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is the IC of the circle on the right at the bottom of the inner circle or the outer circle, in what i've labeled as F or in what i've labeled as G? is that outer circle moving? thanks.
The wheel is rolling without slipping, with point F in contact with a stationary surface. Therefore, F has zero velocity (hence, the IC for the wheel).
The wheel on the left moves as a single rigid body (a single angular velocity and a single angular acceleration for the body).
It turns out that v_O is NOT equal to v_C. What you need in order to relate the motion of the two wheels is that v_B = v_A. (From this, you can see that v_O < v_C.)
I found omega for both of the circles, but am having trouble finding values for acceleration. Right now I have Ac = Aic + (alpha x r) - (omega^2*r). I know r, omega, and Ac. After I substitute those in I still need to find Aic and alpha. Any suggestions?
I am having a hard time finding the angular acceleration of the small wheel. I've found the angular velocity to be 2.66. I wrote an equation for each relating A_a-->A_c, then A_c-->A_instant center, then A_a-->A_instant center. But when i subsitute 2 equations into the third, i get the angular acceleration to be 0. Any Help?
scott- I believe that A_c and A_instant center are the same point, so that makes sense that the angular acceleration becomes zero. i believe that you need to relate A-a to A-b and A_b to A_d and then you should get a system of equations to solve.
The way I worked it out was the V_A = V_B and a_A i = a_B i. That is, only the i components of the accelerations of A and B are equal. The perpendicular components are different because the wheel's have different radii. I was able to get a complete expression for a_B, however I don't know any other accelerations on that left wheel....so I have too many unknowns and can't solve for alpha, and therefore the acceleration of D. My hunch is that a_O=a_C, but I'm not sure why since their velocities aren't equal. Please correct me if anything I said was incorrect because I've kind of hit a wall.
I was assuming that a_A = a_O because both are only moving in the +i direction. Can anyone validate that the distance x between pt. A and pt. O is not changing. This would at least narrow down possibilities for solving the problem.
We have had quite a few comments posted on this since I last checked. I will respond to all of the new comments here:
--to mburr-- As stated above, the IC for the wheel on the left is at point F; thus, the r that you use for omega for this wheel is 0.3 meters (distance from F to B). The IC for the wheel on the right is at its contact with ground. For this wheel the r value to use to find omega is 0.15 meters (distance from IC to C). To find the speed of A, you use 0.3 meters (distance from IC to A).
However, I am not sure that I am answering the question that you are wanting to ask since you began asking about acceleration. If I am not answering your question, let me know.
--to mburr-- The acceleration of A is NOT the same as the acceleration of C. Use the acceleration equation between the IC of this wheel to C to find alpha (note that the acceleration of the IC for this wheel points in the y-direction). Then use an acceleration equation between C and A to find a_A.
--to tommy woroszylo-- You are almost there. With the acceleration equation that you give, you can solve for the two unknowns of a_IC and alpha (note that a_IC has only a y-component).
--to scott n-- See my comment above to tommy woroszylo. That should lead you to the answer. If not, let us know.
--to mburr-- You need to use the vector from the IC to A: this will be 0.3*j.
--to kgjohnso-- All that you know about the acceleration of the no-slip contact point here is that its horizontal component is zero. There will be a non-zero vertical component as we talked about in class.
--to mburr-- Yes!
--to waltgrove-- See tommy woroszylo's comment and my response above.
--to boilerup-- a_C and a_IC are not equal. a_C is to the right (+x-direction). a_IC is vertical (+j-direction).
--to l. bashover-- The horizontal components of a_A and a_B are the same (since the cable does not stretch). The vertical components might not be the same.
--to kglohnso-- See comment above to l. bashover.
--to scott n-- First relate the accelerations of the IC and point C. See comment to tommy woroszylo above.
--to elaine-- See comment to l. bashover above.
--to peter griffen-- The fundamental point is that v_A = v_B. Use this to relate the omegas for the two wheels. (It turns out that the omegas are also equal; however, this is true for the one example and not true in general.)
--to andrew crandall-- You need to relate a_B to a_F using the x-component of the acceleration equation. This will give you alpha for the wheel on the left.
You are correct in doubting the validity of saying that a_C equals a_O -- they are not the same.
--to w_groff-- From the acceleration equation, you know: a_A = a_C + alpha_AC x r_A/C -omega_AC^2*r_A/C From this you can see that a_A cannot equal a_C. Although the distance between these two points is constant, there accelerations are NOT the same.
To all: Let me know if I missed your questions with my answers.
I have found alpha for the body on the left but now don't know how to relate a_B to a_D. I have too many unknowns (ie. a_F & a_B). Does anyone have any suggestions?
--to scott-- Sorry for the delay in responding to you comment. I did not have time to check the blog before class today.
Knowing alpha for the left wheel and the x-component of a_B, you can write an acceleration equation between O and B in order to find the y-component of a_B. Then you would be in a position to write an acceleration equation between B and D to get a_D.
22 comments:
Good question (and nice artwork!).
The wheel is rolling without slipping, with point F in contact with a stationary surface. Therefore, F has zero velocity (hence, the IC for the wheel).
The wheel on the left moves as a single rigid body (a single angular velocity and a single angular acceleration for the body).
G is just "another point on the wheel".
