# Mecflu - mec?nica newtoniana- ?nica part?cula

(Parte 1 de 2)

Newtonian Mechanics— Single Particle

() ( )i dx mf x gdt gtdx dt

fx m

2-2. Using spherical coordinates, we can write the force applied to the particle as rFθθφφ=++Feee (1)

But since the particle is constrained to move on the surface of a sphere, there must exist a reaction force that acts on the particle. Therefore, the total force acting on the particle is rrF−e

rR=re (3) where R is the radius of the sphere and is constant. The acceleration of the particle is

We must now express in terms of , re reθe, and φe. Because the unit vectors in rectangular coordinates, e, , e, do not change with time, it is convenient to make the calculation in

sin cos sin sin cos cos cos cos sin sin sin cos r θ φ e e e e e e e e (5)

Then ( ) ( )12 sin sin cos cos cos sin sin cos sin

sinr φθ

3φ θφ θ θ φ θ θφ φ θ φ θ

which is the only second time derivative needed. The total force acting on the particle is

2 sin cos 2c os sin Fm R

Fm Rθφ

NEWTONIAN MECHANICS—SINGLE PARTICLE 31

2-3. y v P

The gravitational force is the only applied force; therefore,

Fm x

Integrating these equations and using the initial conditions,

0c os 0s in xt v

We find

cos sin xt v

So the equations for x and y are cos xt v t

Suppose it takes a time t to reach the point P. Then, 0 0 cos cos

1 sin sin

Eliminating A between these equations,

2s in 21 cos tan 0 v v t

from which

2-4. One of the balls’ height can be described by 2002yyvtgt=+−. The amount of time it takes to rise and fall to its initial height is therefore given by 02vg. If the time it takes to cycle the ball through the juggler’s hands is 0.9sτ=, then there must be 3 balls in the air during that time τ. A single ball must stay in the air for at least 3τ, so the condition is 023vgτ≥, or . 1013.2msv−≥⋅ mg N point of maximum acceleration flightpath plane e a) From the force diagram we have ()2rmmvR−=Nge. The acceleration that the pilot feels is ()2rmmvR=+Nge, which has a maximum magnitude at the bottom of the maneuver.

b) If the acceleration felt by the pilot must be less than 9g, then we have

v R

A circle smaller than this will result in pilot blackout.

2-6.

Let the origin of our coordinate system be at the tail end of the cattle (or the closest cow/bull).

a) The bales are moving initially at the speed of the plane when dropped. Describe one of these bales by the parametric equations

NEWTONIAN MECHANICS—SINGLE PARTICLE 3

where , and we need to solve for . From (2), the time the bale hits the ground is 080 my=0x

b) She could drop the bale earlier by any amount of time and not hit the cattle. If she were late by the amount of time it takes the bale (or the plane) to travel by 30 m in the x-direction, then

2-7. Air resistance is always anti-parallel to the velocity. The vector expression is:

W wρ va

where 2wAmbcρ=. Solving with a computer using the given values and , we find that if the rancher drops the bale 210 m behind the cattle (the answer from the previous problem), then it takes 4.4 s to land 62.5 m behind the cattle. This means that the bale should be dropped at 178 m behind the cattle to land 30 m behind. This solution is what is plotted in the figure. The time error she is allowed to make is the same as in the previous problem since it only depends on how fast the plane is moving.

With air resistance No air resistance

34 CHAPTER 2 2-8.

P Q x y v α h

In order to calculate the time when a projective reaches the ground, we let y = 0 in (2):

Substituting (4) into (1) we find the relation between the range and the angle as

v xt

v xt gt

Eliminating t between these equations yields

We can find the x-coordinate of the projectile when it is at the height h by putting y = h in (7):

This equation has two solutions:

v xv

v xv

NEWTONIAN MECHANICS—SINGLE PARTICLE 35 where corresponds to the point P and to Q in the diagram. Therefore, 1x2x

21 0 4v dx x v ghg

2-9.

