Homework Problem No. 6/110

Nov 5, 2008

Homework Problem No. 6/110

Suggestions:

FBD
Newton/Euler: Use summation of forces in both x- and y-directions. Use the center of mass G for Euler's equation. [Note that the center of mass is NOT at the geometric center O of the wheel. See comment below.]
Kinematics: Point C is the no-slip contact point for the wheel. Therefore, the x-component of aC is zero; however, the y-component is NOT zero. Suggest that you do the kinematics in two steps. First, use: aO = aC + alpha x rO/C - omega^2*rO/C. This equation will give you the y-component of acceleration of C. Second, use: aG = aC + alpha x rG/C - omega^2*rG/C. This equation will give you the acceleration of G in terms of the known omega and the unknown alpha.
Solve: Substitute the kinematics from 3. into your three Newton/Euler equations of 2.

Comment: As discussed above, aC has only a y-component. This y-component is NOT zero and NOT parallel to the vector rG/C. Therefore, you cannot use C as you summation of moments point in the Euler equation (you are stuck with point G).

This is a very good problem. It helps make the points that you need to be careful with your choice of point for Euler's equation and with your kinematics. This is the problem that we will collect on Friday ...

Let us know if you have questions on this.

10 comments:

garrettmocas said...: For my moment equation, I used the radius of G as r and the radius of the whole wheel as R, and go the equation: N*R+f*r=iG*alpha+m*rG/O x aG

Does that seem correct?; November 6, 2008 at 9:21 AM
rpmccaul said...: I think N acts at the 40mm, which i think your calling r not R, so you should switch your R's.

does that make sense? Force N acts along Y-axis 40mm away from G, so the moment is N*r; November 6, 2008 at 11:01 AM
Dale Szul said...: I am unable to find the answer, and my thought process has been this thus far:

(1) sum_Fx = F = m*a_Gx (F is friction)

(2) sum_Fy = N - mg = m*a_Gy

(3) sum_MG = N(.04) + F(.1) = I_G*alpha

For kinematics I used G/O, O/C and C/G to find relations between a_Gx, a_Gy and alpha. With these relations I solved for alpha using (3) and then (1) and (2) to find the unknown N and F. Is this process correct?; November 6, 2008 at 4:05 PM
Byron said...: Your equation 3) is off due to the fact that there is an acceleration at G. So your eq 3 should be

(3) sum_MG = N(.04) + F(.1) = I_G*alpha + m*rG/O x aG

If you follow the kinematics steps that the professor outlined, you should get the right overall answer.; November 6, 2008 at 4:10 PM
Adam Watson said...: byron,
there is an acceleration in G, but it's still just I*alpha because since you're summing the moments about G and not the moments about O, you would use rG/G, not rG/O, which is zero. The r is from whatever point you're summing the moment about to the center of gravity G, so if you're doing it at G, rG/G is zero.

Did you use rG/O and get the right answer?; November 6, 2008 at 6:10 PM
Adam Watson said...: nvm, it's odd, so the answers not in the book. I'm pretty sure what I said is still right though.; November 6, 2008 at 6:12 PM
Byron said...: Ah, that's right, thanks. I'm used to G being the center of the uniform mass.; November 6, 2008 at 6:51 PM
CMK said...: FYI, the answers for this problem are:
N = normal force = 87.7 newtons
f = friction force = 24.4 newtons (to left); November 6, 2008 at 7:43 PM
gbaker said...: For the FBD on this problem, it has a counter clockwise moment. Why would this not produce a friction force in the -x direction, opposite of what you have shown; November 25, 2008 at 1:28 PM
CMK said...: The direction of the friction force changes as the disk rolls; half of the time it acts to the left and the other half it acts to the right (see the animation for the problem that is posted on the blog).

Since we never set friction = mu*N, we can arbitrarily draw in the direction of friction. The mathematics will take care of the sign.

Good question.; November 25, 2008 at 5:05 PM