Mecflu - din?mica de um sistema de part?culas

Mecflu - din?mica de um sistema de part?culas

(Parte 1 de 2)

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, 0xy==.

sin sin r r zrd r d d

z rd r d d

Using z = r cos θ and doing the integrals gives r z

9-2. z y a z h

By symmetry, 0xy==.

Use cylindrical coordinates ρ, φ, z. 0mass densityρ= h ah

h ah zd d dz h z d dz

The center of mass is on the axis 3 of the cone from the vertex.4 h

9-3. z y a

By symmetry, 0xy==.

From problem 9-2, the center of mass of the cone is at 1

From problem 9-1, the center of mass of the hemisphere is at

So the problem reduces to zhmahρπ==i

zamaρπ=−=i

mz m z h a z m h aρρ ρρ+−==++

ha z

DYNAMICS OF A SYSTEM OF PARTICLES 279 9-4.

x a

By symmetry, 0y=. If masslengthσ= then Maσθ= So

Using Maσθ= and cosxaθ=′,

1 cos sin sin

2s in 2 a xa d

a x

9-5.

r – r r r th 1 position of the particle mass of the particle total mass constant gravitational field i i i mi

g Calculate the torque about 0r i i i i

i i m m r F r g rg r g rg r rg r g g

Now if the total torque is zero, we must have

0iimM=∑r or

which is the definition of the center of mass. So

0CM0 about or center of gravitycenter of mass. τ == =r

Integrating twice with gives 0==rv

rt m =x

Fmm t rx

DYNAMICS OF A SYSTEM OF PARTICLES 281

0 CM

0 CM

F t

F t

F m

vxax 9-7.

y a

a xm a

9-8. By symmetry, 0x=. Also, by symmetry, we may integrate over the x > 0 half of the triangle to get y. σ = mass/area a x

a x yd yd x a y dy dxσσ

282 CHAPTER 9 9-9.

v v

Let the axes be as shown with the projectile in the y-z plane. At the top just before the explosion,

mmvE v where and are the masses of the fragments. The initial momentum is 1m2m

The final momentum is

:y yp Em m m v or v Em m +==+ or :0zzzm p mv m v v vm

The energy equation is y zE

or

Substituting for yv and v gives 1

DYNAMICS OF A SYSTEM OF PARTICLES 283

Em m m v

zm v m

=− gives

2Em m v

So travels straight down with speed =1m1v travels in the y-z plane 2m tan tanyz z

Em m v v m m

m mvvm θ −−

9-10.

x x y B

First, we find the time required to go from A to B by examining the motion. The equation for the y-component of velocity is

At B, v; thus 0y=0sinBtvgθ=. The shell explodes giving m and horizontal velocities v and (in the original direction). We solve for and using conservation of momentum and

energy.

mvEm (3)

Solving for in (2) and substituting into (3) gives an equation quadratic in v. The solution is 2v1 cos mE v

and therefore we also must have cos mE v

Now we need the positions where m and m land. The time to fall to the ocean is the same as the time it took to go from A to B. Calling the location where the shell explodes x = 0 gives for the positions of and upon landing:

sinv x v v

Using (4) and (5) and simplifying gives sin 2v m mE gm m m m θ 2xx

9-1. The term in question is

For n = 3, this becomes

But by Eq. (9.1), each quantity in parentheses is zero. Thus 3

a) 00lnm v u m

DYNAMICS OF A SYSTEM OF PARTICLES 285 b) Relative to Stumblebum’s original frame of reference we have: Before throwing tank

After throwing the tank we want Stumblebum’s velocity to be slightly greater than 3 m/s (so that he will catch up to the orbiter).

(This velocity is relative to Stumblebum’s original reference frame; i.e., before he fires his pressurized tank.) Since Stumblebum is traveling towards the orbiter at 2.02 m/s, he must throw the tank at v = 9 m/s + 2.02 m/s

9-13. From Eq. (9.9), the total force is given by

()e α αβ

As shown in Section 9.3, the second term is zero. So the total force is

It is given that this quantity is zero. Now consider two coordinate systems with origins at 0 and 0′ m r ′r

where is a vector from 0 to 0′ 0r αr is the position vector of mα in 0 α′r is the position vector of mα in 0′

The torque in 0 is given by α αα τ=×∑rF

The torque in 0′ is

e e r F rF r F

9-14. Neither Eq. (9.1) or Eq. (9.31) is valid for a system of particles interacting by magnetic forces. The derivations leading to both of these equations assumes the weak statement of Newton’s Third Law [Eq. (9.31) assumes the strong statement of the Third Law also], which is αββα=−f

That this is not valid for a system with magnetic interactions can be seen by considering two particles of charge and moving with velocities and : 1q2q1v2v v f vf

Now ijiiijq=×fvB where ijB is the magnetic field at due to the motion of iqjq.

