Selected Hw Problems for Ch. 9

Selected Hw Problems for Ch. 9 - 9.6 The Figure shows a...

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9.6 The Figure shows a cubical box that has been constructed from uniform metal plate of negligible thickness. The box is open at the top and has edge length L = 40 cm. Find (a) the x coordinate, (b) the y coordinate, and (c) the z coordinate of the center of mass of the box. A homogeneous symmetric body has its center of mass (com) at its center of symmetry. This means that the plates making up each side have their coms at their respective centers. We can then calculate the com of the entire box as the com due to five point masses (each the mass of a side) at the center point of each plate. Now, formally, com i i i 1 rm r M = GG i i Mm = where If the mass of the entire box is designated M then the mass of each side is M/5. The vector equation above is 3 separate eqns, one for each coordinate axis.
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For the x axis, xi x i i 11 M M M M M r m r [ ( 0 )( 0 . 2 m 0 . 2 m 0 . 2 m 0 . 4 m ) ] MM 5 5555 = =++++ x 31 r [ (0.2m) (0.4m)] 55 =+ x r0 . 2 0 m = Note that by the symmetry of the box and the location of its center along the x direction (at x = 0.20 m) we could have anticipated this result. We do this for the y direction finding that, y . 2 0 m = For the z axis, zi z i i 1 1 M MMMM r m r [ ( 0 0 . 2 m 0 . 2 m 0 . 2 m 0 . 2 m ) ] 5 =
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z 4 r[ ( 0 . 2 m ) ] 5 = z 4 ( 0 . 2 m ) ] 5 = z r0 . 1 6 m =
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9.15 The time taken to get to the top of the trajectory is given by solving for t, yy o vv g t =− Where and at the top of the trajectory v y = 0 so, yo o o s i n oo 0v s i n g t vs i n t g θ = The x directed velocity of the shell in getting there is, xo o o c o s = θ So the horizontal distance traveled by the shell in getting to the top of its trajectory, to just before it explodes is, 1x o xv t = 2 o 1o o o i n v x (v cos ) cos sin gg θ = θ= θ θ
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Conservation of momentum requires that momentum of the shell before the explosion equals the momentum of its pieces after the explosion. At the top of the trajectory just before and just after the explosion, along the y direction we have, yi y1f y2f MM Mv v v 22 =+ Which gives, since and that after the explosion . v0 = y1f = = Along the x direction, xi x1f x2f Mv v v Then with and given that xi xo o o vv v c o s == θ x1f = oo x 2 f M Mv cos 0 v 2 θ= + o o v2 v c o s = θ Since there was no y velocity just after the explosion the time taken to drop to the ground is the same as the time it took the shell to get to the top of the trajectory.
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Hence the distance traveled by the 2 nd piece of shell is, oo 1o o1o o vs i n xx ( 2 v c o s) tx ( 2 v c o g θ =+ θ=+ θ 22 vv xs i n c o s 2 s i n c o s gg θ + θ θ 2 2 o 2 m 20 v s x 3 sin cos 3 sin(60 )cos(60 ) 53 m m g 9.8 s ⎛⎞ ⎜⎟ ⎝⎠ θ = =
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9.20 The Figure gives an overhead view of the path taken by a 0.165 kg cue ball
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This note was uploaded on 04/22/2008 for the course PHY 2048 taught by Professor Field during the Spring '08 term at University of Florida.

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Selected Hw Problems for Ch. 9 - 9.6 The Figure shows a...

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