Mar 25, 2008

Homework Hint: Problem 6/110


  • FBD: normal and friction forces (N and f) at C; weight at G (NOT at O)
  • Newton-Euler: Summation of forces in x- and y-directions. Suggest summation of moments about the cm G. [Note that if you choose to sum moments about C, you need the FULL form of the RHS of Euler's equation: I_C*alpha + m*(r_G/C) x (a_C). The last term does NOT vanish since neither a_C is zero nor is (r_G/C) parallel to (a_C).]
  • Kinematics: Write down the kinematics equation relating the acceleration of G to the acceleration of C (a_C points toward O). Also need an additional kinematics equation relating acceleration of G to the acceleration of O (a_O is in the x-direction)
  • Solve: You have seven equations in terms of seven unknowns (N, f, a_Gx, a_Gy, alpha, a_O, a_C). [7 equations/7 unknowns sounds worse than it actually is -- some of the resulting kinematics equations are pretty simple.]
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10 comments:

  1. mheidaMarch 25, 2008 at 2:44 PM

    I have come up with 5 equations: 3 newton-euler and the 2 kinematic equations listed. What are the other 2? Also, would the omega and alpha values be the same for bot of the kinematic equations i have so far?

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  2. Purdue FSAEMarch 25, 2008 at 3:18 PM

    I am also having a lot of trouble with this one. I have 7 unknowns, but only 5 equations. Where are the other two coming from??

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  3. n.c.kempfertMarch 25, 2008 at 4:54 PM

    This comment has been removed by the author.

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  4. n.c.kempfertMarch 25, 2008 at 4:55 PM

    Even though Dr. Krousgrill suggests summing moments about G for this problem, I found it a lot simpler to sum them about C as we did in class on Monday with the spool problem. It's true that the last couple of terms don't go away as simply as they did for the one in class (due to the funky imbalance of this thing), but I think that doing so reduces some of the work in the kinematics department.

    Here's what I did:

    - Sum forces in x and y as normal, that's simple.

    - Sum moments about C (point of contact) including the m*r_G/C crossed with the acceleration of C (a_C). Also, don't forget to use the parallel axis theorem to get the correct Ic.

    ^^Doing all of this should get you alpha.

    - Kinematics: just relate the acceleration of G to that of C. If you assume no slip, then acc of C is only in the j-direction, regardless of the ever-changing motion of this thing. Furthermore, because you know the angular velocity of the whole body at this moment, finding the upward acceleration of C should be a breeze. Also, they were kind enough to give all of the needed dimensions for your r_G/C vector.

    ^^Doing all of this should net you the acceleration of G.

    - Once you have the acceleration of G, you are in good shape. If your force summations in the x and y look like mine, then that should be all you need.

    Hope this helps!

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  5. CMKMarch 25, 2008 at 6:22 PM

    --to mheida and amccallu--

    Keep in mind that the two kinematics equations are planar VECTOR equations; each vector equation produces two scalar equations (x- and y-components) for a total of 4 scalar equations. Combined with the 3 Newton-Euler equations, you have 7 equations total.

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  6. CMKMarch 25, 2008 at 6:28 PM

    I really like n.c.kempfert's solution method. He has skillfully used the Euler equation in order to easily find the angular acceleration of the wheel. Good job!

    If you choose to use his approach, give careful attention to the details that he has provided. Please add comments to this post if you have questions on the details of his method.

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  7. rhardistMarch 25, 2008 at 6:54 PM

    I'm having trouble deciding just by looking.. Should there be terms for (a_G/C)rel and (v_G/C)rel? I can see how there would be since technically point C is moving with respect to G, but at the same time I can see how there wouldn't be since they're both on the same rigid body.. I will try both I suppose unless someone responds before I finish trying this out.

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  8. n.c.kempfertMarch 25, 2008 at 7:06 PM

    No, you don't have to use the relative velocity and acceleration terms, but I think they would both be zero anyway.

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  9. ME BoilerMarch 26, 2008 at 12:42 AM

    n.c.kempfert,

    when I try and do this step:
    - Sum moments about C (point of contact) including the m*r_G/C crossed with the acceleration of C (a_C). Also, don't forget to use the parallel axis theorem to get the correct Ic.

    ^^Doing all of this should get you alpha.

    I get alpha=(3.924-.4a(c))/.041 Is that right? should alpha be in terms of a(c)? This makes is extremely messy when I try and relate a_G to a_C. I know a_c is only in the y direction but is there a way to calculate it?

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  10. CMKMarch 26, 2008 at 1:15 AM

    I'll jump in here since NCK might have already retired for the evening (and maybe I should too).

    If you write the acceleration of C (only in y) in terms of the acceleration of O (only in x):

    a_C = a_O + alpha x r_C/O - omega^2*r_C/O

    and equate "j" components, you get:

    a_C = + R*omega^2 = (0.1)*(2)^2 = 0.4 m/sec^2

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