Suggestions:
- FBD
- Work-Energy: Write the kinetic energy in terms of the no-slip contact point (call this point C). Don't forget to use the parallel axis theorem to find the mass moment of inertia about C using the known information of the mass moment of inertia about the center of mass G. The change in potential energy comes about in two parts: from the 12-inch long straight-line path, and from the curved part of the path (this 2nd part requires some thought). Don't forget that the potential energy depends on the vertical position of the center of mass G.
- Kinematics: Use the fact that C is the IC of the wheel to relate the speed of G to the angular speed of the wheel.
- Solve: Find the speed of the center of mass G of the wheel.
Now draw an FBD of the wheel at the bottom position A. From this, sum forces in the y-direction to relate the normal force to the y-component of the acceleration of G (you will use the above result from Step 4 to find the y-component of the acceleration of G).
Let us know if you have any questions on this problem.
10 comments:
When I find the acceleration in the y-direction of G at A, my answer is in the negative-y direction. This in turn makes my answer for the normal force negative. What am I doing wrong? I use this equation to get a_Gy:
a_G=0+(alpha)X(r_G/C)-omega^2*r_G/C
where: omega=v_G/r
That might be too complicated an equation to use, since we do not know alpha. I used the fact that since it is moving on a curve, the only acceleration in the y direction should be v^2 / r, with r being the radius of the turn. This guarantees an acceleration toward the center of the curve.
I like Sheng's suggestion; the path description for point G is the easiest to use.
In response to Erik's question, please note that the acceleration of C is NOT zero; it has a y-component. You can use this equation to find aC once you know aG (which you can get from Sheng's path description).
I am not sure how to find the change in potential energy. I have the datum at the starting position making v1=0 but for v2 all I can get is that the change in height is 6 in for the straight part. I am not sure how to take the curved path into account. Does anyone have any suggestions? I could just be going about this the entirely wrong way. Thanks in advance!
The picture for this homework problem shows the change in height for the curved part to be 18 - 18cos(30).
Where did that come from? And I guess more importantly, how could we get it without that hint?
I added a couple more dimensions to the drawing. Does that help?
I'm having troubles with the kinetic energy. Using point C as the instant center, I get T1=1/2*Ic*w^2, but I don't see how that would change at the bottom of the hill. So is that the right equation to use for T1? and what is different about T2?
Sure, T = (1/2)*IC*omega^2 is valid for BOTH positions. What changes between positions 1 and 2 is the value of omega.
You are given the speed of G at position 1 (from which you can find omega1). You determine omega2 using the work-energy equation. You then find the speed of G at position 2 from omega2.
Does that help?
yes sir, thanks I appreciate it.
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