DYNAMICS OF A SYSTEM OF PARTICLES 287

Since ijf is perpendicular to both and ivijB (which is either in or out of the paper), ijf can only be parallel to jif if and ivjv are parallel, which is not true in general.

Thus, equations (9.1) and (9.31) are not valid for ystem of particles with magnetic interactions.a s

9-15. σ = mass/length

F dt = becomes where m is the mass of length x of the rope. So ;mxmxσσ== dv xg x x v dv dx xg x v

dx dt dv xg xv v

Try a power law solution:

Substituting,

Since this must be true for all x, the exponent and coefficient of x must be the same on both sides of the equation.

Thus we have: 1 = 2n or 1

g gana=+=

So

3 3 gx g gxdv dv dx dv

gL mgL

2f L Umghmg==−

Energy lost 6 mgL=

9-16.

T x

The equation of motion for the falling side of the chain is, from the figure, bx b x

From Example 9.2, we have for the energy conservation case gb x x x xg g

Substitution gives us 2

To find the tension on the other side of the bend, change to a moving coordinate system in which the bend is instantaneously at rest. This frame moves downward at a speed 2ux= with respect to the fixed frame. The change in momentum at the bend is

DYNAMICS OF A SYSTEM OF PARTICLES 289

Equating this with the net force gives 2

Using equation (3), we obtain 2

as required. Note that equation (5) holds true for both the free fall and energy conservation cases.

9-17. As the problem states, we need to perform the following integral

Our choice of ε is 10 for this calculation, and the results are shown in the figure. We plot the natural velocity

dα τ

9-18. Once we have solved Problem 9-17, it becomes an easy matter to write the expression for the tension (Equation 9.18):

This is plotted vs. the natural time using the solution of Problem 9-17.

T mg

9-19.

released at t = 0 at time t ceiling table b

The force that the tabletop exerts on the chain counteracts the force due to gravity, so that we may write the change in momentum of the center of the mass of the chain as

which has a time derivative

. Setting this last expression equal to (1) gives us

Although M. G. Calkin (personal communication) has found that experimentally the time of fall for this problem is consistently less than the value one would obtain in the above treatment by about 1.5%, he also finds evidence that suggests the free fall treatment is more valid if the table is energy absorbing.

DYNAMICS OF A SYSTEM OF PARTICLES 291

9-20.

a – x a + x

Let ρ = mass/length The force on the rope is due to gravity

F ax g a x gxg ρ ρρ

2 dp dv dv ma dt dt dt ρ==

So dp

F dt = becomes xg a dt =

Now dv dv dx dv v dt dx dt dx =⋅=

So xg av dx = or vdv xdxa =

Integrating yields

Since v = 0 when x = 0, c = 0. Thus vx a =

When the rope clears the nail, x = a. Thus vga=

9-21. Let us call the length of rope hanging over the edge of the table, and the total length of the rope. The equation of motion is x L mgx gxdx mx

Let us look for solution of the form ttxAeBeωω−=+ Putting this into equation of motion, we find gL ω=

9-2. Let us denote (see figure) m and 2m mass of neutron and deuteron respectively

0v velocity of deuteron before collision 1v and v velocity of neutron and deuteron, respectively after collision 2

m α ψ a) Conservation of energy:

Conservation of momentum is

Solving these equations, we obtain 2 sets of solutions

DYNAMICS OF A SYSTEM OF PARTICLES 293

1 19.79km/sv = 2 5.12km/sv = b) Let us call α the lab scattering angle of the neutron, then from the sine theorem we have sin sin mv mv v αψψα=⇒=

c) From a) we see that 2 02 02

8 v vv

9-23. Conservation of momentum requires fv to be in the same direction as u (component of 1 fv ⊥ to must be zero). 1u

if f m

The fraction of original kinetic energy lost is if i mu mu m m

K m

K mu

m m m m

294 CHAPTER 9 9-24.

ω b x ω

The energy of the system is, of course, conserved, and so we have the following relation involving the instantaneous velocity of the particle:

