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# HW-H - HW-H Chapter 15 Chapter 18 43 49 53 34 58 65 Problem...

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HW-H Chapter 15 43, 49, 53 Chapter 18 34, 58, 65 ____________________________________________________________ Problem 15-43 (( WileyPlus )) 43. (a) A uniform disk pivoted at its center has a rotational inertia of 2 1 2 Mr , where M is its mass and r is its radius. The disk of this problem rotates about a point that is displaced from its center by r + L , where L is the length of the rod, so, according to the parallel-axis theorem, its rotational inertia is 2 2 1 1 2 2 ( ) Mr M L r + + . The rod is pivoted at one end and has a rotational inertia of mL 2 /3, where m is its mass. The total rotational inertia of the disk and rod is 2 2 2 2 2 2 2 1 1 ( ) 2 3 1 1 (0.500kg)(0.100m) (0.500kg)(0.500m 0.100m) (0.270kg)(0.500m) 2 3 0.205kg m . I Mr M L r mL = + + + = + + + = × (b) We put the origin at the pivot. The center of mass of the disk is = + = 0.500 m+0.100 m = 0.600 m d L r l

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away and the center of mass of the rod is l r L = = = / ( . ) / . 2 0 500 2 0 250 m m away, on the same line. The distance from the pivot point to the center of mass of the disk-rod system is = + + = 0.500 0.600 + 0.270 0.250 0.500 + 0.270 = 0.477 . d M m M m d r l l kg m kg m kg kg m a fa f a fa f (c) The period of oscillation is ( 29 2 2 0.205 kg m 2 2 1.50 s . (0.500 kg 0.270 kg)(9.80 m/s )(0.447 m) I T M m gd π π × = = = + + _______________________________________________________________________ _ Problem 15-49 (( My solution )) L = 1.85 m
2 2 0 0 12 1 sin Mx ML I Mgx I + = - = = θ θ τ For θ ≈ 0, θ ϖ θ θ 2 2 2 12 1 sin - + - = x L gx where 2 2 12 1 x L x g + = ϖ The period T is x L x g T 12 2 2 2 + = = π ϖ π Note that 3 12 2 12 2 12 2 2 L L x L x x L x = = + where the function x L x 12 2 + has a minimum at m L L L x 53 . 0 289 . 0 3 2 12 = = = =

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Then T is s L g T 074 . 2 747 . 4 = = (( Mathematica ))
T 1 2 1 g x 2 1 1 2 L 2 x 2 L 2 1 2 x 2 x g r u l e 1 x L y x L y

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(( WileyPlus )) 49. This is similar to the situation treated in Sample Problem 15-5, except that O is no longer at the end of the stick. Referring to the center of mass as C (assumed to be the geometric center of the stick), we see that the distance between O and C is h = x . The parallel axis theorem (see Eq. 15-30) leads to 2 2 2 2 1 .
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