“JUST THE MATHS” SLIDES NUMBER 13.10 INTEGRATION APPLICATIONS 10 (Second moments of an arc) by A.J.Hobson 13.10.1 Introduction 13.10.2 The second moment of an arc about the y -axis 13.10.3 The second moment of an arc about the x -axis 13.10.4 The radius of gyration of an arc
UNIT 13.10 - INTEGRATION APPLICATIONS 10 SECOND MOMENTS OF AN ARC 13.10.1 INTRODUCTION Let C denote an arc (with length s ) in the xy -plane of cartesian co-ordinates. Let δs denote the length of a small element of this arc. Then the “second moment” of C about a fixed line, l , in the plane of C is given by C h 2 δs, lim � δs → 0 where h is the perpendicular distance, from l , of the ele- ment with length δs . 1
✡ ✡ ✡ ✡ ✡ ◗◗◗◗◗◗◗◗ ✡ ✡ ✡ h ✡ ✡ l ◗ δs r ✡ r ✡ ✡ C 13.10.2 THE SECOND MOMENT OF AN ARC ABOUT THE Y-AXIS Consider an arc of the curve whose equation is y = f ( x ) , joining two points, P and Q, at x = a and x = b , respec- tively. Q y ✻ δs δy r r P ✲ x O a δx b 2
The arc may be divided up into small elements of typical length, δs , by using neighbouring points along the arc, separated by typical distances of δx (parallel to the x - axis) and δy (parallel to the y -axis). The second moment of each element about the y -axis is x 2 δs . The total second moment of the arc about the y -axis is given by C x 2 δs. lim � δs → 0 But, by Pythagoras’ Theorem � 2 � δy � ( δx ) 2 + ( δy ) 2 = � � � 1 + δs ≃ δx. � � δx Thus, the second moment of the arc becomes � 2 � δy x = b � x = a x 2 � lim � 1 + δx � � � δx δx → 0 � 2 � d y � b � � a x 2 = � 1 + d x. � � d x 3
Note: If the curve is given parametrically by x = x ( t ) , y = y ( t ) , then, d y d y d t d x = . d x d t Hence, � 2 + � � 2 � � � d y � d x 2 � � d y � � d t d t � � 1 + = , � � d x d x d t provided that d x d t is positive on the arc being considered. If d x d t is negative on the arc, then the above formula needs to be prefixed by a negative sign. Thus, the second moment of the arc about the y -axis is given by � 2 2 � d x d y � t 2 � t 1 x 2 � + d t, ± � � � d t d t according as d x d t is positive or negative. 4
13.10.3 THE SECOND MOMENT OF AN ARC ABOUT THE X-AXIS (a) For an arc of the curve whose equation is y = f ( x ) , contained between x = a and x = b , the second moment about the x -axis will be � 2 � d y � b � a y 2 � � 1 + d x. � � d x Note: If the curve is given parametrically by x = x ( t ) , y = y ( t ) , then, the second moment of the arc about the x -axis is given by � 2 2 � d x d y � t 2 � � t 1 y 2 + d t, ± � � � d t d t according as d x d t is positive or negative. 5
(b) For an arc of the curve whose equation is x = g ( y ) , contained between y = c and y = d , we may reverse the roles of x and y in section 13.10.2 so that the second moment about the x -axis is given by � 2 � d x � d � � c y 2 � 1 + d y. � � � d y y ✻ S d δy δs R c ✲ x O rr δx 6
Note: If the curve is given parametrically by x = x ( t ) , y = y ( t ) , then, the second moment of the arc about the x -axis is given by � 2 2 � d x d y � t 2 � t 1 y 2 � + d t, ± � � � d t d t according as d y d t is positive or negative and where t = t 1 when y = c and t = t 2 when y = d . EXAMPLES 1. Determine the second moments about the x -axis and the y -axis of the arc of the circle whose equation is x 2 + y 2 = a 2 , lying in the first quadrant. Solution Using implicit differentiation, 2 x + 2 y d y d x = 0 . 7