a) Zero resisting force (): 0rF= The equation of motion for the vertical motion is:

where v is the initial velocity of the projectile and t = 0 is the initial time. 0

The time required for the projectile to reach its maximum height is obtained from (2). Since corresponds to the point of zero velocity, mt

0m v t g b) Resisting force proportional to the velocity ()rFkmv=−: The equation of motion for this case is:

where –kmv is a downward force for mtt<′ and is an upward force for mtt>′. Integrating, we obtain

() 0 ktgk v g vt ekk

For t, v(t) = 0, then from (6), mt=

which can be rewritten as

0ln 1m kv kt

Since, for small z (z 1) the expansion

is valid, (8) can be expressed approximately as vk v kv t

which gives the correct result, as in (4) for the limit k → 0.

2-10. The differential equation we are asked to solve is Equation (2.2), which is . Using the given values, the plots are shown in the figure. Of course, the reader will not be able to distinguish between the results shown here and the analytical results. The reader will have to take the word of the author that the graphs were obtained using numerical methods on a

computer. The results obtained were at most within 10

10 v vs t t (s) v (m/s) v vs x x (m) v (m/s)

100 x vs t t (s) x (m)

2-1. The equation of motion is

22dx m kmv mgdt

This equation can be solved exactly in the same way as in problem 2-12 and we find

NEWTONIAN MECHANICS—SINGLE PARTICLE 37 gk v x

1 log gk v sv v

2-12. The equation of motion for the upward motion is

22dx m kvdt

Using the relation d x dv dv dx dv v

we can rewrite (1) as vd v dx

Integrating (3), we find

log2 kv g x Ck where the constant C can be computed by using the initial condition that when x = 0: 0vv=

log2 C kvk

Therefore, kv g x

Now, the equation of downward motion is

22dx m kvdt

This can be rewritten as vd v dx

Integrating (8) and using the initial condition that x = 0 at v = 0 (w take the highest point as the origin for the downward motion), we find g x

At the highest point the velocity of the particle must be zero. So we find the highest point by substituting v = 0 in (6):

kv g x

Then, substituting (10) into (9), kv g g

Solving for v, g v

kv g v

We can find the terminal velocity by putting x → ∞ in(9). This gives t g v k

Therefore,

2-13. The equation of motion of the particle is

Integrating, dv k dt

and using Eq. (E.3), Appendix E, we find v kt C

Therefore, we have

2 Atv

NEWTONIAN MECHANICS—SINGLE PARTICLE 39 and where C′ is a new constant. We can evaluate C′ by using the initial condition, at t = 0:

Substituting (5) into (4) and rearranging, we have aC e dx v

Now, in order to integrate (6), we introduce At−≡ue so that du = –Au dt. Then,

1 AtAt aC e a C u du xd t

Ce A C u u

aC du A Cu u

Using Eq. (E.8c), Appendix E, we find

Again, the constant C″ can be evaluated by setting x = 0 at t = 0; i.e., x = 0 at u = 1:

Therefore, we have

Using (4) and (5), we can write

1 sin sin vava

From (6) we see that v → 0 as t → ∞. Therefore,

Also, for very large initial velocities,

lim sin sin 1

Therefore, using (1) and (12) in (10), we have

() 2xt ka and the particle can never move a distance greater than 2kaπ for any initial velocity.

y x d α β a) The equations for the projectile are 0 cos

1 sin xv t yv t gtαα= =−

Solving the first for t and substituting into the second gives

1 tan

2c os gx yxvαα=−

Using x = d cos β and y = d sin β gives cos sin cos tan

2c os cos 0c os tan

2c os gd d gd d β sinβ αβα

2c os tan sin cos cos

2c os sin cos cos sin cos v d g v g

2c os sin cos v d

NEWTONIAN MECHANICS—SINGLE PARTICLE 41 b) Maximize d with respect to α

0s in sin cos cos cos 2 cos

c) Substitute (2) into (1) max 2 cos sin

Using the identity siABABA−=+−B

we have max 2 sin sin 1s in2 2

cos 2 1 sinvv g d π β β

0 max 1s in v dgβ=+

2-15.

