How can i know that i should take the formula (1/3) m l^2 not (1/12)ml^2 ? I don

Question

How can i know that i should take the formula (1/3) m l^2 not (1/12)ml^2 ? I dont understand how to determine that could anyone explain this for me please in a simple way please dont make it more complicated 17.1 MASS MOMENT OF INERTIA EXAMPLE 17.4 of two thin rods, each having a weight of 10 lb. Determine the moment f inertia of the pendulum about an axis passing through (a) point O, and (b) the mass center G of the pendulum. The pendulumin Fig 17-7is suspended from the pin at O and consists 2 t SOLUTION Part (a). Using the table on the inside back cover, the moment of inertia of rod OA about an axis perpendicular to the page and passing tbrough point O of the rod is lo mf Hence Fig. 17-7 (halo- me-K3210fbsì)(2ft 2-0.414 slug-ft2 This same value can be obtained using lG1 t parallel- 3332.2 ft/s axis theorem. 10 lb 1ob322f 0.414 slug-ft2 For rod BC we have 10 lb 2ft 120 (22)Jen (10 lb -1346 slug.ft le moment of inertia of the pendulum about O is therefore lo 0.414+1.346 1.76 slug f2 Ans The mass center G will be located relative to point O distance to be y, Fig, 17-7, and using the formula for ng the mass center, we have yym 110/322)+2(10/32.2) (10/32.2) + (10/32.2) (10/32.2)150ft ich t of inertia Ig may be found in the same manner as lo shich requires es successive applications of the parallel-axis theorem to oments of inertia of rods OA and BC to G. A more however, involves using the result for lo. i.e., 3285 0 501) 20 lb a.so fto 21.50t)

Explanation / Answer

(1/3) m l^2 will be used if a rod is hinged or supported at one end or if we want to calculate the moment of inertia about an end.However the basic formula for moment of inertia of a rod about center of gravity is IG=(1/12)ml^2.

By using the formula (1/12)ml^2 and parallel axis theorm we can find moment of inertia at any location on the rod, so use the basic formula always.

Hope it is clear .

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