Dynamics of Mechanical Systems 2009 Part 7 pps

9.4.7 we obtain theaddition theorem for angular momentum for a rigid body:9.4.14 Consider again a particle P with mass m moving in an inertial reference frame R as in Figure 9.5.1.. From

282 Dynamics of Mechanical Systems

where the final expression is obtained by recognizing M as the total mass of the particles

of B and by recalling that G is the mass center of B so that:

(9.3.6)

(See Section 6.8.)

Equation (9.3.5) shows that for a rigid body the computation of the linear momentum

is in essence as simple as the computation of linear momentum for a single particle as in

Eq (9.3.1)

Consider again a particle P having mass m and velocity v in a reference frame R as in

Figure 9.4.1 Let Q be an arbitrary reference point Then, the angular momentum AQ of P

about Q in R is defined as:

(9.4.1)

where p locates P relative to Q Observe that AQ has the units mass–length–velocity, or

mass–(length)2 per unit time

Angular momentum is sometimes called moment of momentum Indeed, if we recognize

mv as the linear momentum L of P, we can write Eq (9.4.1) in the form:

(9.4.2)

Consider next a set S of N particles P i (i = 1,…, N) having masses m i and velocities in

R as in Figure 9.4.2 Again, let Q be an arbitrary reference point Then, the angular

momentum of S about Q in R is defined as:

(9.4.3)

where pi locates P i relative to Q.

Finally, consider a rigid body B with mass m and mass center G moving in a reference

frame R as in Figure 9.4.3 Consider B to be composed of N particles P i with masses m i

FIGURE 9.4.1

A particle P with mass m, velocity v, and

reference point Q.

m i i i

(i = 1,…, N) Again, let Q be an arbitrary reference point Then, as in Eq (9.4.3), the angular momentum of B about Q in R is defined as:

(9.4.4)

where pi locates P i relative to Q.

Observe that because P i and mass center G are both fixed in B their velocities in R are

related by the expression (see Eq (4.9.4)):

(9.4.5)

where ωω is the angular velocity of B in R and where r i locates P i relative to G (see Figure

9.4.3) Observe further from Figure 9.4.3 that pi and ri are related by the expression:

N

i G

i

i G

i N

i N

i i G

i N

i N

i i

N

G

i i i N

1 1

ωωωωωω

N

G

i N

where the middle two terms in the last line are zero because G is the mass center of B (see

Section 6.8) The first term may be recognized as the moment of the linear momentum of

a particle with mass M at G about Q That is,

(9.4.8)

The last term of Eq (9.4.7) may be expressed in terms of the central inertia dyadic (see

Sections 7.4 and 7.6) as follows: Let nω be a unit vector with the same direction as ωω Then,ω

ω may be expressed as:

N

P

i i i

N

G

B G

i i

m i i i

Finally, by substituting from Eqs (9.4.8) and (9.4.13) into Eq (9.4.7) we obtain theaddition theorem for angular momentum for a rigid body:

(9.4.14)

Consider again a particle P with mass m moving in an inertial reference frame R as in

Figure 9.5.1 Let P be acted upon by a force F as shown Then, by Newton’s law (see Eq (8.3.1)), F is related to the acceleration a of P in R by the expression:

(9.5.1)

Recalling that the acceleration is the derivative of the velocity we can use the definition

of linear impulse of Eqs (9.2.1) and (9.2.3) to integrate Eq (9.5.1) That is,

(9.5.2)

or

(9.5.3)

where I represents the impulse applied between t1 and t2

Equation (9.5.3) states that the linear impulse is equal to the change in linear momentum

This verbal statement is often called the principle of linear momentum.

This principle is readily extended to systems of particles and to rigid bodies Consider

first the system S of N particles P i with masses mi (i = 1,…, N) and moving in an inertial

FIGURE 9.5.1

A particle P with mass m moving in

an inertial reference frame.

t t

t t

1 2

1 2

1 2

I= ∆L

frame R as in Figure 9.5.2 Let the particles be acted upon by forces F i (i = 1,…, N) as

shown Then, from Newton’s law, we have for each particle:

(9.5.4)

where ai is the acceleration of P i in R.

