121616949-math.216 - function σ x = x-19 Find the center...

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202 Chapter 9 Applications of Integration The numerator is the moment of the system: Z 10 0 x (1 + x ) dx = Z 10 0 x + x 2 dx = 1150 3 , and the denominator is the mass of the beam: Z 10 0 (1 + x ) dx = 60 , and the balance point, officially called the center of mass, is ¯ x = 1150 3 1 60 = 115 18 6 . 39 . It should be apparent that there was nothing special about the density function σ ( x ) = 1 + x or the length of the beam, or even that the left end of the beam is at the origin. In general, if the density of the beam is σ ( x ) and the beam covers the interval [ a, b ], the moment of the beam around zero is M 0 = Z b a ( x ) dx and the total mass of the beam is M = Z b a σ ( x ) dx and the center of mass is at ¯ x = M 0 M . EXAMPLE 9.19 Suppose a beam lies on the
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Unformatted text preview: function σ ( x ) = x-19. Find the center of mass. This is the same as the previous example except that the beam has been moved. Note that the density at the left end is 20-19 = 1 and at the right end is 30-19 = 11, as before. Hence the center of mass must be at approximately 20 + 6 . 39 = 26 . 39. Let’s see how the calculation works out. M = Z 30 20 x ( x-19) dx = Z 30 20 x 2-19 x dx = x 3 3-19 x 2 2 fl fl fl fl 30 20 = 4750 3 M = Z 30 20 x-19 dx = x 2 2-19 x fl fl fl fl 30 20 = 60 M M = 4750 3 1 60 = 475 18 ≈ 26 . 39 ....
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