{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–7. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 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 × = = = + + _______________________________________________________________________ _ Problem 15-49 (( My solution )) L = 1.85 m
2 2 0 0 12 1 sin Mx ML I Mgx I + = - = = θ τ For ≈ 0, ϖ 2 2 2 12 1 sin - + - = x L gx where 2 2 12 1 x L x g + = The period T is x L x g T 12 2 2 2 + = = 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 = = = =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(( 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 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 15

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

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online