mg θ mg sin θ

The equation of motion along the plane is

Rewriting this equation in the form

dtgk vk θ =−

We know that the velocity of the particle continues to increase with time (i.e., 0dvdt>), so that

()2singkvθ>. Therefore, we must use Eq. (E.5a), Appendix E, to perform the integration. We find

1 tanh

sin sin v t C k gk θθ

The initial condition v(t = 0) = 0 implies C = 0. Therefore,

( )sin tanh sin g dx v gkkd θ θ= t t

We can integrate this equation to obtain the displacement x as a function of time:

( )sin tanh sing xg k θθ=∫tdt

Using Eq. (E.17a), Appendix E, we obtain

( )ln cosh sin sin

sin gk tg

The initial condition x(t = 0) = 0 implies C′ = 0. Therefore, the relation between d and t is

( )1 ln cosh sind gkk

From this equation, we can easily find dke

2-16. The only force which is applied to the article is the component of the gravitational force along the slope: mg sin α. So the acceleration is g sin α. Therefore the velocity and displacement along the slope for upward motion are described by:

At the highest position the velocity becomes zero, so the time required to reach the highest position is, from (1),

At that time, the displacement is

NEWTONIAN MECHANICS—SINGLE PARTICLE 43

For downward motion, the velocity and the displacement are described by

where we take a new origin for x and t at the highest position so that the initial conditions are v(t = 0) = 0 and x(t = 0) = 0.

We find the time required to move from the highest position to the starting position by substituting (4) into (6):

Adding (3) and (7), we find

2-17. v0

Fence35˚ 0.7 m60 m

The setup for this problem is as follows:

R = 60 m. It must be at least h = 2 m high, so we also need 200sinvgτθτ−2hy−=. Solving for , we obtain 0v gR v

2-18.

a) The differential equation here is the same as that used in Problem 2-7. It must be solved for many different values of v in order to find the minimum required to have the ball go over the fence. This can be a computer-intensive and time-consuming task, although if done correctly is

, and the trajectory is shown in Figure (a). (We take the density of air as 13 kgρ−=.⋅.)

With air resistance No air resistance fence height fence range x (m) y (m) b) The process here is the same as for part (a), but now we have v fixed at the result just obtained, and the elevation angle θ must be varied to give the ball a maximum height at the fence. The angle that does this is 0.71 rad = 40.7°, and the ball now clears the fence by 1.1 m.

This trajectory is shown in Figure (b).

Flight Path fence height fence range x (m) y (m)

NEWTONIAN MECHANICS—SINGLE PARTICLE 45 2-19. The projectile’s motion is described by cos xv t

where v is the initial velocity. The distance from the point of projection is 0

0 x y r

Using (1), we have

v v t

For small values of α, the second term in (5) is imaginary. That is, r = 0 is never attained and the value of t resulting from the condition 0r= is unphysical.

Only for values of α greater than the value for which the radicand is zero does t become a physical time at which does in fact vanish. Therefore, the maximum value of α that insures for all values of t is obtained from r 0r >

or,

2-20. If there were no retardation, the range of the projectile would be given by Eq. (2.54):

The angle of elevation is therefore obtained from

Rgv θ =

kV R

vk vgg θ θ

Since we expect the real angle θ to be not too different from the angle 0θ calculated above, we can solve (4) for θ by substituting 0θ for θ in the correction term in the parentheses. Thus,

4s in 1 kv v

Next, we need the value of k. From Fig. 2-3(c) we find the value of km by measuring the slope of the curve in the vicinity of v = 140 m/sec. We find ()()110 N500 m/s0.2 kg/skm≅≅. The curve is that appropriate for a projectile of mass 1 kg, so the value of k is

