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i assume the an is the same as the radial line. I solve for the rb/a=.6-.1=.5. I figure I could use vb=va+omega x rb/a . but i don't know what omega could be. so i need so help.
My suggestion is to work the problem in two steps: * First use the rigid body velocity equation relating points O and A. This equation will give you the angular velocity (omega) for the disk. * Then use the rigid body acceleration equation relating points O and B. For this equation, you will write out aB in terms of its x-y components. The omega vector for this equation is the same as you found in the first step (angular velocity is a property of the motion of the entire body). This vector equation produces two scalar equations (one for the x-component, the other for the y-component) with two unknowns (the magnitude of the acceleration of B and the angular acceleration of the body).
I recommend aligning your x-y axes with x along line OA and y along line OB. This will simplify the geometry somewhat. However, any choice of alignment will work.
I found omega and wrote out the rigid body acceleration eq. for O and B. However, what I don't understand is how we are supposed to write out aB in x-y components, and how we are supposed to find rB/O from this.
Can anyone explain to me what the following sentence from the problem is saying?
At a certain instant point A has a velocity vA = .8m/s in the direction shown, and at the same instant the tangent of the angle theta made by the total acceleration vector of any point B with its radial line to O is 0.6.
I get what vA is, but what is the second part of the sentence saying.
Angular velocity and angular acceleration do not depend on the length of rB/O. So, you can pick any length for rB/O you would like as long as it is less than 100mm.
So i've worked out my equation and now have a_b and alpha (the angular acceleration) left. One equation with 2 unknowns. Any suggestions on how to find a_b?
You should end up with 2 equations (1 in the i direction and 1 in the j direction) and 2 unknowns from the rigid body acceleration equation relating O and B. Vector a_B equals the magnitude of a_B times the unit vector along a_B. You are given that tan(theta) = 0.6. With this you should be able to right a_B in vector form.
Great job. You guys have nailed it. Pay close attention to the last two comments by scott: the length of rB/O is not important (any length works) and that the single vector equation is actually two equations in disguise (x- and y-components).
10 comments:
My suggestion is to work the problem in two steps:
* First use the rigid body velocity equation relating points O and A. This equation will give you the angular velocity (omega) for the disk.
* Then use the rigid body acceleration equation relating points O and B. For this equation, you will write out aB in terms of its x-y components. The omega vector for this equation is the same as you found in the first step (angular velocity is a property of the motion of the entire body). This vector equation produces two scalar equations (one for the x-component, the other for the y-component) with two unknowns (the magnitude of the acceleration of B and the angular acceleration of the body).
I recommend aligning your x-y axes with x along line OA and y along line OB. This will simplify the geometry somewhat. However, any choice of alignment will work.
Let me know if I did not answer your question.
I found omega and wrote out the rigid body acceleration eq. for O and B. However, what I don't understand is how we are supposed to write out aB in x-y components, and how we are supposed to find rB/O from this.
Can anyone explain to me what the following sentence from the problem is saying?
At a certain instant point A has a velocity vA = .8m/s in the direction shown, and at the same instant the tangent of the angle theta made by the total acceleration vector of any point B with its radial line to O is 0.6.
I get what vA is, but what is the second part of the sentence saying.
Is it saying that vB = .6vA?
I figured it out. It is saying:
tan(theta) = 0.6
I to do not understand how to find rB/O. We have no distance to work with in terms of B.
Angular velocity and angular acceleration do not depend on the length of rB/O. So, you can pick any length for rB/O you would like as long as it is less than 100mm.
So i've worked out my equation and now have a_b and alpha (the angular acceleration) left. One equation with 2 unknowns. Any suggestions on how to find a_b?
You should end up with 2 equations (1 in the i direction and 1 in the j direction) and 2 unknowns from the rigid body acceleration equation relating O and B. Vector a_B equals the magnitude of a_B times the unit vector along a_B. You are given that tan(theta) = 0.6. With this you should be able to right a_B in vector form.
Great job. You guys have nailed it. Pay close attention to the last two comments by scott: the length of rB/O is not important (any length works) and that the single vector equation is actually two equations in disguise (x- and y-components).
If you do not follow this, leave a comment.
Thanks to all who participated!
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