Explanation The below figure represents the schematic diagram of system. Figure-(1) Write the expression for the relation for the length of bar from triangle Δ O A B . l 2 = O A ¯ 2 + O B ¯ 2 − 2 ( O A ¯ ) ( O B ¯ ) cos ( 90 ° + 30 ° ) l 2 = r 2 + r 2 − 2 ( r ) ( r ) cos ( 90 ° + 30 ° ) (I) Here, length of the bar is l , the distance from O to centre of roller is r , length between O and A is O A ¯ , L and length between O and B is O B ¯ . Write the expression for the distance between O and G , and r . sin 30 ° = d r (II) Here, distance between O and G is d . Write the expression for moment of inertia of bar about point O . I O = 1 12 m l 2 + m d 2 (III) Here, mass of the bar is m . Write the expression for the angular velocity of the bar. ω = v A r (IV) Here, the speed of the roller is v A . The bar is released from the rest; thus the initial kinetic energy is zero. T 1 = 0 Write the expression for the final kinetic energy. T 2 = 1 2 I O ω 2 (V) Here, mass moment of bar about point O is I O . Write the expression for the initial gravitational potential energy. V 1 g = − m g l 2 cos θ (VI) Here, initial position of the uniform bar is θ and acceleration due to gravity is g . Write the expression for the initial gravitational potential energy. V 2 g = − m g ( r + l 2 sin θ ) (VII) Write the expression for the potential energy and the work energy equation. T 1 + V 1 g + U ′ 1 − 2 = T 2 + V 2 g (VIII) Here, work done by friction on system is U ′ 1 − 2 . Conclusion: Substitute 700 mm for l in Equation (I). ( 700 mm ) 2 = r 2 + r 2 − 2 ( r ) ( r ) cos ( 90 ° + 30 ° ) ( 700 mm ( 1 m 1000 mm ) ) 2 = r 2 + r 2 − 2 ( r ) ( r ) cos ( 90 ° + 30 ° ) ( 0.7 m ) 2 = 2 r 2 − 2 r 2 cos 120 ° ( 0.7 m ) 2 = 3 r 2 r = ( 0.7 m ) 2 3 r = 0.40415 m Substitute 0.40415 m for r in Equation (II). sin 30 ° = d 0.40415 m d = sin 30 ° × 0.40415 m d = 0.20207 m Substitute 5 kg for m , 0.20207 m for d and 700 mm for l in Equation (III). I O = 1 12 5 kg ( 700 mm ) 2 + 5 kg ( 0.20207 m ) 2 = 1 12 5 kg ( 700 mm ( 1 m 1000 mm ) ) 2 + 5 kg ( 0.20207 m ) 2 = 0.40833 kg .m 2 Substitute 3.25 m / s for v A and 0

8th Edition

ISBN: 9781118885840

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

Publisher: WILEY

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Chapter 6.9, Problem 206RP

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The work done by friction on the system over this portion of motion.

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Chapter 6.9, Problem 205RP

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