Hence, d y d x = − x y. y ✻ ✡ ✡ a ✡ ✡ ✡ ✲ x O The second moment about the y -axis is therefore given by � � 1 + x 2 � � a � 0 x 2 � y 2 d x � � x 2 � a x 2 + y 2 d x. � = 0 y But x 2 + y 2 = a 2 . Therefore, ax 2 � a Second moment = y d x. 0 8
Making the substitution x = a sin u , � π a 3 sin 2 u d u Second moment = 2 0 1 − cos 2 u = a 3 � π d u 2 0 2 π 0 = πa 3 2 − sin 2 u u 2 = a 3 4 . 4 By symmetry, the second moment about the x -axis will also be πa 3 4 . 2. Determine the second moments about the x -axis and the y -axis of the first quadrant arc of the curve with para- metric equations x = a cos 3 θ, y = a sin 3 θ. Solution d x d θ = − 3 a cos 2 θ sin θ and d y d θ = 3 a sin 2 θ cos θ. 9
y ✻ ✲ x O The second moment about the y -axis is given by � 0 � 2 x 2 9 a 2 cos 4 θ sin 2 θ + 9 a 2 sin 4 θ cos 2 θ d θ. − π On using cos 2 θ + sin 2 θ ≡ 1, this becomes � π a 2 cos 6 θ. 3 a cos θ sin θ d θ 2 0 = 3 a 3 � π cos 7 θ sin θ d θ 2 0 π − cos 8 θ = 3 a 3 2 = 3 a 2 8 . 8 0 Similarly, the second moment about the x -axis is given by � 2 2 � d x d y � � π � y 2 + d θ 2 � � 0 � d θ d θ � π a 2 sin 6 θ. (3 a cos θ sin θ ) d θ 2 0 10
= 3 a 3 � π sin 7 θ cos θ d θ 2 0 π sin 8 θ = 3 a 3 2 = 3 a 3 8 . 8 0 This second result could be deduced, by symmetry, from the first. 13.10.4 THE RADIUS OF GYRATION OF AN ARC Having calculated the second moment of an arc about a certain axis it is possible to determine a positive value, k , with the property that the second moment about the axis is given by sk 2 , where s is the total length of the arc. We simply divide the value of the second moment by s in order to obtain the value of k 2 and, hence, the value of k . The value of k is called the “radius of gyration” of the given arc about the given axis. 11
Note: The radius of gyration effectively tries to concentrate the whole arc at a single point for the purposes of considering second moments; but, unlike a centroid, this point has no specific location. EXAMPLES 1. Determine the radius of gyration, about the y -axis, of the arc of the circle whose equation is x 2 + y 2 = a 2 , lying in the first quadrant. Solution y ✻ ✡ ✡ a ✡ ✡ ✡ ✲ x O From Example 1 in Section 13.10.3, the second mo- ment of the arc about the y -axis is equal to πa 3 4 . 12
Also, the length of the arc is πa 2 . Hence, the radius of gyration is � � πa 3 4 × 2 πa = a � � √ � 2 . � 2. Determine the radius of gyration, about the y -axis, of the first quadrant arc of the curve with para- metric equations x = a cos 3 θ, y = a sin 3 θ. Solution y ✻ ✲ x O From Example 2 in Section 13.10.3, d x d θ = − 3 a cos 2 θ sin θ and d y d θ = 3 a sin 2 θ cos θ. 13
Also, the second moment of the arc about the y -axis is equal to 3 a 3 8 . The length of the arc is given by � 2 2 � d x d y � a � � + d θ − � π � � d θ d θ 2 � π � 9 a 2 cos 4 θ sin 2 θ + 9 a 2 sin 4 θ cos 2 θ d θ. = 2 0 This simplifies to � π 3 a cos θ sin θ d θ 2 0 π sin 2 θ 2 = 3 a = 3 a 2 . 2 0 Thus, the radius of gyration is � � 3 a 3 8 × 2 3 a = a � � � 2 . � 14
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