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CDIM09 - CHAPTER 9 Dynamics of a System of Particles 9-1...

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CHAPTER 9 Dynamics of a System of Particles 9-1. Put the shell in the z > 0 region, with the base in the x - y plane. By symmetry, 0 xy == . 2 1 2 1 22 2 00 2 sin sin r rr r z rd r dd z r ππ φθ ρ θθφ == = === = ∫∫∫ Using z = r cos θ and doing the integrals gives ( ) ( ) 44 21 33 3 8 z = 9-2. z h x y a z h a h =− + By symmetry, 0 . Use cylindrical coordinates , φ , z . 0 mass density = 277

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278 CHAPTER 9 2 0 000 2 4 0 4 h ah a z h z zddd z h z ddd z πρ φρ ρρρφ −+ === == ∫∫∫ The center of mass is on the axis 3 of the cone from the vertex. 4 h 9-3. z h x y a By symmetry, 0 xy . From problem 9-2, the center of mass of the cone is at 1 4 z h = . From problem 9-1, the center of mass of the hemisphere is at () 21 3 ,0 8 za r a r = −= = So the problem reduces to 2 11 1 ; 43 z hm a h ρπ i 3 22 2 32 ; 83 z am a ρ π =− = i 1 2 12 1 2 3 42 mz h a z mm h a ρρ +− ++ for = 23 3 ha z = +
DYNAMICS OF A SYSTEM OF PARTICLES 279 9-4. θ′ θ /2 θ y x a a By symmetry, 0 y = . If mass length σ = then M a σθ = So 2 2 1 x xdm M θ =− = 2 2 1 xx M a d = Using M a = and cos xa = , 2 2 1 cos sin sin 22 2sin 2 a d a θθ  == ′′   = 2 sin 2 0 a x y = = 9-5. r i r 0 r 0 r i m i

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280 CHAPTER 9 th th 1 position of the particle mass of the particle total mass constant gravitational field i i i mi Mm = = == = r g Calculate the torque about 0 r () ( ) ( ) ( ) 0 0 10 0 0 i ii i m mm mM ττ = =− × × =×−× × × ∑∑ rr F rr g rgrg rg r rg rg g Now if the total torque is zero, we must have 0 = or 0 1 m M = which is the definition of the center of mass. So 0C M 0 about or center of gravity center of mass. τ = 9-6. Since particle 1 has F = 0, , then 00 0 rv 1 0 = r . For particle 2 0 20 ˆˆ then F Fr m Fx ±± x Integrating twice with gives 0 2 0 2 ˆ 2 F rt m = x 2 0 11 22 CM 12 ˆ 4 F t m + + rx
DYNAMICS OF A SYSTEM OF PARTICLES 281 2 0 CM 0 CM 0 CM ˆ 4 ˆ 2 ˆ 2 F t m F t m F m = = = rx vx ax 9-7. O y a a H H x 52˚ 52˚ By symmetry 0 y = 0 16 H mm = Let 0 ,1 6 H m == m Then 3 1 1 ii xm M = x () cos 52 1 2c o s 5 2 18 9 a a m ° = 0.068 xa = 9-8. By symmetry, 0 x = . Also, by symmetry, we may integrate over the x > 0 half of the triangle to get y . σ = mass/area 22 00 32 aa x xy x ydydx a y dy dx ∫∫ a y =

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282 CHAPTER 9 9-9. POW! 45˚ z y m 1 m 2 v 1 v y v z Let the axes be as shown with the projectile in the y - z plane. At the top just before the explosion, the velocity is in the y direction and has magnitude 0 0 2 y v v = . 0 12 00 0 2 22 y E mm vE v + == = + where and are the masses of the fragments. The initial momentum is 1 m 2 m () 0 0, , 0 i E pmm =+ + The final momentum is f p pp = + ( ) 11 1 0,0, p mv = ( ) ,, x y z p mvvv = The conservation of momentum equations are 2 :0 xx v o r v 0 x pm 01 2 2 2 2 1 : yy y p Em m m v o r v m += = + 2 2 1 1 o r z zz m p v v m = The energy equation is ( ) 0 0 1 1 2 1 2 2 y z E E m v mvv ++ = + + + or ( ) 3 2 y z Em vm vv =+ + Substituting for y v and v gives 1
DYNAMICS OF A SYSTEM OF PARTICLES 283 () 01 2 1 2 21 2 2 z Em m m v mm m = + 2 1 1 z m vv m =− gives 021 1 11 2 2 Emm v + So travels straight down with speed = 1 m 1 v travels in the y - z plane 2 m ( ) 12 012 22 2 2 121 4 2 tan tan yz z y vvv mmm v vm θ −− + =+ = + == + m The mass

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CDIM09 - CHAPTER 9 Dynamics of a System of Particles 9-1...

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