Substituting the values of the various quantities into (5) we find 17.1θ=°. Since this angle is somewhat greater than 0θ, we should iterate our solution by using this new value for 0θ in (5). We then find 17.4θ=°. Further iteration does not substantially change the value, and so we conclude that

If there were no retardation, a projectile fired at an angle of 17.4° with an initial velocity of 140 m/sec would have a range of

9.8 m/sec

NEWTONIAN MECHANICS—SINGLE PARTICLE 47

2-21. x α v x Assume a coordinate system in which the projectile moves in the 2xx3− plane. Then, cos xv t

or, re e

Using the property of the unit vectors that 3ijijkε×=e, we find

This gives

1 cos sin

cos vt vt gt mg which is the same result as in (6).

48 CHAPTER 2 2-2.

x y

a) Note that when E = 0, the force is always perpendicular to the velocity. This is a centripetal acceleration and may be analyzed by elementary means. In this case we have also so that ⊥vB vB×=vB.

centripetal mv ma qvBr

Solving this for r

withcqBmω≡/.

b) Here we don’t make any assumptions about the relative orientations of v and B, i.e. the velocity may have a component in the z direction upon entering the field region. Let

, with xyz=++rijk=vr and . Let us calculate first the v × B term. =ar

(0 xy z B y xB

The Lorentz equation (1) becomes

cqB xy m

y c qE EqB yx x zqE

NEWTONIAN MECHANICS—SINGLE PARTICLE 49

The z-component equation of motion (8) is easily integrable, with the constants of integration given by the initial conditions in the problem statement.

zqE

c) We are asked to find expressions for and , which we will call and x y xvyv, respectively. Differentiate (6) once with respect to time, and substitute (7) for yv xc y c x E

or xc x c E v B

This is an inhomogeneous differential equation that has both a homogeneous solution (the solution for the above equation with the right side set to zero) and a particular solution. The most general solution is the sum of both, which in this case is

() ()12cos sin y xc c E vC t C t B

where K is yet another constant of integration. It is found upon substitution into (6), however, that we must have K = 0. To compute the time averages, note that both sine and cosine have an average of zero over one of their periods 2cTπω≡.

d) We get the parametric equations by simply integrating the velocity equations.

() ()12sin cos y c c

() ()12cos sinc c y c where, indeed, D and xyD are constants of integration. We may now evaluate all the C’s and

(() sin c c

These cases are shown in the figure as (i) yAEB>, (i) yAEB<, and (ii) yAEB=. (i)

(i) (i)

 2-23. ()() atFtmatkte−== (1)

with the initial conditions x(t) = v(t) = 0. We integrate to get the velocity. Showing this explicitly,

Integrating this by parts and using our initial conditions, we obtain

vt t em α αα α

By similarly integrating v(t), and using the integral (2) we can obtain x(t).

xt t em α αα α α

NEWTONIAN MECHANICS—SINGLE PARTICLE 51

2-24. mg d=length of incline s=distance skier travels along level ground θ x mg sin θ mg cos θ y x

While on the plane: FN∑ so Nmcos0ymgmyθ=−== cosgθ= sinxfFm∑; gFθ=−cosfFNmgµµθ== sin cosmg mg mxθ µθ− =

So the acceleration down the plane is:

()1sincos constantagθµθ=−=

While on level ground: ; Nm=′ggfFmµ=−

2constantagµ=−=

For motion with constant acceleration, we can get the velocity and position by simple integration:

Solving (1) for t and substituting into (2) gives:

v v v v x

or

Using this equation with the initial and final points being the top and bottom of the incline respectively, we get:

Using the same equation for motion along the ground:

()sincosgdgsθµθµ−=

Solving for µ gives sin cosdds θµθ=+

Substituting θ = 17°, d = 100 m, s = 70 m gives 0.18µ=

Substituting this value into (3):

NEWTONIAN MECHANICS—SINGLE PARTICLE 53

2-25.

a) At A, the forces on the ball are: N mg The track counters the gravitational force and provides centripetal acceleration

(Parte 1 de 2)