p0342 - 388 CHAPTER 3. EFFECTS OF GRAVITY ON PRESSURE 3.42...

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Unformatted text preview: 388 CHAPTER 3. EFFECTS OF GRAVITY ON PRESSURE 3.42 Chapter 3, Problem 42 Problem: Use the parallel-axis theorem to determine the moment of inertia about the point at which the square and circle touch. .. ........ ....... . . .. . .. . . . . . . h . . . . . . .. . .. . . . .. . .. . . . . . . . . 2R . . . . . . . . .. . .. . . h .. ........ ....... ............................................................ . . .................................... ... . . ... ... ... . .. . . ....... .... .................. ............................................................... . . ............................................................... . . ............................................................... . . .. ........................................................... . . ................................ ...... . ............................................................. . ......... ... .... ............. . . ............................................................... .................. .... .... ...... . ............................................................... ............................................................... . . . . . . .... .. . ............................................................ . .............................................. ................................................. .. . . ........ . ............................................................ . ......... . ......... ......................... .. .............................................................. . . ............................................................... . . ............................................................... ............................................................... . . . ............................................................... . . ......... ........................... .. ......... .... .................. .. ............................ .... ...... . . . . . . ............................................................... . . . . . ...... . ......... . ............................................................. . . .............................................. .. . . ... ................ . ..... .................................... .. . ................................................. .. . ... . . . .... . ......... .. .. . ........... ... .................................... ....... ................................................... ......................................................... ........................................................ . . .. .. ......................................................... .. .. . . .. ............................................................... . ................................................................... ............................................................ . .. ................................................. . ............................................................. . .. .................................................... ..... . ..................................................................... . . . .. . ......................................................... . . ............ ........................... ... ....................................................... ........ .... ........ . .. .... ................... .. .. .................................................................... .. . ................................................................. ............................................................... .................................................. . ......................................................... .................................................... .......... .. .............................................. ...... ...................................... ....... ... ....... ........................ ... .............. .......... . .. .. . . . . . . . . . . .. .. . . . .. .. . . . . . . . . . . . .. .. . . h/2 h= √ πR R ..... Solution: As shown in the figure, the centroid of the circle is R below the common vertex, while the centroid of the square is h/2 above it. The moment of inertia of the square and circle, according to the parallel-axis theorem, is equal to the sum of the moment of inertia of the circle shifted by R and the moment of inertia of the square shifted by h/2. So, first note that for the circle, the area and moment of inertia are Ac = π R2 Also, for the square, since h = √ and 1 Ic = π R4 4 π R, we have As = h2 = π R2 and Is = 1 4 π2 4 h= R 12 12 Hence, using the parallel-axis theorem for the circle, 1 5 Ioc = Ic + z 2 Ac = π R4 + R2 · π R2 = π R4 c 4 4 Similarly, for the square, Ios = Is + z 2 As = s π2 4 π 2 1 R + R (π R2 ) = π 2 R4 12 4 3 This, summing these results, the moment of inertia for the square and circle is I= π (15 + 4π) R4 12 ...
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

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