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The T-shaped link ABO is pinned to ground at point O. Link CD is pinned to ABO at point C. Link ED is pinned to link CD at D. Link ED is pinned to ground at E. Link ED has known angular velocity (0.4 rad/sec, CCW) and angular acceleration (0.06 rad/sec^2, CCW).
My suggestion is to write down the velocity and acceleration equations for EACH link. (For link ABO, use the kinematics equations between points O and C.) From these three vector equations (six scalar equations), you can solve for all of the unknowns. In particular, you will need the angular velocity and angular acceleration of link ABO.
Once the omega and alpha for ABO are found from above, then use these to answer the question about the acceleration of A with respect to B: a_A - a_B = alpha_ABO x r_A/B - omega_ABO^2*r_A/B.
It we're supposed to write down the velocity and acceleration equations for each link wouldn't that be 6 vector equations with 12 scalar equations? Is that the shortest way to do it?
It's not as daunting of a task as it might seem -- many of the unknowns you don't really care to find. Say we are looking at the velocity equations on this problem (one for each link):
v_C = v_O + omega_ABO x r_C/O v_D = v_E + omega_ED x r_D/E v_D = v_C + omega_CD x r_D/C
You would substitute the first two vector equations into the third (since you don't really care to find the v_C and v_D vector components). This (along with v_O = 0 and v_E = 0) would leave you with:
omega_ED x r_D/E = omega_ABO x r_C/O + omega_CD x r_D/C
This one vector equation is two scalar equations in terms of unknowns omega_ABO and omega_CD. So, instead of being six equations for six unknowns, it is actually two equations in terms of two unknowns.
The same thing holds for the acceleration equations.
I am having trouble with the three equations for each link. For link RD I get a -j component numerical answer for vD which seems right since ED is CCW. For CD link my equation gives me a final equation all in the j component for vC with vD plugged in from first equation. My last equation for link CO gives me an equation then that has vC equaling an equation in the i compoment only. So I am stuck with omegaCO being 0 since my other two equations do not have i components. I know i made an error somewhere but I cannot seem to find it?
Your process is correct, and your answer is correct!
To check out this, find the IC for link CD: Construct perpendiculars to v_D and v_C. The IC for CD is at the intersection of these two perpendiculars (point C itself). And, importantly, the velocity of that point is ZERO. So, if v_C is zero, then omega_OC must also be zero. All this agrees with your vector calculations.
Be careful, however: omega_CD is zero for only this one instant in time. alpha_CD will NOT be zero, as you will find out from your acceleration equation.
I after substituting equations, I get (-.4j)=(-3omega_ABCi)+(3omega_CDj)
After going through the acceleration equations, I found omega_OC to be .14 rad/sec. I don't know if this is correct and even if it is, it doesn't seem to help me much.
ok I understand your post CMK. Thank you. That points out that I have made a mistake in the acceleration equations somewhere. I will continue to dig for the mistake.
I am having trouble with this problem. I have read all of the posts and I am getting even more confused. Everyone has called everything something different. Also, I think there are a lot of errors in the statements. Can anyone clearly explain how this equation helps me solve this problem (in the same variables stated by cmk in his first comment): -.4j = -3omega_ABOi + 3omega_CDj
This equation states that omega_ABO is zero which I believe is right, but it also says that omega_CD is -.1333 rad/s. How is this possible if v_C is zero? Does anyone have any insight?
I am a tad bit confused on the final equation a_A - a_B = alpha_ABO x r_A/B - omega_ABO^2*r_A/B. It comes from a_A = a_B + alpha_CD x rA/B - w_cd^2 * r_A/B, right? THey are asking for the acceleration of A with respect to B, which would be a_A/B, correct? So a_A/B the same thing as a_A - a_B, right?
As your analysis shows, and, as confirmed by IC analysis, point C is the instant center of link CD.
This says two things: the speed of C is zero and C is the center of rotation for link CD. So, at this instant, CD appears to rotate about C. There is no contradiction in having v_C = 0 and omega_CD not zero.
Your first question had to do with how you could use this equation:
-.4j = -3omega_ABOi + 3omega_CDj
to get the answers. It looks like you have that part down, since you are able to get the right answers. However, if you have questions as to what it all means, let me know.
Does this add any insight? If not, let me know and I will try a different approach.
Yes, all that you say is correct. They are asking for the acceleration of A relative to B, a_A/B. By definition (a week or so ago in class), this is: a_A/B = a_A - a_B.
i am trying to find alpha_cd using instant centers, but i confusing myself. i know v_c=0 and i think the equation to use is alpha_cd= a_e/r_cd. is that right?
The method of instant centers works only for velocities, not accelerations. The reason for this is that the acceleration of a point is not perpendicular to the line connecting the point back to the center of rotation.
For accelerations, you need to use the full vector equations: one vector equation for each rigid body.
Yes, once you know omega_ABO and alpha_ABO, you can find a_A and a_B. The final answer to your question is a_A - a_B. This can be found directly from omega_ABO and alpha_ABO (see first comment in this post).
16 comments:
The T-shaped link ABO is pinned to ground at point O. Link CD is pinned to ABO at point C. Link ED is pinned to link CD at D. Link ED is pinned to ground at E. Link ED has known angular velocity (0.4 rad/sec, CCW) and angular acceleration (0.06 rad/sec^2, CCW).
