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Unformatted text preview: Homework 5 Solution CHANG 1 The sphere swings around the vertical rod, so its motion is a horizontal uniform circular motion. The FBD of the sphere is shown here: w = mg P 1 P 2 P 1 cos30 P 2 cos60 P 2 sin60 P 1 sin30 y n a n w = mg P 1 P 2 P 1 cos30 P 2 cos60 P 2 sin60 P 1 sin30 w = mg P 1 P 2 P 1 cos30 P 2 cos60 P 2 sin60 P 1 sin30 y n a n The n y coordinate system is set up with naxis pointing to the left toward the center of the circle. In the figure P 1 and P 2 represent the tensions in the upper and lower cords, respectively. Apply Newtons 2nd law with a y 0 (no vertical motion): F y ma y P 1 cos30 P 2 cos60 mg F n ma n P 1 sin30 P 2 sin60 m 4 2 R T 2 With the given information, m 10 Kg, R 0.5 m, and P 1 2 P 2 , the period and the tensions can then be solved to be 2 P 2 cos30 P 2 cos60 10 9.8 P 2 79.54 N 2 P 2 sin30 P 2 sin60 10 4 2 .5 T 2 T 2 10 0.5 2sin30 sin60 79.54 1.15 s CHANG 2 Because block A is on the horizontal surface, the only possible motion is B sliding down the ramp while A sliding to the right, and from kinematics we know a A a B a . Draw the individual FBDs for blocks A and B : m B g T N B f kB x y a f kA N A T m A g x y a m B g T N B f kB x y a m B g T N B f kB x y a f kA N A T m A g x y a f kA N A T m A g x y a Note that xaxes for both blocks are in the directions of motion (accelerations of the blocks), and friction forces are drawn in the opposite directions of the motion. We now apply Newtons 2nd law for both blocks along the coordinate axes with a y 0: Block A : F y ma y N A m A g N A m A g f kA kA N A kA m A g F x ma x T f kA...
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This note was uploaded on 10/17/2009 for the course PHYS 2305 taught by Professor Tschang during the Spring '08 term at Virginia Tech.
 Spring '08
 TSChang
 Physics, Circular Motion, Work

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