Explanation Given information : The mass m A of the pulley at A is 3.0 kg. The mass m B of the pulley at B is 10 kg. The mass ( m ) of the block is 2.0 kg. The radius r 1 of the disk A is 30 mm. The radius r 2 of the disk B is 100 mm. The angular velocity ω B of pulley B is 6 rad / s . Determine the block’s speed v C after it descends 0.5 m starting form rest using the formula. v C = r 2 ω B (I) Here, the angular velocity of disk B is ω B . Determine the block’s speed v B after it descends 0.5 m starting form rest using the formula. v C = r 1 ω A (II) Here, the angular velocity of disk A is ω A . Express the Equation for principle of work and energy as follows: ( T 1 + T A + T B ) + V 1 = T 2 + V 2 ( T 1 + T A + T B ) + V 1 = ( 1 2 [ 1 2 m A ( r 1 ) 2 ] ω A 2 + 1 2 [ 1 2 m B ( r 2 ) 2 ] ω B 2 + 1 2 m ( v C ) 2 − m g ( s C ) ) (III) Here, the kinetic energy at position 1 is T 1 , the kinetic energy of pulley A is T A , the kinetic energy of pulley B is T B , and the acceleration due to gravity is g . Conclusion : Assumption: At rest condition kinetic energy at position 1, at pulley A m and at pulley B are equal to zero at the same time the velocity at position 1 is also equal to zero (due to rest condition). T 1 = T A = T B = 0 V 1 = 0 Substitute 100 mm for r 2 and 6 rad / s for ω B in Equation (I). v C = ( 100 mm × 1 m 1,000 mm ) 6 = 0.6 m / s Substitute 30 mm for r 1 and 0.6 m / s for v C in Equation (II)

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Chapter 18.5, Problem 37P

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The distance the block must descend, starting from rest, in order to cause B to have an angular velocity of 6 rad/s .

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Chapter 18.5, Problem 36P

Chapter 18.5, Problem 40P

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The distance the block must descend, starting from rest, in order to cause B to have an angular velocity of 6 rad / s .

Solution Summary: The author determines the distance the block must descend, starting from rest, in order to cause B to have an angular velocity.

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The distance the block must descend, starting from rest, in order to cause B to have an angular velocity of 6 rad/s .

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Chapter 18 Solutions