JUST THE MATHS SLIDES NUMBER 13.12 INTEGRATION APPLICATIONS 12 (Second moments of n re (B)) b A.J.Hobson 13.12.1 The prllel xis theorem 13.12.2 The perpendiculr xis theorem 13.12.3 The rdius of grtion of n re
UNIT 13.12 - INTEGRATION APPLICATIONS 12 SECOND MOMENTS OF AN AREA (B) 13.12.1 THE PARALLEL AXIS THEOREM Let M g denote the second moment of given region, R, bout n xis, g, through its centroid. Let M l denote the second moment of R bout n xis, l, which is prllel to the first xis, in the sme plne s R nd hving perpendiculr distnce of d from the first xis. d h δa l centroid R g We hve M l = R (h + d)2 δa = R (h2 + 2hd + d 2 ) 1
Tht is, M l = R h2 δa + 2d R hδa + d2 R δa = M g + Ad 2 Note: The summtion, R hδa, is the first moment bout the n xis through the centroid nd, therefore, zero. The Prllel Axis Theorem sttes tht EXAMPLES M l = M g + Ad 2. 1. Determine the second moment of rectngulr region bout n xis through its centroid, prllel to one side. Solution b O 2
For rectngulr region with sides of length nd b, the second moment bout the side of length b is 3 b 3. The perpendiculr distnce between the two xes is 2. Hence, giving 3 b 3 = M g + b 2 2 = Mg + 3 b 4, M g = 3 b 12. 2. Determine the second moment of semi-circulr region bout n xis through its centroid, prllel to its dimeter. 3
Solution O The second moment of the semi-circulr region bout its dimeter is π 4 8. The position of the centroid is distnce of 4 3π from the dimeter long the rdius which perpendiculr to it. Hence, Tht is, π 4 8 = M g + π2 2. 4 3π M g = π4 8 84 9π 2. 2 = M g + 84 9π 2 4
13.12.2 THE PERPENDICULAR AXIS THEOREM Let l 1 nd l 2 denote two stright lines, t right-ngles to ech other, in the plne of region R with re A. Let h 1 nd h 2 be the perpendiculr distnces from these two lines respectivel of n element δa in R. l h 1 1 h 2 l 2 δa The second moment bout l 1 is given b M 1 = R h2 1δA. The second moment bout l 2 is given b M 2 = R h2 2δA. 5
Adding these two together gives the second moment bout n xis perpendiculr to the plne of R nd pssing through the point of intersection of l 1 nd l 2. This is becuse the squre of the perpendiculr distnce, h 3 of δa from this new xis is given, from Pthgors s Theorem, b EXAMPLES h 2 3 = h 2 1 + h 2 2. 1. Determine the second moment of rectngulr region, R, with sides of length nd b bout n xis through one corner, perpendiculr to the plne of R. Solution b O R The required second moment is 1 3 3 b + 1 3 b3 = 1 3 b(2 + b 2 ). 6
2. Determine the second moment of circulr region, R, with rdius, bout n xis through its centre, perpendiculr to the plne of R. Solution O R The second moment of R bout dimeter is π 4 4. Tht is, twice the vlue of the second moment of semi-circulr region bout its dimeter. The required second moment is thus π 4 4 + π4 4 = π4 2. 7
13.12.3 THE RADIUS OF GYRATION OF AN AREA Hving clculted the second moment of two-dimensionl region bout certin xis it is possible to determine positive vlue, k, with the propert tht the second moment bout the xis is given b Ak 2, where A is the totl re of the region. We divide the vlue of the second moment b A in order to obtin the vlue of k 2 nd hence the vlue of k. The vlue of k is clled the rdius of grtion of the given region bout the given xis. Note: The rdius of grtion effectivel tries to concentrte the whole re t single point for the purposes of considering second moments. Unlike centroid, this point hs no specific loction. EXAMPLES 1. Determine the rdius of grtion of rectngulr region, R, with sides of lengths nd b bout n xis through one corner, perpendiculr to the plne of R. 8
Solution b O R The second moment is 1 3 b(2 + b 2 ). Since the re itself is b, we obtin k = 2 + b 2. 2. Determine the rdius of grtion of circulr region, R, bout n xis through its centre, perpendiculr to the plne of R. 9
Solution O R The second moment bout the given xis is π 4 2. Since the re itself is π 2, we obtin k = 2. 10