The angular momentum about the center of the cylinder is not conserved since the tension in the string causes a torque. Note that although the velocity of the particle has both radial and angular components, there is only one independent variable, which we chose to be θ. Here ωθ= is the angular velocity of the particle about the point of contact, which also happens to be the rate at which the point of contact is rotating about the center of the cylinder. Hence we may write

9-25. The best elements are those that will slow down the neutrons as much as possible. In a collision between (the neutron) and (moderator atom), we would thus want to minimize

(kinetic energy of the neutron after the collision); or alternatively, maximize T (kinetic energy of the moderator atom after the collision). From Eq. (9.8)

4 cos Tm m

Since one cannot control the angle ζ, we want to maximize the function

DYNAMICS OF A SYSTEM OF PARTICLES 295 with respect to . (m = constant) 2m1

0 when m mdf m

By evaluating 2 one can show that the equilibrium point is a maximum. Thus, is a

maximum for . Back to reactors, one would want elements whose mass is as close as possible to the neutron mass (thus, as light as possible). Naturally, there are many other factors to consider besides mass, but in general, the lower the mass of the moderator, the more energy is lost per collision by the neutrons.

where f is the force acting on the first particle due to the second particle. Now 12

r k kr v

Nr r f r r r r internal torque vanishes only if the internal force is directed along the line joining two particles. The system is not conservative.

9-27. The equation for conservation of yp in the lab system is (see fig. 9-10c):

1 2 20s in simv m v nψζ= − Thus sin sin mvmv ζψ= or sin sin mTmT ζψ=

9-28. Using the notation from the chapter: 10:,ifmTTTT1==

If we want the kinetic energy loss for to be a maximum, we must minimize 1m10T T or,

From Eq. (9.8):

4 cos Tm m

To maximize this, ζ = 0 (it can’t = 180°).

T T TT 2 0T

4TT m

ζ = 0 implies ψ = 0, 180° (conservation of ). So the reaction is as follows vp Before:

After:

m v m m v m v

m v

So travels in + x direction 2m

DYNAMICS OF A SYSTEM OF PARTICLES 297 direction if

9-29. From Eq. (9.69) sin tan cos m θψθ=+

Substituting gives sin 2 tan

or sin 2 tan cos 2mm ζψζ=−

9-30. Before:

After:

m v m m v m v

=− ⋅ The impulse P is the change in momentum.

So

9-31. From Eq. (9.69) sin tan

cos m

Substituting gives sin

tan cos m

9-32. 1ipmu=

Tm u mv mv mu m u u v v mv mu v mv

0 implies 26or 3 dT u uv v

0, so this is a maximum

DYNAMICS OF A SYSTEM OF PARTICLES 299 9-3. From Eq. (9.87b) in the text, we have cos sin

1 cos sin 2 cos sin

Tm m Tm m m m

Substituting 21mmα≡ and cos ψ ≡ y we have

α = 12

α = 4 α = 2α = 1

9-34. Before After m m x m u θ

Cons. of zp: 112cos45cosmumvmvθ=°+ (1)

Cons. of yp: 120sin45simvmvnθ=°− (2) Cons. of energy (elastic collision)

Solve (1) for cos θ :

Solve (2) for sin θ :

sin 2v v θ=

Combining this with (3) gives

112vu= Substitute into (3) and the result is

9-35. From the following two expressions for 10TT,

TvTu = cossinEq. (9.87b) m

we can find the expression for the final velocity of in the lab system in terms of the scattering angle ψ : 1v 1m cos sin mu m

m m v ψ ψ

If time is to be constant on a certain surface that is a distance r from the point of collision, we have

cos sin mu t m

m m r ψ ψ

This is the equation of the required surface. Let us consider the following cases:

DYNAMICS OF A SYSTEM OF PARTICLES 301 i) : 21mm=

(The possibility r = 0 is uninteresting.) iI) : 212mm=

mu t

m m m m and taking the limit , we find 2m→∞

v t–v t m = 2m m = ∞ m = m m O

− vt

This result is useful in the design of a certain type of nuclear detector. If a hydrogenous material is placed at 0 then for neutrons incident on the material, we have the case . Therefore, neutrons scattered from the hydrogenous target will arrive on the surface A with the same time delay between scattering and arrival, independent of the scattering angle. Therefore, a coincidence experiment in which the time delay is measured can determine the energies of the incident neutrons. Since the entire surface A can be used, a very efficient detector can be constructed.

