Explanation The initial and final position of the can when door is slammed is shown below. The dotted lines indicate the final position of the can. Figure-(1) Write the expression for the conversion of energy between initial and final position of the can. 1 2 I ω 2 = m g h (I) Here, the moment of inertia of can is I , the angular velocity of can is ω , the mass of can is m , the gravitational acceleration is g and the height of the corner of rack is h . Write the expression for the moment of inertia of can. I = 3 2 m r 2 (II) Here, the radius of can is r . The schematic diagram of the can and door is shown below. Figure-(2) Write the expression for the linear velocity of the can. v = Ω × R (III) Here, the linear velocity of the can is v , the angular velocity of the door is Ω and the distance between can and corner of rack is R . Write the expression for the conservation of angular momentum of the system. H initial = H final I ω = m v ( r − h ) (IV) Here, the initial angular momentum is H initial and the final angular momentum is H final . Conclusion: Substitute 3 2 m r 2 for I in Equation (I). 1 2 ( 3 2 m r 2 ) ω 2 = m g h ω 2 = 4 g h 3 r 2 ω = 4 g h 3 r 2 (V) Substitute 9.81 m / s 2 for g , 7 mm for h and 35 mm for r in Equation (V). ω = 4 ( 9.81 m / s 2 ) ( 7 mm ) 3 ( 35 mm ) 2 = 4 ( 9.81 m / s 2 ) ( 7 mm ) ( 10 − 3 m 1 mm ) 3 { ( 35 mm ) ( 10 − 3 m 1 mm ) } 2 = 0

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ISBN: 9781119047315

Author: Bolton

Publisher: VST

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Chapter 6.9, Problem 203RP

To determine

The maximum angular velocity Ω of the door.

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Chapter 6.9, Problem 202RP

Chapter 6.9, Problem 204RP

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