Explanation The below figure represent the schematic diagram of the system. Figure-(1) Write the expression of centroidal distance of the body in x axis. x ¯ = L 1 x ¯ 1 + L 2 x ¯ 2 L 1 + L 2 (I) Here, the length of the vertical slender rod is L 1 , the length of horizontal slender rod is L 2 , the centroidal distance of the vertical slender rod is x ¯ 1 and the centroidal distance of the horizontal slender rod is x ¯ 2 . Write the expression of centroidal distance of the body in y axis. y ¯ = L 1 y ¯ 1 + L 2 y ¯ 2 L 1 + L 2 (II) Here, the centroidal distance of the vertical slender rod is y ¯ 1 and the centroidal distance of the horizontal slender rod is y ¯ 2 . Write the expression of mass. m = ρ l (III) Here, the mass per unit length is ρ and the length is l . Write the expression of moment of inertia of the body about the mass centre. I G = 2 [ 1 12 m l 2 + m d 2 ] (IV) Here, the distance between point C and the mass centre is d . Write the expression of the distance between point C and the mass centre. d = 2 l 4 Write the expression of conservation of linear impulse. ∫ P d t i = 2 m v G (V) Here, the linear impulse is ∫ P d t , the unit vector in x direction is i and the velocity of centre of mass G is v G . Substitute ρ l for m in Equation (V). ∫ P d t i = 2 ( ρ l ) v G (VI) Write the expression of conservation of angular momentum. ( l − y ¯ ) ∫ P d t = I G ω (VII) Here, the angular velocity is ω . Write the expression of position vector of the point B with respect to centre of mass G . r B / G = − B C i + y ¯ j (VIII) Here, the distance between the point B and C is B C . Write the expression of velocity of corner B . v B = v G + ω × r B / G (IX) Write the expression of distance between the point B and C . B C = l − x ¯ (X) Conclusion: Substitute l for L 1 , l for L 2 , 0 for x ¯ 1 and l / 2 for x ¯ 2 in Equation (I). x ¯ = ( l ) ( 0 ) 1 + ( l / 2 ) l l + l = ( l 2 / 2 ) 2 l = 0.25 l (XI) Substitute 500 mm for l in Equation (XI). x ¯ = 0.25 ( 500 mm ) = 0.25 ( 500 mm ) ( 1 m 1000 mm ) = 0.125 m Substitute l for L 1 , l for L 2 , l / 2 for y ¯ 1 and 0 for y ¯ 2 in Equation (II). y ¯ = ( l / 2 ) l + ( l ) ( 0 ) l + l = ( l 2 / 2 ) 2 l = 0.25 l (XII) Substitute 500 mm for l in Equation (XII). y ¯ = 0.25 ( 500 mm ) = 0.25 ( 500 mm ) ( 1 m 1000 mm ) = 0.125 m Substitute ρ l for m and 2 l 4 for d in Equation (IV). I G = 2 [ 1 12 ( ρ l ) l 2 + ( ρ l ) ( 2 l 4 ) 2 ] = 2 [ ρ l 3 12 + ( 2 ρ l 3 16 ) ] = 2 ρ l 3 ( 1 12 + 1 8 ) = 5 ρ l 3 12 Substitute 500 mm for l , 3 kg / m for ρ and 8 N ⋅ s for ∫ P d t in Equation (VI)

8th Edition

ISBN: 9781118885840

Author: James L. Meriam, L. G. Kraige, J. N. Bolton

Publisher: WILEY

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Chapter 6.9, Problem 185RP

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The velocity vB of corner B.

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Chapter 6.9, Problem 184RP

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The velocity v B of corner B .

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The velocity vB of corner B.

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