JUST THE MATHS UNIT NUMBER INTEGRATION APPLICATIONS 12 (Second moments of an area (B)) A.J.Hobson

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1 JUST THE MATHS UNIT NUMBE 13.1 INTEGATION APPLICATIONS 1 (Second moments of n re (B)) b A.J.Hobson The prllel xis theorem The perpendiculr xis theorem The rdius of grtion of n re Exercises Answers to exercises

2 UNIT INTEGATION APPLICATIONS 1 SECOND MOMENTS OF AN AEA (B) THE PAALLEL AXIS THEOEM Suppose tht M g denotes the second moment of given region,, bout n xis, g, through its centroid. Suppose lso tht M l denotes the second moment of bout n xis, l, which is prllel to the first xis, in the sme plne s nd hving perpendiculr distnce of d from the first xis. l d centroid h δa g We hve M l = (h + d) δa = (h + hd + d ). Tht is, M l = h δa + d hδa + d δa = M g + Ad, since the summtion, hδa, is the first moment bout the n xis through the centroid nd therefore zero; (see Unit 13.7, section ). 1

3 The Prllel Axis Theorem sttes tht M l = M g + Ad. EXAMPLES 1. Determine the second moment of rectngulr region bout n xis through its centroid, prllel to one side. Solution b O For rectngulr region with sides of length nd b, the second moment bout the side of length b is 3 b from Exmple 1 in the previous Unit, section The perpendiculr distnce between the two xes is then, so tht the required second moment, M g is given b Hence, 3 ) b 3 = M g + b( = Mg + 3 b 4 M g = 3 b 1.. Determine the second moment of semi-circulr region bout n xis through its centroid, prllel to its dimeter.

4 Solution O The second moment of the semi-circulr region bout its dimeter is π4, from Exmple 8 in the previous Unit, section Also the position of the centroid, from Exmple in Unit 13.7, section , is distnce of 4 from the dimeter, long the rdius which perpendiculr to it. 3π Hence, Tht is, π 4 ( ) 8 = M g + π 4. = M g π 9π. M g = π π THE PEPENDICULA AXIS THEOEM Suppose l 1 nd l re two stright lines, t right-ngles to ech other, in the plne of region with re A nd suppose h 1 nd h re the perpendiculr distnces from these two lines, respectivel, of n element δa in. 3

5 l h 1 1 h l δa The second moment bout l 1 is given b M 1 = h 1δA nd the second moment bout l is given b M = h δa. Adding these two together gives the second moment bout n xis, perpendiculr to the plne of nd pssing through the point of intersection of l 1 nd l. This is becuse the squre of the perpendiculr distnce, h 3, of δa from this new xis is given, from Pthgors s Theorem, b h 3 = h 1 + h. EXAMPLES 1. Determine the second moment of rectngulr region,, with sides of length nd b, bout n xis through one corner, perpendiculr to the plne of. 4

6 Solution b O Using Exmple 1 in the previous Unit, section , the required second moment is b b3 = 1 3 b( + b ).. Determine the second moment of circulr region,, with rdius, bout n xis through its centre, perpendiculr to the plne of. Solution O The second moment of bout dimeter is, from Exmple in the previous Unit, section , equl to π4 ; tht is, twice the vlue of the second moment of semicirculr region bout its 4 dimeter. 5

7 The required second moment is thus π π4 4 = π THE ADIUS OF GYATION OF AN AEA 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, where A is the totl re of the region. We simpl divide the vlue of the second moment b A in order to obtin the vlue of k 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; but, unlike centroid, this point hs no specific loction. EXAMPLES 1. Determine the rdius of grtion of rectngulr region,, with sides of lengths nd b bout n xis through one corner, perpendiculr to the plne of. Solution b O 6

8 Using Exmple 1 from the previous section, the second moment is nd, since the re itself is b, we obtin 1 3 b( + b ) k = + b.. Determine the rdius of grtion of circulr region,, bout n xis through its centre, perpendiculr to the plne of. Solution O From Exmple in the previous section, the second moment bout the given xis is π4 nd, since the re itself is π, we obtin k =. 7

9 EXECISES Determine the rdius of grtion of ech of the following regions of the x-plne bout the xis specified: 1. Bounded in the first qudrnt b the x-xis, the -xis nd the lines x =, = b. Axis: Through the point (, ) b, perpendiculr to the x-plne.. Bounded in the first qudrnt b the x-xis, the -xis nd the lines x =, = b. Axis: The line x = Bounded in the first qudrnt b the x-xis, the -xis nd the curve whose eqution is x + =. Axis: Through the origin, perpendiculr to the x-plne. 4. Bounded in the first qudrnt b the x-xis, the -xis nd the curve whose eqution is Axis: The line x =. x + = ANSWES TO EXECISES ( + b )

JUST THE MATHS SLIDES NUMBER INTEGRATION APPLICATIONS 12 (Second moments of an area (B)) A.J.Hobson

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