9-36. Since the initial kinetic energies of the two particles are equal, we have

or,

Now, the kinetic energy of the system is conserved because the collision is elastic. Therefore,

Substituting the second equality in (4) into (3), we find m mu m u

or,

Using 212mmα=, (6) becomes

solving for α, we obtain

Impulse

Fd t

kgm

Since the initial velocity is zero, fvv=∆ ImpulsePmv=∆=∆

So

DYNAMICS OF A SYSTEM OF PARTICLES 303

muzzle m 480 s

Tm u

Thus,

01 T vTu

Now, from the diagram above, we have

cos cosm

Thus, cos v mV

Using Eq. (9.84) in the text,

Therefore, we find cos cos Tm

Tm m θψψ

If we define cos cosS

we have

Tm S

as desired.

9-39.

θ′θ u y x

As explained in Section 9.8, the component of velocity parallel to the wall is unchanged. So sinxvuθ= yv is given by cos y v uεθ== or cosyvuεθ= Thus

sin tan cosuu θθεθ=′ or

DYNAMICS OF A SYSTEM OF PARTICLES 305

9-40. Because of the string, is constrained to move in a circle of radius a. Thus, initially, will move straight up (taken to be the y direction). Newton’s rule applies to the velocity component along . The perpendicular component of velocity (which is zero) is unchanged. Thus will move in the original direction after the collision.

From conservation of yp we have

1s in straight up sinmummεαα+=+v

(2) then gives

sin along sin ummmmαεα−=+uv yvtgt=− and v, we can get the velocities before and after the

where 2 ugthg=−=1t

hv t gt

v vg

Thus ghvv

lostifTTT=−

Fraction lost1 if iTT T

uv h h h uh

9-42.

5 m/sθ

As explained in Section 9.8, the velocity component in the y-direction is unchanged.

For the x component we have

m 3m5c os30 5 s 2s x x v vvu m 23sxv=

DYNAMICS OF A SYSTEM OF PARTICLES 307 9-43.

αT v

Conservation of gives xp

or mT mv

Conservation of yp gives mvmvsinα=−

or sin 42v v α=

mT m v mT mvv

vm v

Simplifying gives vT mTv

The equation for conservation of energy is

or

Substituting (1) into (2) gives a quadratic in : 1v

Substituting this into the previous expressions for cos α and sin α and dividing gives

cos ααα==

Thu s , the recoil angle of the helium, is 5.8.α° v x grav impulse 0 0 where mass/length , since ,0.

Fm g xg

F mvm vm v v v v µ µ== =

d x mx

So the total force is

9-45. Since the total number of particles scattered into a unit solid angle must be the same in the lab system as in the CM system [cf. Eq. (9.124) in the text],

()()2sin2sinddσθπθθσψπψψ=⋅ (1) Thus,

sin d ψψσθσψθθ= (2)

DYNAMICS OF A SYSTEM OF PARTICLES 309

The relation between θ and ψ is given by Eq. (9.69), which is sin tan

where 1xmm=2. Using this relation, we can eliminate ψ from (2):

x 2 siθψ== (4) θθ

tan cos cos sin cos

tan cosdx d d x d ψ θθψψ ψθψ θ θ++== + θ

(5)

1t an ψcoψ=+, (5) becomes

1c os 1c os1

sin 12 coscos1 cos x x dx x x d x θ θψ

= (6)

Substituting (4) and (6) into (2), we find

1c os

9-46. The change in angle for a particle of mass µ moving in a central-force field is [cf. Eq. (9.121)]. Let ψ = capital θ here.

rd r

b a r θ

In the scattering from an impenetrable sphere, is the radius of that sphere. Also, we can see from the figure that minr 2θπψ=−.

(Parte 1 de 2)

Comentários