer to this question so we know if we did it right or not?

CMK said...

For the axes shown in the post:

a_P = (35.8*j - 80*k) in/sec^2
alpha = (-1.2*k) rad/sec^2

jonbrueck said...

thanks

Josh said...

When I do the cross product of alpha and r, I keep getting a resultant in the i direction. My alpha is in the k direction and my r has a j and a k component. Can anyone tell me if I have my r in the wrong direction or why there is no i component in the answer.

Mackenzie said...

I am getting alpha to be in the i direction. when I use alpha in the k direction I am getting the wrong answer for the acceleration. however, when I use alpha in the i direction for acceleration I am getting the right answer. it is posted that alpha is in the k direction....is the answer wrong or am I wrong and my math just so happens to be working out?

ccb056 said...

I think alpha is in the -i (negative i) direction.

CMK said...

Sorry. I typed the comment too quickly as I was trying to get out the door this afternoon.

As you are finding out, alpha should be-1.2*i, not -1.2k. Apologies if that wasted your time this afternoon.