Let G be the mass center of S Then,

(9.5.5)

where ri locates P i relative to G as in Figure 9.5.2.

Let the system of forces Fi be represented by an equivalent force system (see Section 6.5)

consisting of a single force F passing through G together with a couple with torque T.

Then, F and T are:

A set of particles moving in an

iner-tial reference frame R.

N

i G i i

N

i i

1

2 2 1

m i i i

N

r

=

∑1

Finally, by integrating in Eq (9.5.7) we have:

(9.5.8)

or

(9.5.9)

where L represents the linear momentum of S as in Eq (9.3.2).

We can develop a similar analysis for a rigid body B as in Figure 9.5.3 where, as before,

we consider B to be composed of a set of N particles P i having masses m i (i = 1,…, N) Let

G be the mass center of B and let R be an inertial reference frame in which B moves Let

ri locate P i relative to G Then, because P i and G are both fixed in B, their accelerations are

related by the expression (see Eq (4.9.6)):

(9.5.10)

where as before ai represents the acceleration of P i in R and where αααα and ωωωω are the angular

acceleration and angular velocity of B in R.

Let P i be acted upon by a force Fi as shown in Figure 9.5.3 Let the set of forces Fi (i =

1,…, N) be represented by an equivalent force system consisting of a single force F passing through G together with a couple with torque T Then, F and T are:

t

G t

t

1 2

1 2

1 2

By substituting from Eqs (9.5.10) and (9.5.12) into Eq (9.5.11) we have:

(9.5.13)

where M is the mass of B.

Finally, by integrating in Eq (9.5.13) we obtain (as in Eqs (9.5.8) and (9.5.9)):

(9.5.14)

where now L represents the linear momentum of B.

Observe the identical formats of Eqs (9.5.3), (9.5.9), and (9.5.14) for a single particle, aset of particles, and a rigid body

We can develop expressions analogous to Eqs (9.5.3), (9.5.9), and (9.5.14) for angularimpulse and angular momentum The development here, however, has the added feature

of involving a reference point (or object point) Because angular momentum is alwayscomputed relative to a point, the choice of that point may affect the form of the relationbetween angular impulse and angular momentum

Consider again a particle P with mass m moving in an inertial reference frame R as in

Figure 9.6.1 Let P be acted upon by a force F as shown Let Q be an arbitrarily chosen reference point Consider a free-body diagram of P as in Figure 9.6.2, where F* is the inertia

force on P given by (see Eq (8.3.2)):

(9.6.1)

where vP and aP are the velocity and acceleration of P in R.

From d’Alembert’s principle, we have:

(9.6.2)

FIGURE 9.6.1

A particle P moving in an inertial reference frame R

with applied force F and reference point Q.

By setting moments about Q equal to zero we have:

where JQ is the angular impulse of F about Q during the time interval (t1, t2) That is, the

angular impulse about a point Q fixed in an inertial reference frame is equal to the change

in angular momentum about Q This verbal statement of Eq (9.6.8) is called the principle

of angular momentum Note, however, it is valid only if Q is fixed in the inertial reference

frame

Consider next a set S of N particles P i with masses mi (i = 1,…, N) moving in an inertial

reference frame R as in Figure 9.6.3 Let the particles be acted upon by forces F i (i = 1,…, N) as shown Consider a free-body diagram of a typical particle P i as in Figure 9.6.4 where

is the inertia force on P i given by:

where and are the velocity and acceleration of P i in R Then, from d’Alembert’s

A set S of particles moving in an inertial reference

frame R with reference point Q.

i i P

By integrating over the time interval in which the forces are applied, we have:

or

(9.6.16)

where here JQ represents the sum of the angular impulses of the applied forces about Q during the time interval (t1, t2) Hence, as with a single particle, the angular impulse about

a point Q fixed in an inertial reference frame is equal to the change in angular momentum

of the set of particles about Q.

