Homework9-F09

# Homework9-F09 - Homework 9 Chapt 12 14 help 12.11 Model The...

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Homework 9 Chapt. 12 & 14 help 12.11. Model: The disk is a rigid body rotating about an axis through its center. Visualize: Solve: The speed of the point on the rim is given by rim . v R ω = The angular velocity ω of the disk can be determined from its rotational kinetic energy which is 2 1 2 0.15 J. K I ω = = The moment of inertia I of the disk about its center and perpendicular to the plane of the disk is given by 2 2 2 2 1/ 2 2 1/ 2 , 1/2 K K I I K I mR so mR ω ω ω = = = = Now, We get rim 2 4 4 4 0.15/0.10 2.4 / K K v R R m s mR m ω = = = = = 12.70. Model: The pulley is a rigid rotating body. We also assume that the pulley has the mass distribution of a disk and that the string does not slip. Visualize: Because the pulley is not massless and frictionless, tension in the rope on both sides of the pulley is not the same. Solve: Applying Newton’s second law to 1 , m 2 , m and the pulley yields the three equations: ( ) ( ) 1 G 1 1 G 2 2 2 2 1 1 2 0.50 N m T F m a F T m a T R T R I α - = - + = - - = Noting that 2 1 , a a a - = = 2 1 p 2 , I m R = and / , a R α = the above equations simplify to 2 1 1 1 2 2 2 2 1 p p 1 1 0.50 N m 1 0.50 N m 2 2 0.060 m a T m g m a m g T m a T T m R m a R R R  - = - = - = + = +   Adding these three equations,

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2 1 1 2 p 2 2 2 1 1 1 2 p 2 1 ( ) 8.333 N 2 ( ) 8.333 N (4.0 kg 2.0 kg)(9.8 m/s ) 8.333 N 1.610 m/s 2.0 kg 4.0 kg (2.0 kg/2) m m g a m m m m m g a m m m - = + + + - - - - = = = + + + + We can now use kinematics to find the time taken by the 4.0 kg block to reach the floor: 2 2 2 1 0 0 1 0 2 1 0 1 1 2 1 1 ( ) ( ) 0 1.0 m 0 ( 1.610 m/s )( 0 s) 2 2 2(1.0 m)
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