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I'm not really sure where to start with this problem. What is going on at point B? We are solving for the angular velocity of AB; does that mean the velocity of its rotation about its center?
You are asked to find the angular velocity (omega) of the bar. This is the usual definition of angular velocity: the time derivative of the angle formed by any line on the body (here, line AB, for example).
Write the usual rigid body kinematics equation, here between points A and C (where C is the point on the bar in contact with the edge). Consider the hint provided on the blog about the motion of C for Problem 5/37.
Point B does not influence the rotation of the bar. If you need to describe the motion of B, you can do so after you have found the omega for the bar.
I'm not sure if it is correct but the equation I get for Va is Va=Vc+Wac x Ra/c. And from the homework hint you can split Vc up into its cartesian components and you know that Va does not have a j component so you can then set that equal to zero but after that I'm pretty lost.
Yes, you need to use the rigid body velocity equation relating the motion of points A and C (as you wrote);
vA = vC +omega_AC x r_A/C
The direction of vA is known (to the left along the x-axis) AND its speed is also known (r*omega_0). Substituting that into the LHS of the equation and substituting the resolution of vC in its x- and y-components into the RHS of the equation will give you TWO scalar equations in terms of two unknowns (vC and omega_AC). Solve these two equations for the two unknowns.
Just looking at my solution, I see where you will have some cancellation of terms like: sqrt(x^2+h^2). However, you will still have some terms like x^2/h and h that add together -- those are the ones that produce the h/( x^2+h^2) in the final answer.
Professor, can i assume w_bar to be in the -k direction since the bar will rotate CW? I did that and got w = rhw_o/(x^2-h^2)... which differs from the answer in the book. I used v_c = v_a + w_bar x r_c/a.
* w_bar = -w_bar k * r_c/a = (-x i + h j)
The value of v_a i used was the one discussed before, and for v_c i used the unknown speed v_c times the sine and cosine of the angle that the bar makes with the horizontal.
If you want, you can use the insight that you have and write the omega vector as being in the -k direction. In the end, you will get the same magnitude for omega as you would with omega assumed in the +k, only the sign will be different. You just need to remember what you assumed at the beginning.
So anyway, this is not the cause of your sign problem. Let me take a guess as to what went wrong. Let's use your -k omega vector in writing out the omega x r term:
(-omega*k) x (-xi + hj) = omega*h*i + omega*x*j
Here you see that the two terms have the same sign.
I will speculate that you did not end up with the same sign on those two terms. It is easy to forget that k x j = -i, or something like that. Could that be your problem?
yes!!! one little sign threw me off, and it was the omega*h*i term (which you accurately predicted) that i incorrectly wrote as negative.... I will be more careful with signs next time. Thanks a ton professor! Good night.
13 comments:
You are asked to find the angular velocity (omega) of the bar. This is the usual definition of angular velocity: the time derivative of the angle formed by any line on the body (here, line AB, for example).
Write the usual rigid body kinematics equation, here between points A and C (where C is the point on the bar in contact with the edge). Consider the hint provided on the blog about the motion of C for Problem 5/37.
Point B does not influence the rotation of the bar. If you need to describe the motion of B, you can do so after you have found the omega for the bar.
I'm not sure if it is correct but the equation I get for Va is Va=Vc+Wac x Ra/c. And from the homework hint you can split Vc up into its cartesian components and you know that Va does not have a j component so you can then set that equal to zero but after that I'm pretty lost.
Yes, you need to use the rigid body velocity equation relating the motion of points A and C (as you wrote);
vA = vC +omega_AC x r_A/C
The direction of vA is known (to the left along the x-axis) AND its speed is also known (r*omega_0). Substituting that into the LHS of the equation and substituting the resolution of vC in its x- and y-components into the RHS of the equation will give you TWO scalar equations in terms of two unknowns (vC and omega_AC). Solve these two equations for the two unknowns.
Let me know if this does not help.
(This is a comment that was posted by jonbruek. Somehow it never appeared on the blog.)
"Is Va = -rWoi + 0j ? I still can't get this to work. "
Response (by CMK): Yes, that is correct. A is moving to the left with a speed of r*omega_0.
my x^2 + h^2 is canceling out. Has anyone else had this problem? What am I doing wrong?
Just looking at my solution, I see where you will have some cancellation of terms like: sqrt(x^2+h^2). However, you will still have some terms like x^2/h and h that add together -- those are the ones that produce the h/( x^2+h^2) in the final answer.
Does that help?
Can we do the equation the other way around and still get it to work? (i.e. v_c = v_a + omega x r_c/a)
... or will that affect the correct answer?
--to dpinto--
Yes, either way works.
The two equations are:
v_C = v_A + omega x r_C/A
and
v_A = v_C + omega x r_A/C
Since r_C/A = - r_A/C, you can see that the two equations are equivalent. (Just be careful with signs!)
Professor, can i assume w_bar to be in the -k direction since the bar will rotate CW? I did that and got w = rhw_o/(x^2-h^2)... which differs from the answer in the book. I used v_c = v_a + w_bar x r_c/a.
* w_bar = -w_bar k
* r_c/a = (-x i + h j)
The value of v_a i used was the one discussed before, and for v_c i used the unknown speed v_c times the sine and cosine of the angle that the bar makes with the horizontal.
What am I doing wrong?
thanks everyone. turns out i don't know how to do algebra... but i finally got it to work.
If you want, you can use the insight that you have and write the omega vector as being in the -k direction. In the end, you will get the same magnitude for omega as you would with omega assumed in the +k, only the sign will be different. You just need to remember what you assumed at the beginning.
So anyway, this is not the cause of your sign problem. Let me take a guess as to what went wrong. Let's use your -k omega vector in writing out the omega x r term:
(-omega*k) x (-xi + hj) = omega*h*i + omega*x*j
Here you see that the two terms have the same sign.
I will speculate that you did not end up with the same sign on those two terms. It is easy to forget that k x j = -i, or something like that. Could that be your problem?
-- to jonbrueck --
Happy to hear that it worked out. Perseverance wins out in the end!
yes!!! one little sign threw me off, and it was the omega*h*i term (which you accurately predicted) that i incorrectly wrote as negative.... I will be more careful with signs next time. Thanks a ton professor! Good night.
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