# Direction is directly in the center of the two

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direction is directly in the center of the two vertical rods because the object has a mirror flip symmetry about this central verti- cal axis. However, this symmetry does not exist for the ˆ y direction because there is no top rod. So the mass to creep closer to the bottom center of mass in the ˆ y will creep closer the bottom of the configuration. Because point 3 is the center of the object, our choices are then 2 and 1. We can quickly see that the weighted average of the masses below point 1 will not equal the weighted average of the points above, there is simply far more mass above the point than below it. So this leaves only choice 2. 3

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2. The figure shows a cylinder of uniform mass density and three possible axes of rotation, labeled “1,” “2” and “3,” that go through the center of the cylinder. Assume the length of the cylinder is much larger than its radius. Axes “1” and “3” are in the plane of the page, while axis “2” is perpendicular to the page. 2 1 3 Rank order the moments of inertia I 1 , I 2 and I 3 about the three axes 1, 2 and 3, respectively, smallest to largest. (a) I 1 I 2 I 3 (b) I 3 I 2 I 1 (c) I 1 I 3 I 2 (d) I 2 I 1 I 3 (e) I 1 I 2 I 3 Imagine you are given this rod in real life with the axes attached, but not labeled, and you are asked to find the moments of inertia. If you were to look at this diagram and try to label the axes, you could label the axis 1, but which axis was 2 and which was 3 would be impossible to determine because simply spinning the object by 90 degrees turns axis 2 into axis 3 but leaves the object in an identical configuration otherwise. This symmetry means that the moment of inertia through axis 2 and 3 must be identical because if you got an answer for the moment of inertia about axis 3 and then rotated the object 90 degrees, you would have to get the same moment of inertia for 2 because you would have to do the exact same calculation. Thus we can say that I 2 I 3 . We also know that I m i r 2 i , where r i is the radius from the axis of rotation. Because the object is much longer than it is thick, the r 2 i terms about axis 2 and 3 will be much larger than the r 2 i about axis 1 for the majority of the points of mass in the object. So if the length 4
is much greater than the width, it is safe to assume the sum of these terms should also be greater. So we choose I 1 I 2 I 3 . 3. A toy bicycle has two wheels and a frame, each of mass M (so the total mass of the bicycle is 3 M ). The front wheel has smaller radius than the back wheel and the bicycle is moving with speed V with the wheels rolling without slipping. Rank order, from smallest to largest, the kinetic energies K f of the front wheel, K b of the back wheel and K r of the frame (“ r ” is for the “rest”).

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