Assignment 9 solution 2009

# Assignment 9 solution 2009 - PROBLEM 16.25 A 10-kg uniform...

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PROBLEM 16.25 A 10-kg uniform disk is placed in contact with an inclined surface and a constant 11 N m couple M is applied as shown. The weight of the link AB is negligible. Knowing that the coefficient of kinetic friction at D is 0.4, determine ( a ) the angular acceleration of the disk, ( b ) the force in the link AB . SOLUTION Free-body diagram: 2 2 1 mr I = 0 10 cos60 cos30 4 0 : o o = - + = Σ + g N N . a m F y y ( 0 81 9 10 cos60 cos30 4 0 o o = - + . N N . N 90 115 . N = ( a ) ( 29 ( 29( 29 α α 2 225 0 10 2 1 225 0 4 0 11 : . . N . I M B = - = Σ + rad/s 247 2 . = α 247 2 . = α rad/s ( b ) 0 sin60 sin30 4 0 : o o = - + = Σ + N N . F a m F AB x x N 19 77 . F AB = 19 77 . AB = F N (compression) N F = μ k N = 0.4 N 10 g F AB = α I

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PROBLEM 16.67 A uniform sphere of radius r and mass m is placed with no initial velocity on a belt that moves to the right with a constant velocity v 1 . Denoting by μ k the coefficient of kinetic friction between the sphere and the belt, determine ( a ) the time t 1 at which the sphere will start rolling without sliding, ( b ) the linear and angular velocities of the sphere at time t 1 . SOLUTION Free-body diagram: 2 5 2 mr I = 0 : = - = Σ + mg N a m F y y mg N = a m N a m F k x x = = Σ + μ : g a k μ = α μ α I r N I M k G = = Σ + : α μ 2 5 2 mr r mg k = r g k μ α 2 5 = Kinematics: gt t a v k μ = = + (1) t r g t k μ α ϖ 2 5 = = + (2) N = C G r F = μ k N mg C G α I a m
C = Point of contact with belt r v v C ϖ + = + r t r g gt k k + = μ μ 2 5 gt v k C μ 2 7 = ( a ) When sphere starts rolling ( t = t 1 ), we have 1 v v C = 1 1 2 7 gt v k μ = g v t k μ 1 1 7 2 = ( b

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