Write mathematical expression of M.I.of a triangle about horizontal axis passing through its apex. ​

ZakjexZK ZakjexZK Best Answer 1 year ago

Moment of inertia of a triangle can be expressed in various ways. There are usually three MOMENTS that can be considered. They are;Axis passing through the centroid.Axis passing through the base.The axis perpendicular to its base.We will look at each expression below.1. Axis passing through the centroidThe moment of inertia of a triangle having its axis passing through the centroid and parallel to its base is expressed as;I = bh3 / 36Here, b = base width and h = HEIGHT2. Axis passing through the baseIf we TAKE the axis that passes through the base, the moment of inertia of a triangle is given as;I = bh3 / 12We can further use the parallel axis theorem to prove the expression where the triangle centroid is located or found at a distance equal to h/3 from the base.3. The axis perpendicular to its baseWhen we want to determine the moment of inertia of a triangle when its axis is perpendicular to its base we have to first consider that axis y’-y’ is used in dividing the whole triangle into two right triangles respectively A and B. However, these triangles will have a common base equal to h, and heights b1 and b2. The moment of inertia for both will be:Iy’ = hb13 / 12 + hb13 / 12If we consider b2 = b – b1 where the parallel axis y-y through the centroid is at a distance ⅔ ( b2 / 2 – b1) from y’-y’ then we can easily find or CALCULATE the moment of inertia ly. We can use the parallel axis theorem to do so. In any case, after algebraic substitutions, we get the expression as:Iy’ = hb / 36 (b2 – b1 b + b12)Moment Of Inertia Of A TriangleCalculating Moment Of Inertia Of A TriangleWe will take the case where we have to determine the moment of inertia about the centroid y. We will consider the moment of inertia y about the x-axis.Moment Of Inertia Of A TriangleWe will use the parallel axis theorem and we will take the centroid as a reference in this case.Here,Iy = ab3 / 12ỹ = b / 3A = ab / 2Iy = ỹ 2 A + Iỹab3 / 12 = (b / 3)2 (ab / 2) + IỹIỹ = (b / 3)2 (ab / 2) – ab3 / 12Iỹ = ab3 / 36Next, we will find the moment of inertia when the axis passes through its base.Moment Of Inertia Of A TriangleWe will consider the differential strip which is parallel to the x-axis for da. It is given as;Dlx = y2 dAdA = l dyIf we take similar triangles we will have,l / b = h – y / hl = b (h – y / h)dA = b (h – y / h) dyNow we will integrate dlx from y = 0 to y = h.Ix = ∫ y2 dAIx = O ∫h y2 b (h – y / h) dyIx = b/h O ∫h (hy2 – y3) dyIx = b/h [h (y3/ 3 + y4 / 4) ]0hTherefore, Ix = bh3 / 12⇒ Check Other Object’s Moment of Inertia:Moment Of Inertia Of An EllipseMoment Of Inertia Of A ConeMoment Of Inertia Of A SquareMoment Of Inertia Of SemicircleMoment Of Inertia Of Solid ConeEnroll for a FREE 60 minute demo class todayJEE Preparation 2020NEET Preparation 2020Join BYJU'S Learning ProgramNameMobile NumberCity / TownEmail AddressCOURSESCBSEICSECATIASJEENEETCommerceBank ExamNCERTEXAMSCAT ExamIAS ExamUPSC SyllabusUPSC 2020Government ExamsJEE MainRESOURCESBlogVideosCBSE Sample PapersCBSE Question PapersDSSLEXAM PREPARATIONFree CAT PrepFree IAS PrepMathsPhysicsChemistryBiologyCOMPANYAbout UsContact UsStudent FeedbackInvestorsCareersBYJU'S in MediaStudents Stories - The Learning TreeFaces of BYJU'S – Life at BYJU'SSocial Initiative - Education for AllBYJU'S APPFOLLOW USFree Textbook SolutionsNCERT SolutionsNCERT ExemplarNCERT Solutions for Class 6NCERT Solutions for Class 7NCERT Solutions for Class 8NCERT Solutions for Class 9NCERT Solutions for Class 10NCERT Solutions for Class 11NCERT Solutions for Class 12RD Sharma SolutionsRS Aggarwal SolutionsICSE Selina SolutionsState BoardsMaharashtraGujaratTamil NaduKarnatakaKeralaAndhra PradeshTelanganaUttar PradeshBiharRajasthanMadhya PradeshDisclaimer | Privacy Policy | Terms of Services | Sitemap© 2020, BYJU'S. All rights reserved.

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