- #1
Sam_Goldberg
- 46
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Homework Statement
Hi guys, this question is from Kleppner and Kolenkow's mechanics book, problem 7.2:
A flywheel of moment of inertia I rotates with angular velocity w at the middle of an axle of length 2L. Each end of the axle is attached to a support by a spring which is stretched to length L and provides tension T. You may assume that T remains constant for small displacements of the axle. The supports are fixed to a table which rotates at constant angular velocity, W, where W << w. The center of mass of the flywheel is directly over the center of rotation of the table. Neglect gravity and assume that the motion is completely uniform so that nutational effects are absent. The problem is to find the direction of the axle with respect to a straight line between the supports.
I'm sorry that I could not provide a picture, but you can find one by searching inside on amazon. The page is 334.
Homework Equations
The Attempt at a Solution
Okay, here goes. There will be three axes. The z axis points to the sky; the y-axis points along the springs; and the x-axis is perpendicular to the springs. All axes meet at the center of the flywheel. I will describe the orientation of the axle which holds the flywheel by two angles, A and B. If A changes wrt time, there is an angular velocity in the x direction, and if B changes, in the z direction. I am not completely sure about this, but from the problem statement, it seems as if the angles A and B are small, that is, they can be dealt with in the first order.
I will now find the torques about the x and z axes and relate them to the changes in angular momentum wrt time. For the x axis:
Torque(x axis) = - 2Lsin(A)T = -2LTA
The change in angular momentum wrt time comes from three sources. First, the flywheel itself can rotate about the x axis, it produces (1/2)IA'', since the moment of inertia about a perpendicular axis is (1/2)I. Second, the angular momentum from the flywheel is a vector, and it can produce an x component due to the rotation of the table at angular velocity W. The value of this is IwW. Finally, the rotation of the axle itself about the z axis changes the vector component; this contribues IwB'. So, we have:
(1/2)IA'' + IwW + IwB' = -2LTA
The torque about the z axis is similar, except for a sign difference for a term:
(1/2)IB'' - IwA' = -2LTB
We have two coupled differential equations, which presents a problem. I could simplify these equations by eliminating a term(s), but I'm not sure which one(s) is too small to be considered. I know that W << w, but that's about it.
Perhaps I made a mistake in coming up with these equations. Let me know if that's the case as well. Thanks in advance.