Greenwood D.T. Advanced Dynamics

Подождите немного. Документ загружается.

59 Impulse response

Then substitution into (1.315) yields

θ =−

mlv

√

(1.335)

resulting in the angular velocity

θ =−

√

2 v

(1.336)

The velocity of particle 1, from (1.331), is

=−

i (1.337)

Second method Let the reference point P be inertially ﬁxed at the impact point on the

surface. Any external impulse on the system acts through P so

= 0 (1.338)

and we have conservation of momentum about P.

The angular momentum before impact is due entirely to translation of particle 2.

=−

mlv

√

k (1.339)

After impact, the center of mass moves vertically downward, while particle 1 moves hori-

zontally away from P due to the inelastic impact assumption. Thus, from the geometry, the

velocity of particle 2 immediately after impact is

√



−

i + j



(1.340)

The resulting total angular momentum about P is

= ρ

× mv

(1.341)

where

√

(i + j) (1.342)

Thus we obtain

θk (1.343)

Finally, setting H

= H

, we ﬁnd that

θ =−

√

2 v

(1.344)

60 Introduction to particle dynamics

in agreement with (1.336). The center of mass has no horizontal motion, so v

is the negative

of the x component of v

√

i =−

i (1.345)

Third method Consider the effect of the vertical reaction impulse

R acting on particle 1

and due to its impact with the ﬁxed frictionless surface. Take longitudinal and transverse

components of

R, relative to the dumbbell, where each component is of magnitude

√

The transverse impulse results in a sudden change v

in the transverse velocity of particle

1, as shown in Fig. 1.25b.

v

√

(1.346)

The velocity of particle 2 is not changed by this transverse impulse. On the other hand,

the longitudinal component of

R affects both particles having a total mass 2m.

v

= v

√

v

(1.347)

The ﬁnal velocity of particle 1 is equal to its initial velocity plus the changes due to the

reaction impluse

R, as shown in Fig. 1.25b. From the vertical components we obtain

v

√

(1.348)

Then, from the horizontal components we obtain

=−

v

√

i =−

i (1.349)

The angular velocity

θ is zero before impact. During impact, the only impulse acting on

particle 2 is longitudinal in direction, so its transverse velocity is unchanged. Therefore, the

angular velocity immediately after impact is

θ =−

v

=−

√

2 v

(1.350)

The most straightforward method discussed here is the second method. It uses conserva-

tion of angular momentum about a ﬁxed point chosen so that the applied angular impulse

equals zero.

Generalized impulse and momentum

The conﬁguration of a system of N particles is given by the position vectors r

(q, t). The

corresponding velocities are



j=1

∂r

∂q

∂r

∂t

(k = 1,...,N ) (1.351)

61 Impulse response

Recall from (1.234) that the velocity coefﬁcients are

∂r

∂q

, γ

∂r

∂t

(1.352)

Thus we obtain



j=1

+ γ

(k = 1,...,N ) (1.353)

The equation of linear momentum for the kth particle is

v

(1.354)

where

is the total impulse acting on the particle. Now take the dot product with γ

and

sum over the particles to obtain



k=1

v

· γ



k=1

· γ

(1.355)

Let us assume that all the

impulses occur during an inﬁnitesimal interval t, implying

that the conﬁguration and the values of the velocity coefﬁcients remain constant during this

interval. From (1.353) we see that

v



j=1



(1.356)

Substitute this expression for v

into (1.355) and obtain



j=1



k=1

· γ





k=1

· γ

(1.357)

Recall from (1.268) that the generalized mass coefﬁcient m

can be expressed as

= m



k=1

· γ

(1.358)

Furthermore, similar to (1.240), the generalized impulse

is given by



k=1

· γ

(i = 1,...,n) (1.359)

Finally, we obtain the equation of generalized impulse and momentum:



j=1

(q, t) 

(i = 1,...,n) (1.360)

62 Introduction to particle dynamics

In general,

includes the contributions of applied and constraint impulses, both internal

and external to the system. However, if the qsareindependent,the

Qs are due to the applied

impulses only.

Example 1.15 Two particles with masses m

and m

are connected by a rigid massless

rod of length l. Initially the system is motionless with a vertical orientation, as shown in

Fig. 1.26. Then a horizontal impulse

F is applied to the upper particle of mass m

.Wewish

to solve for the velocity

x and the angular velocity

θ immediately after the impulse.

Figure 1.26.

