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hw12solution

# hw12solution - Homework 12 Solution CHANG 1 Consider the...

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Homework 12 Solution CHANG 1 Consider the disk, the rod, and the sphere as one system and use the energy equation with U e 0 (no spring) and W 1 2 0 (no friction): K 1 U g 1 K 2 U g 2 . The gravitational potential energy is for the entire system, including all three parts. However, since the disk is pivoted at the center, its center of mass does not move and we only need to consider the vertical positions for the rod and the sphere to calculate U g . The kinetic energy is due to the rotation of the assembly, and hence K 2 1/2 I 2 while K 1 0 because the system is initially at rest. The total moment of inertia for the assembly is calculated to be I I d I r I s 1 2 m d R 2 1 3 m r L 2 m s r 2 1 2 15 0.4 2 1 3 8 0.6 2 4 0.6 2 3.6 Kgm 2 With the reference chosen through the pivot, U g 1 0 since the initial vertical positions for the center of mass of the rod and the sphere are zero (on the datum). When the rod swings down to the vertical position, we have y cm 2 L /2 for the rod and y s 2 L for the sphere. Note that both are negative because they are below the reference line. The final angular velocity can then be found using the energy equation: 0 0 1 2 I 2 m r gy cm 2 m s gy s 2 0 1 2 3.6   2 8 9.8 0.3 4 9.8 0.6 5.11 rad/s CHANG 2 Consider the block and the pulley together as one system. Take the starting point as position1 and position 2 represents the configuration after 5 revolutions of pulley’s rotation. The energy equation is written as K 1 U e 1 W 1 2 K 2 U e 2 . Note that U e 1 0 (no stretching in the spring) and the gravitational potential energy is not included because the pulley is stationary and the block slides horizontally. The kinetic energy includes the translational kinetic energy from the block and the rotational kinetic energy from the pulley, and K 1 0 since the system starts from rest. The work term W 1 2 has two parts, the work done by friction, k Ns

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hw12solution - Homework 12 Solution CHANG 1 Consider the...

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