Moment of Inertia of an Equilateral Triangle with Pivot at one Vertex

Transcription

1 oment of nertia of an Equilateral Triangle with Pivot at one Vertex There are two wa (at leat) to derive the expreion f an equilateral triangle that i rotated about one vertex, and ll how ou both here. One wa i direct: divide the triangle into hizontal trip, write the expreion f the moment of inertia of one trip, and integrate the expreion to get the um of moment of inertia of all the trip (i.e. the entire triangle.) The econd wa i ver clever and i much hter. Sometime a bit of thinking can help u ee a hter wa to olve a problem. The onl danger i that a hter route to a olution tpicall involve ome logic, and if one tep in the logic i fault the entire olution end up increct. Ftunatel, the logic we need f thi problem i prett imple. And now, without further ado... The long, direct wa (which wk f an ituation): Conider the thin hizontal trip in the triangle below. The trip ha width w and height d. ve defined the -axi to go from 0 at the top of the triangle (at the pivot) to at the bottom of the triangle. So the ditance from the pivot to the center of the trip i impl. Note that each hizontal trip of the triangle will have a different and a different width w.

2 ve alo defined the ma of the trip a dm. We can connect d and dm b recognizing that: ma of trip / ma of triangle area of trip / area of triangle dm / w d / (/) (Note that the bae of the triangle i, the length of the ide.) w d dm We will ue thi below when we write the expreion f the moment of inertia of the trip. We can alo make two me connection (which we will need later): Connect to Thi i a imple matter of uing either the left half right half of the big triangle. The lower angle in each cner i 60 degree, o: in60 O / Connect w to Thi one i alo imple: ue a triangle created b and either the left right half of the trip. The angle in the upper cner i 0 degree, o: tan0 O (w/) / w Notice that if ou ue thee two expreion together, when, w equal... jut a it hould (i.e. the width of the triangle at the bottom i the length of the ide.) Now we can write the moment of inertia of the trip a it i rotated about the pivot at the top. Uing the parallel axi theem (the ame thing we did to derive the moment of inertia f the rectangle in lab): d dm w + dm f we replace dm d w d w with the appropriate expreion: w d + w d + 6 w d

3 And then replace w with it expreion: d d d d d Note that both term include a contant with d. We can implif the expreion with a bit of facting: d d d d Now we can add the d f all the hizontal trip that make up the triangle. Note that the trip tart at 0 and end at, o: d We re almot done! We want to expre the moment of inertia uing, the ma of the triangle, and, the length of the ide of the triangle. Which mean we need to replace in our expreion uing the connection we wrote earlier f and : Or: oment of inertia f an equilateral triangle rotated about one cner And that it! Notice how the final expreion came out ver imilar to the expreion that we ued f the quare in lab. thi coincidence? Ye and no: obvioul triangle and quare hare ome of the ame geometr, and the definition we ued f both are imilar; but remember that the pivot we ued f the quare in lab wa arbitrar (i.e. choe it to be a ditance of / from the center.) So there i ome coincidence involved here. And now the hter, clever wa... The big equilateral triangle can be divided into four maller equilateral triangle, a hown in the figure below. 5

4 The center of ma of each triangle i hown b the blue dot. Note that the center of ma of the central triangle i alo the center of ma of the big triangle. The ide length of each mall triangle i /, a hown. And the ditance from the center of ma of the big triangle to it pivot i d. We can firt write d in term of. We can ue the triangle haded in green on the picture to write that: co0 O (/) / x x Recall that, from the earlier derivation, the height of the triangle i Then we can write that d - x So: d Now f the clever part: the ma of each mall triangle i onl one-fourth the ma of the big triangle. And the ide length of the mall triangle i onl half the ide length of the big triangle. We know that the moment of inertia of the big triangle, if it wa rotated about it center of ma, mut be the ame expreion a that of each mall triangle... but appropriatel caled f it ma and ize dimenion.

5 Since moment of inertia i proptionate to the ma of an object and proptionate to the quare of the linear dimenion, we know that Due to the ma, f the big triangle mut be four time a much a the f the mall triangle and due to the length of ide, f the big triangle mut be four time a much a the f the mall triangle. Thi mean that due to both fact, big 6 mall Thi expreion impl recognize how the ize of the triangle affect moment of inertia. But now we can alo write how the four mall triangle together mut equal the big triangle! But we have to be careful: our mall i the moment of inertia f a mall triangle if it i rotated about it own center of ma. Three of the four mall triangle are not rotated about their own center of ma, o we mut ue the parallel axi theem f thee three: big four mall plu additional f three triangle uing parallel axi theem Recall that the additional f the parallel axi theem i the ma of the object time the ditance from it center of ma to the center of rotation quared. Note that thi ditance in the picture i two time z (the pink line). We can find z in term of, then double it and then ue thi ditance in our expreion f the moment of inertia: z x 6 6 So the ditance we want i twice thi, : And now: big mall + m mall + mall + 6

6 Next, we can replace mall with the connection we developed above: big big mall + + big + 6 Now ubtract ¼ big from each ide, and then olve f big : big 6 big Note that thi i the moment of inertia of the big triangle if it i rotated about it center. Alo note that we didn t have to ue an integral to get it... we impl ued a geometr argument to relate the mall and big triangle. Alo note... thi onl wked becaue we could contruct the large triangle from identical maller triangle! We have one me tep: ince we want to place the pivot f our big triangle at the top cner, we need to ue the parallel axi theem to get the expreion f the moment of inertia about that pivot. pivot big + d pivot + + : oment of inertia f an equilateral triangle rotated about one cner 5 Of coure we arrive at the ame expreion a we did uing the firt method. Thi econd method i ueful f two reaon: it help to confirm that our expreion i probabl crect, ince we derived it two completel different wa and got the ame reult; and it demontrate a different wa (other than the direct wa) of deriving a moment of inertia expreion.