Is the velocity of O equal to the velocity of C?
It turns out that v_O is NOT equal to v_C. What you need in order to relate the motion of the two wheels is that v_B = v_A. (From this, you can see that v_O < v_C.)
I know you cannot use instant centers for the radius in the acceleration equation. What do you use for r after you find omega of the first wheel?
Or is acceleration of A equal to acceleration of C?
I found omega for both of the circles, but am having trouble finding values for acceleration.
Right now I have Ac = Aic + (alpha x r) - (omega^2*r). I know r, omega, and Ac. After I substitute those in I still need to find Aic and alpha.
Any suggestions?
I am having a hard time finding the angular acceleration of the small wheel. I've found the angular velocity to be 2.66. I wrote an equation for each relating A_a-->A_c, then
A_c-->A_instant center, then
A_a-->A_instant center. But when i subsitute 2 equations into the third, i get the angular acceleration to be 0. Any Help?
tom what r did you use?
Is the acceleration of the non-slipping point of wheel C equal 0?
no it does not equal zero
How do you solve the acceleration equation if the IC doesn't equal 0.
scott- I believe that A_c and A_instant center are the same point, so that makes sense that the angular acceleration becomes zero. i believe that you need to relate A-a to A-b and A_b to A_d and then you should get a system of equations to solve.
I still don't understand how to relate the acceleration of point A to that of point B. Could someone explain this?
At the instant, isn't Aa = Ab?
To find A_a, i need the angular acceleration of the wheel on the right. This i still can't get.
Is the acceleration of point A same as point B?
So would I be correct in saying that wa = wc (yes those "w"'s are omegas)
The way I worked it out was the V_A = V_B and a_A i = a_B i. That is, only the i components of the accelerations of A and B are equal. The perpendicular components are different because the wheel's have different radii. I was able to get a complete expression for a_B, however I don't know any other accelerations on that left wheel....so I have too many unknowns and can't solve for alpha, and therefore the acceleration of D. My hunch is that a_O=a_C, but I'm not sure why since their velocities aren't equal. Please correct me if anything I said was incorrect because I've kind of hit a wall.
I was assuming that a_A = a_O because both are only moving in the +i direction. Can anyone validate that the distance x between pt. A and pt. O is not changing. This would at least narrow down possibilities for solving the problem.
We have had quite a few comments posted on this since I last checked. I will respond to all of the new comments here:
--to mburr--
As stated above, the IC for the wheel on the left is at point F; thus, the r that you use for omega for this wheel is 0.3 meters (distance from F to B). The IC for the wheel on the right is at its contact with ground. For this wheel the r value to use to find omega is 0.15 meters (distance from IC to C). To find the speed of A, you use 0.3 meters (distance from IC to A).
However, I am not sure that I am answering the question that you are wanting to ask since you began asking about acceleration. If I am not answering your question, let me know.
--to mburr--
The acceleration of A is NOT the same as the acceleration of C. Use the acceleration equation between the IC of this wheel to C to find alpha (note that the acceleration of the IC for this wheel points in the y-direction). Then use an acceleration equation between C and A to find a_A.
--to tommy woroszylo--
You are almost there. With the acceleration equation that you give, you can solve for the two unknowns of a_IC and alpha (note that a_IC has only a y-component).
--to scott n--
See my comment above to tommy woroszylo. That should lead you to the answer. If not, let us know.
--to mburr--
You need to use the vector from the IC to A: this will be 0.3*j.
--to kgjohnso--
All that you know about the acceleration of the no-slip contact point here is that its horizontal component is zero. There will be a non-zero vertical component as we talked about in class.
--to mburr--
Yes!
--to waltgrove--
See tommy woroszylo's comment and my response above.
--to boilerup--
a_C and a_IC are not equal. a_C is to the right (+x-direction). a_IC is vertical (+j-direction).
--to l. bashover--
The horizontal components of a_A and a_B are the same (since the cable does not stretch). The vertical components might not be the same.
--to kglohnso--
See comment above to l. bashover.
--to scott n--
First relate the accelerations of the IC and point C. See comment to tommy woroszylo above.
--to elaine--
See comment to l. bashover above.
--to peter griffen--
The fundamental point is that v_A = v_B. Use this to relate the omegas for the two wheels. (It turns out that the omegas are also equal; however, this is true for the one example and not true in general.)
--to andrew crandall--
You need to relate a_B to a_F using the x-component of the acceleration equation. This will give you alpha for the wheel on the left.
You are correct in doubting the validity of saying that a_C equals a_O -- they are not the same.
--to w_groff--
From the acceleration equation, you know:
a_A = a_C + alpha_AC x r_A/C -omega_AC^2*r_A/C
From this you can see that a_A cannot equal a_C.
Although the distance between these two points is constant, there accelerations are NOT the same.
To all: Let me know if I missed your questions with my answers.
I have found alpha for the body on the left but now don't know how to relate a_B to a_D. I have too many unknowns (ie. a_F & a_B). Does anyone have any suggestions?
--to scott--
Sorry for the delay in responding to you comment. I did not have time to check the blog before class today.
Knowing alpha for the left wheel and the x-component of a_B, you can write an acceleration equation between O and B in order to find the y-component of a_B. Then you would be in a position to write an acceleration equation between B and D to get a_D.
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