Finally, consider a rigid body B moving in an inertial reference frame R as in Figure 9.6.5 Let G be the mass center of B, let Q be a reference point, and let O be the origin of

R Then, from Eqs (9.4.8), (9.4.12), and (9.4.13), the angular momenta of B about O, Q, and G are:

(9.6.17)

(9.6.18)

(9.6.19)

where PG and QG locate G relative to O and Q as in Figure 9.6.5, and where as before ω

is the angular velocity of B in R, I B/G is the central inertia dyadic of B, and m is the mass

of B Consider the derivatives of these momenta For A B/O we have:

(9.6.20)

FIGURE 9.6.5

A rigid body B moving in an inertial reference

frame R and a reference point Q.

where we have used Eq (4.6.6) If IB/G is expressed in terms of unit vectors fixed in B, its

components relative to these vectors are constant Hence, B dI B/G /dt is zero Also, the third

term is zero Therefore, R dA B/O /dt becomes:

Let B be subjected to a system of applied forces that may be represented by a single

force F passing through G together with a couple with torque M G Similarly, let the inertia

forces on B be represented by a single force F* passing through G together with a couple

with torque T* Then, a free-body diagram of B may be constructed as in Figure 9.6.6.

Recall from Eqs (8.6.5) and (8.6.6) that F* and T* may be expressed as:

From d’Alembert’s principle, the entire force system of a free-body diagram is a zerosystem Thus, the moment of the force system about any and all points is zero Hence,

consider setting moments about O, Q, and G equal to zero: for O we have:

(9.6.25)

By substituting for F* and T* from Eq (9.6.24) we have:

(9.6.26)

where MO is defined as MG + P × F and is identified as the moment of the applied forces

about O By comparing Eqs (9.6.21) and (9.6.26) we have:

where MQ is defined as QG × F + MG and is identified as the moment of the applied forces

about Q By comparing Eqs (9.6.22) and (9.6.29), we have:

By inspecting Eqs (9.6.27), (9.6.30), and (9.6.33), we see that if the object point is either

fixed in the inertia frame R or is the mass center of the body, then the moment of the

applied forces about the object point is equal to the derivative of the angular momentumabout the object point

By integrating Eqs (9.6.27) and (9.6.33) over the time interval when the forces areapplied, we obtain:

where JO and JG are the angular impulses of the applied forces about O and G.

Eqs (9.6.36) and (9.6.37) are, of course, analogous to Eqs (9.6.8) and (9.6.16) for a singleparticle and for a set of particles They all state that the angular impulse is equal to thechange in angular momentum when the object point is fixed in an inertial reference frame.For a rigid body, the object point may also be the mass center Other object points do notproduce the simple relation between angular impulse and angular momentum as in Eqs.(9.6.8), (9.6.16), (9.6.36), and (9.6.37)

Before looking at examples illustrating the impulse momentum principles, it is helpful toconsider also the conservation of momentum principles Simply stated, these principlesassert that if the impulse is zero, the momentum is unchanged — that is, the momentum

t1 and t2 are arbitrary, and the linear momentum is unchanged throughout all time

inter-vals That is, the linear momentum is constant, or conserved Expressions analogous to

Eq (9.7.1) may also be obtained for sets of particles and for a rigid body using Eqs (9.5.9)and (9.5.14)

B O t

Similarly, from Eq (9.6.8), if the angular impulse JQ on a particle is zero, we have:

(9.7.2)

Therefore, if the angular impulse is zero, the angular momentum is the same at thebeginning and end of the impulse time interval and thus throughout the time interval;hence, the angular momentum is constant, or conserved Similar statements can be madefor sets of particles and for a rigid body using Eqs (9.6.16), (9.6.36), and (9.6.37)

Observe that because linear and angular impulses are vectors, they may be nonzero butstill have zero projection in some directions In such cases, the projection of the momenta

along those directions is conserved For example, if n is a unit vector such that:

Example 9.8.1: A Sliding Collar

Consider a sliding block (or collar) of mass m moving on a smooth rod as in Figure

9.8.1 Let a force F be exerted on B as shown and let the magnitude of F have a time profile

as in Figure 9.8.2 From Eq (9.2.1), the impulse of F is:

max

ˆ

ˆ 20

0

where n is a unit vector parallel to the support as in Figure 9.8.1 Let the velocity of B before and after the application of F be V On and Vn, respectively Then, from the linear

impulse–momentum principle, we have:

(9.8.2)

or

(9.8.3)

Example 9.8.2: A Braked Flywheel

As a second example, consider a flywheel W rotating about its axis with an angular speed

as in Figure 9.8.3 Let W be supported by a shaft S, and let S be subjected to a brake that

exerts a moment M about the axis of S and W and whose magnitude M is depicted in

Figure 9.8.4 Specifically, let M(t) be sinusoidal such that:

(9.8.4)

Then, from Eq (9.2.2), the angular impulse about O, the center of W, is:

(9.8.5)

where k is a unit vector parallel to the shaft as in Figure 9.8.3.