At the outset we see that the center of mass must move horizontally due to

F, and therefore

neither particle can have any vertical velocity initially. Let us choose (x ,θ) as generalized

coordinates consistent with this motion. The kinetic energy at this time is

T =

(

x −l

θ)

(1.361)

From (1.261), the mass coefﬁcients are

∂

= m

+ m

xθ

= m

θ x

∂

x∂

=−m

θθ

∂

∂θ

= m

(1.362)

The velocity coefﬁcients for particle 1 are

= i, γ

1θ

= 0 (1.363)

where i is a horizontal unit vector. Then (1.359) gives the generalized impulses

= 0 (1.364)

63 Impulse response

A substitution into the equation of generalized impulse and momentum yields the following

equations:

+ m

)

x −m

l

θ =

F (1.365)

−m

l

x +m



θ = 0 (1.366)

The solutions are



x =

,

θ =

(1.367)

which are the values of

x and

θ immediately after the impulse. Note that these results are

independent of the mass m

of the lower particle which remains momentarily motionless.

Impulse and energy

Consider a particle of mass m and velocity v

which is subject to a total impulse

F.From

the principle of linear impulse and momentum we know that

m(v − v

) =

F (1.368)

where v is the velocity of the particle immediately after the impulse. Now take the dot

product with

(v + v

). We ﬁnd that the increase in kinetic energy is



− v



F · (v + v

) (1.369)

Then, by the principle of work and kinetic energy, the work done on the particle is

W =

F · (v + v

) (1.370)

This result is valid even if the force F and the time interval are ﬁnite and arbitrary because

these assumptions apply to the principle of linear impulse and momentum (1.368). For

motion in a straight line, the work done on a particle is equal to the impulse times the mean

of the initial and ﬁnal velocities.

Example 1.16 A ﬂexible inextensible rope of length L and linear density ρ initially lies

straight and motionless on a smooth horizontal ﬂoor (Fig. 1.27). Then, at t = 0, end A is

suddenly moved to the right with a constant speed v

. Considering the rope as a system of

particles, let us look into the applicable impulse and momentum principles, as well as work

and kinetic energy relations.

Assuming that x (0) = 0, y(0) = 0, we see that

y = 2x (1.371)

64 Introduction to particle dynamics

L−x

Figure 1.27.

and therefore

y = 2

x = v

(1.372)

Consider a mass element ρdx which suddenly has its velocity changed from zero to v

Using the principle of linear impulse and momentum, the required impulse is

Fdt = ρv

dx (1.373)

Thus, the force is

F = ρv

x =

ρv

(1.374)

The remainder of the moving rope has zero acceleration, so the force F is constant and is due

to successive elements ρdx undergoing a sudden change in velocity. This continues over

the interval 0 < t < 2L/v

, that is, until the rope is again straight with end A leading B.

For t > 2L/v

, the applied force equals zero. Over the entire interval, the applied impulse

F = Ft =



ρv





= ρ Lv

(1.375)

This is equal to the increase in momentum.

Now consider the work done on the system by the force F.

W = 2FL = ρ Lv

(1.376)

This is equal to twice the ﬁnal kinetic energy, where

T =

ρ Lv

(1.377)

The energy lost in the process is W − T and is due to what amounts to inelastic impact as

each element suddenly changes its velocity from zero to v

Another viewpoint is obtained by considering an inertial frame in which end A is ﬁxed

for t > 0 and end B moves to the left with velocity v

. In this case,

F =

ρv

(1.378)

as before, but F does no work since A is ﬁxed. The lost energy is equal to the initial kinetic

energy

ρ Lv

because the ﬁnal energy is zero. We see that, although the work of F as well

65 Bibliography

as the initial and ﬁnal kinetic energies are different in the two inertial frames, the lost energy

is unchanged. Relative to this second inertial frame, it is more obvious that the lost energy

is due to inelastic impact, since the velocity of each element ρdx has its velocity suddenly

changed from v

to zero.

1.6 Bibliography

Desloge, E. A. Classical Mechanics, Vol. 1. New York: John Wiley and Sons, 1982.

Ginsberg, J. H. Advanced Engineering Dynamics, 2nd edn. Cambridge, UK: Cambridge University

Press, 1995.

Greenwood, D. T. Principles of Dynamics, 2nd edn. Englewood Cliffs, NJ: Prentice Hall, 1988.

1.7 Problems

1.1. A particle of mass m can slide without friction on a semicircular depression of

radius r in a block of mass m

. The block slides without friction on a horizontal

surface. Consider planar motion starting from rest with the initial conditions θ(0) =

π/2, x(0) = 0. (a) Find the values of

θ and

x when the particle ﬁrst passes through

θ = 0. (b) What is the maximum value of x in the entire motion?

