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MAT136H1 Lecture : 8.3 Physics & Engineering Applications Question #4 (Medium)

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ultramarineeel914

22 Mar 2013

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MAT136H1 Full Course Notes

MAT136H1 Full Course Notes

Document Summary

: because about the -axis the distance of is. Question #4 (medium): finding the moment and the center of mass. Moment of the system about the -axis is considered. The -axis shows point-masses located on the -axis. Find the moment of the system about the origin and the center of mass. Number line with points at: with with with. Since the number line only presents one dimensional line with only values to deal with, moment is given by: . The total mass of the system is .

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Convert the above Riemann Sum to an integral by letting delta x rightarrow 0. Since this integral must be zero, we get an equation. Solve the equation for x to show that Suppose that the length of the rod is 5m and that its density is delta (x ) = (0.lx3 + 2) kg/3. Find the center of mass of the rod, and indicate where it is on a diagram. Is your center of mass to the left or right of the geometric center of the rod? Explain why this is so. The same process can be used to compute the center of mass of a 2-D object. The coordinates of the center of mass of such an object will be and . The notation used to refer to the different quantities is (the moment about the y/-axis), (the moment about the x -axis), and . Of course, M is simply the total mass of the object, represents the "vertical" line upon which the solid would balance, and represents the "horizontal" line upon which the solid would balance Therefore, the point (x, y) represents the one point at which the solid would balance perfectly. An exactly analogous set of formulas holds for 3-D objects. Find the center of mass of a lamina defined by the parabola y = 2 - 3x2 and the x-axis if its density is given by delta(x, y ) = x4 y2. For the above region, find the geometric center, and if it differs from the center of mass, explain why this is so and why it makes sense. (Hint: making a judicious choice for delta will ensure that the geometric center and the center of mass are the same point - so finding the center of mass with the above formulas and this choice of delta will actually give you the location of the geometric center of the object.) When engineers design an appliance, one of the concerns is how hard the appliance will be to tip over, when tipped, it will right itself as long as its center of mass lies on the correct side of the fulcrum, the point on which the appliance is riding as it tips. See the figure below. Suppose the profile of an appliance of approximately constant density is parabolic (y = a ( l - x2)), like an old-fashioned radio. What values of a will guarantee that the appliance will have to be tipped more than 45 degree to fall over? Your answer should be exact.