My suggestion is to write down the velocity and acceleration equations for EACH link. (For link ABO, use the kinematics equations between points O and C.) From these three vector equations (six scalar equations), you can solve for all of the unknowns. In particular, you will need the angular velocity and angular acceleration of link ABO.
Once the omega and alpha for ABO are found from above, then use these to answer the question about the acceleration of A with respect to B: a_A - a_B = alpha_ABO x r_A/B - omega_ABO^2*r_A/B.
Let me know if this helps.
It we're supposed to write down the velocity and acceleration equations for each link wouldn't that be 6 vector equations with 12 scalar equations? Is that the shortest way to do it?
Yeah, that's right. My arithmetic failed me.
It's not as daunting of a task as it might seem -- many of the unknowns you don't really care to find. Say we are looking at the velocity equations on this problem (one for each link):
v_C = v_O + omega_ABO x r_C/O
v_D = v_E + omega_ED x r_D/E
v_D = v_C + omega_CD x r_D/C
You would substitute the first two vector equations into the third (since you don't really care to find the v_C and v_D vector components). This (along with v_O = 0 and v_E = 0) would leave you with:
omega_ED x r_D/E = omega_ABO x r_C/O + omega_CD x r_D/C
This one vector equation is two scalar equations in terms of unknowns omega_ABO and omega_CD. So, instead of being six equations for six unknowns, it is actually two equations in terms of two unknowns.
The same thing holds for the acceleration equations.
Does this make sense?
I am having trouble with the three equations for each link. For link RD I get a -j component numerical answer for vD which seems right since ED is CCW. For CD link my equation gives me a final equation all in the j component for vC with vD plugged in from first equation. My last equation for link CO gives me an equation then that has vC equaling an equation in the i compoment only. So I am stuck with omegaCO being 0 since my other two equations do not have i components. I know i made an error somewhere but I cannot seem to find it?
--to rjaneshe--
Your process is correct, and your answer is correct!
To check out this, find the IC for link CD: Construct perpendiculars to v_D and v_C. The IC for CD is at the intersection of these two perpendiculars (point C itself). And, importantly, the velocity of that point is ZERO. So, if v_C is zero, then omega_OC must also be zero. All this agrees with your vector calculations.
Be careful, however: omega_CD is zero for only this one instant in time. alpha_CD will NOT be zero, as you will find out from your acceleration equation.
Does this help?
I am having the exact same problem as rjaneshe...
I after substituting equations, I get (-.4j)=(-3omega_ABCi)+(3omega_CDj)
After going through the acceleration equations, I found omega_OC to be .14 rad/sec. I don't know if this is correct and even if it is, it doesn't seem to help me much.
sure does, i was just thinking that when i reread that ABO was oscillating, thanks
ok I understand your post CMK. Thank you. That points out that I have made a mistake in the acceleration equations somewhere. I will continue to dig for the mistake.
I am having trouble with this problem. I have read all of the posts and I am getting even more confused. Everyone has called everything something different. Also, I think there are a lot of errors in the statements. Can anyone clearly explain how this equation helps me solve this problem (in the same variables stated by cmk in his first comment): -.4j = -3omega_ABOi + 3omega_CDj
This equation states that omega_ABO is zero which I believe is right, but it also says that omega_CD is -.1333 rad/s. How is this possible if v_C is zero? Does anyone have any insight?
I am a tad bit confused on the final equation
a_A - a_B = alpha_ABO x r_A/B - omega_ABO^2*r_A/B.
It comes from
a_A = a_B + alpha_CD x rA/B - w_cd^2 * r_A/B, right?
THey are asking for the acceleration of A with respect to B, which would be a_A/B, correct? So a_A/B the same thing as a_A - a_B, right?
--to scott--
As your analysis shows, and, as confirmed by IC analysis, point C is the instant center of link CD.
This says two things: the speed of C is zero and C is the center of rotation for link CD. So, at this instant, CD appears to rotate about C. There is no contradiction in having v_C = 0 and omega_CD not zero.
Your first question had to do with how you could use this equation:
-.4j = -3omega_ABOi + 3omega_CDj
to get the answers. It looks like you have that part down, since you are able to get the right answers. However, if you have questions as to what it all means, let me know.
Does this add any insight? If not, let me know and I will try a different approach.
--to boilerup--
Yes, all that you say is correct. They are asking for the acceleration of A relative to B, a_A/B. By definition (a week or so ago in class), this is: a_A/B = a_A - a_B.
i am trying to find alpha_cd using instant centers, but i confusing myself. i know v_c=0 and i think the equation to use is alpha_cd= a_e/r_cd. is that right?
--to deborah--
The method of instant centers works only for velocities, not accelerations. The reason for this is that the acceleration of a point is not perpendicular to the line connecting the point back to the center of rotation.
For accelerations, you need to use the full vector equations: one vector equation for each rigid body.
Once we have found omega_ABO and omega_CD.
Do we need to find a_B and a_A for the acceleration equation?
Yes, once you know omega_ABO and alpha_ABO, you can find a_A and a_B. The final answer to your question is a_A - a_B. This can be found directly from omega_ABO and alpha_ABO (see first comment in this post).
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