Figure P 1.1.

1.2. A simple pendulum of mass m and length l is initially motionless in the downward

equilibrium position. Then its support point is given a constant horizontal acceleration

a =

√

3g, where g is the acceleration of gravity. Find the maximum angular deviation

from the initial position of the pendulum.

1.3. Two smooth spheres, each of mass m and radius r , are motionless and touching when

a third sphere of mass m, radius r, and velocity v

strikes them in perfectly elastic

impact, as shown. Solve for the velocities of the three spheres after impact.

66 Introduction to particle dynamics

Figure P 1.3.

1.4. Three particles, each of mass m, can slide without friction on a ﬁxed wire in the

form of a horizontal circle of radius r. Initially the particles are equally spaced with

two particles motionless and the third moving with velocity v

. Assuming that all

collisions occur with a coefﬁcient of restitution e = 0.9, solve for the ﬁnal velocities

of the particles as time t approaches inﬁnity.

1.5. A particle is shot with an initial velocity v

at an angle θ = θ

above the horizontal.

It moves in a uniform gravitational ﬁeld g. Solve for the radius of curvature ρ of the

trajectory as a function of the ﬂight path angle θ.

1.6. A river at 60

◦

north latitude ﬂows due south with a speed v = 10 km/h. Determine

the difference in water level between the east and west banks if the river is

wide. Assume a spherical earth with radius R = 6373 km, acceleration of gravity

g = 9.814 m/s

, and rotation rate ω

= 72.92 × 10

−6

rad/s.

1.7. An elastic pendulum has a spring of stiffness k and unstressed length L. Using the

length l and the angle θ as coordinates, write the differential equations of motion.

Figure P 1.7.

1.8. Point A moves with a constant velocity v

. It is connected to a particle of mass m

by a linear damper whose tensile force is F = c(v

−

x). (a) Assuming the initial

conditions x(0) = 0,

x(0) = 0, solve for the particle velocity

x as a function of time.

(b) Find the work W done on the system by the force F acting on A. (c) Show that

the work W approaches twice the value of the kinetic energy T as time t approaches

inﬁnity.

67 Problems

y = v

Figure P 1.8.

1.9. A linear damper with a damping coefﬁcient c connects two particles, as shown in

Fig. 1.15 on page 31. Show that, for an arbitrary linear motion of the two parti-

cles over some time interval t, the magnitude of the impulse

F applied by the

damper to either particle is directly proportional to the change in length l of the

damper.

1.10. Ten identical spheres are arranged in a straight line with a small space between

adjacent spheres. The spheres are motionless until sphere 1 is given an impulse

causing it to move with velocity v

toward sphere 2. The spheres collide in sequence

with a coefﬁcient of restitution e = 0.9. (a) What is the ﬁnal velocity of sphere 10?

(b) What is the velocity of sphere 9 after its second collision?

1.11. Consider the system of Example 1.14 on page 57 and use (x, y,θ) as generalized

coordinates, where the Cartesian coordinates of particle 1 are (x, y). Use the equation

of generalized impulse and momentum to solve for: (a)

x and

θ immediately after

inelastic impact; (b) the impulse acting on the system at the time of impact due to the

horizontal surface.

1.12. Two particles, each of mass m, are connected by a rigid massless rod of length l. The

system is moving longitudinally with a velocity v

, as shown, and strikes a smooth

ﬁxed surface with perfectly elastic impact, e = 1. (a) Find the velocity components

(

y) of the center of mass and also the angular velocity ω immediately after impact.

(b) Solve for the impulse

F in the rod at impact.

(x, y)

45°

Figure P 1.12.

68 Introduction to particle dynamics

1.13. Particles A, B, and C are each of mass m and are connected by two rigid massless

rods of length l, as shown. There is a joint at B. The three particles are motionless in a

straight conﬁguration when particle D, also of mass m, strikes particle B inelastically

as it moves with velocity v

in a transverse direction. Later, particles A and C collide,

also inelastically, as particles B and D continue to move together. Find: (a) the angular

velocity of rod AB just before A and C collide and (b) the ﬁnal velocity of particle A.

Figure P 1.13.

1.14. A particle of mass m is attached by a string of length l to a point O



which moves in

a circular path of radius r at a constant angular rate

θ = ω

. (a) Find the differential

equation for φ, assuming that the string remains taut and all motion occurs in the

xy-plane. (b) Assume the initial conditions φ(0) = 0,

φ(0) =−ω

and let l = r.

Find φ

max

in the motion which follows.

O′

Figure P 1.14.