Physics II workshop_Part_52

Physics II workshop_Part_52 - x horizontal and y vertical....

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106 Center of Mass of 2-D and 3-D Objects Sometimes we cannot compute the center of mass of a system by treating its components as point masses. In these cases, we cannot avoid calculus. We can still write the center of mass vector (in 2-d) as r R CoM = X CoM ˆ i + Y CoM ˆ j but now the x - and y - components of the center of mass are given by integrals over volume: X CoM = 1 M x ρ dV and Y CoM = 1 M y dV where M = total mass of the object, = density. For regular shapes it is usually possible to identify a plane of symmetry that allows you to convert the integral over volume to one over just a single co-ordinate. Uniform triangle The triangle has a uniform density . (a) Draw the axis of symmetry for the triangle. (b) Where should you place your origin of co-ordinates? Draw co-ordinate axes with
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Unformatted text preview: x horizontal and y vertical. (c) Use symmetry to find the y co-ordinate of the center of mass. What is it? L H 107 To find the x-co-ordinate: (d) You only need to find the X CoM for one half of the triangle. Explain why. (e) What is the total mass of one of these right triangles, in terms of , L and H ? (f) Write down the equation for y ( x ) for the top right triangle. (g) For this 2-d object, the X CoM co-ordinate is given by an integral over area. Divide the top right triangle into small vertical strips of width dx . Write down an equation for the area, dA , of these strips. (h) Now use your answers to (d) and (f) to solve the integral X CoM = 1 M x dA ....
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This note was uploaded on 11/26/2011 for the course PHY 2053 taught by Professor Lind during the Fall '09 term at FSU.

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Physics II workshop_Part_52 - x horizontal and y vertical....

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