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Jul 15, 2008

Problem 6/101

Some thoughts on this problem:

FBD:  as shown above

Newton/Euler:
sum F_x = -T*cos(theta/2) = m*a_Gx
sum F_y = -T*sin(theta/2) - mg + N = m*a_Gy
sum M_O = -N*(L/2)*cos(theta) + [T*cos(theta/2)]*[(L/2)*sin(theta)] - [T*sin(theta/2)]*[(L/2)*cos(theta)] = I_G*alpha

Three equations. Five unknowns (T, N, a_Gx, a_Gy, alpha).


Kinematics:
Need to develop a NET of two additional equations.
First, relate motion of points A and B. A moves on a straight, horizontal path. Therefore, a_A is in the x-direction. B moves on a circular path centered at point C. Therefore, since v_B = 0, we know that a_B is perpendicular to cable BC. For this, use the following vector equation:

a_B = a_A + alpha x r_B/A

This gives us two additional scalar equations but also introduces two additional scalar unknowns: a_A and a_B.

Now, relate the acceleration of G to alpha. For this, use the following vector equation:

a_G = a_A + alpha x r_G/A

This gives us two additional scalar equations with no additional unknowns. Therefore, we have a total of seven equations with seven unknowns. You can now solve.

Let us know if you have any questions.



5 comments:

Anonymous said...

in the definition of your theta, some how angle(ABC)=angle(CB)=theta/2. However, using the parallel axis theorem suggests that angle(CB)should equal theta not theta/2.
In summing F_x and F_y i believe u used theta/2 instead of using theta?

CMK said...

Here is my line of reasoning on the angles:

* Angle CAB is the supplement of theta. Therefore, angle CAB = 180 - theta.

* Angle BCA = angle CBA since x = L (opposite angles of equal sides are equal).

* Since the angles of triangle ABC must add up to 180 and since angle BCA = angle CBA, we therefore have angle BCA = angle CBA = theta/2, as shown in the figure.

* BC makes an angle of theta/2 with the horizontal (angle BCA = theta/2), as shown in the figurer. [Again, it is AB that is at an angle of theta with the horizontal, not BC.]


Let me know if you see the flaw in my reasoning.

Kyle said...

how many points do we stand to lose if we get lost in the algebra of solving for the 7 unknowns?

CMK said...

As we talked about in class today, setting up these problems is the most important part of the solution.

For the exam, solving the equations is worth only about 2 points out of 20.

For this homework problem, you will get FULL credit if you set up the equations correctly and clearly indicate the number of equations and number of unknowns to be found.

Danny said...

In regards to the question of the angle, I used the laws of cosines and law of sines to get an angle of theta/2 as well.