From Eq (9.4.13) the angular momentum of W about O before and after braking is:

0π

where I O is the axial moment of inertia of W (the moment of inertia of W about O for the

axial direction) Then from the angular impulse–momentum principle the angular speed

of W after braking is determined by the expressions:

(9.8.7)

or

(9.8.8)

Example 9.8.3: A Struck Pinned Bar (Center of Percussion)

Next, consider a thin horizontal bar B with length  and mass m and pinned at one end

O about a vertical axis as in Figure 9.8.5 Let B be struck at a point along its length as shown If B is initially at rest it will begin to rotate about O after it is struck.

Consider a free-body diagram of B showing the impulses and momentum changes of

B as in Figure 9.8.6 where P is the impact force magnitude and O x and O y represent

components of the pin reaction force along and perpendicular to B, and where G is the mass center of B The linear impulse change of B and the angular impulse change of B about G are:

Observe that if B is struck at a point such that x is 2/3 the pin reaction is zero That

is, if the impact force is applied at a point 2/3 along the bar length, no pin reaction isgenerated This means that even if the bar is not pinned it will initially move such thatits end away from the impact has zero velocity This point of application of the impact

force is called the center of percussion.

Example 9.8.4: A Pinned Double Bar Struck at One End

As an extension of the foregoing example consider two identical, pin-connected bars eachhaving length  and mass m and resting on a smooth horizontal surface as depicted in Figure 9.8.7 Let the end O of one of the bars be struck with the striking force being perpendicular to the bars and having a magnitude P as shown The objective is to deter-

mine the motion of the bars just after the impact

Consider a free-body diagram of the bars and also a free-body diagram of the left, orunstruck, bar as in Figures 9.8.8 and 9.8.9, where as in the foregoing example the inertia

2ω

v=( )l 2 ω

O y t

forces are not shown Instead, the changes in momenta of the bars are depicted where

v 1x , v 1y and v 2x , v 2y are the velocity components of the mass centers, G1 and G2, of the bars

in the directions shown; ω1 and ω2 are the angular speeds of bars just after impact; and

nx, ny, and nz are mutually perpendicular unit vectors in the directions shown In Figure

9.8.9, Q x and Q y are the components of the reaction force transmitted across the pin Q

when the bars are struck

From these free-body diagrams and from the impulse–momentum principles, we obtainthe following relations:

y

y t

0

1

12

where the third and sixth equations are obtained by evaluating moments about Q and G1,

and where I is the central moment of inertia: (1/12)m2 Observe from the kinematics ofthe bars that we can obtain the expressions:

9.9 Additional Examples: Conservation of Momentum

The most useful impulse–momentum principles are those related to the conservation ofmomentum, where external impulse to a system is absent In this section we will consider

a few simple examples to illustrate these principles

Example 9.9.1: Colliding Collinear Blocks

First, consider two blocks A and B with masses m A and m B moving along the same line

as in Figure 9.9.1 Let the blocks be moving to the right as shown, and let the speed of A

be greater than the speed of B (v A > v B) so that collision occurs Let the collision be “plastic”

so that the blocks move together as a unit after collision Given the masses and speeds ofthe blocks before collision, the objective is to determine their common speed after collision

If we consider the two blocks as a system, we observe that there are no external impulses

on the system This means that the overall momentum of the system is unchanged by thecollision Thus, we have:

(9.9.1)

where v represents the common speed of the blocks after collision Solving for v we obtain:

(9.9.2)

Observe in Eq (9.9.2) that if the masses are equal, v is the average of the speeds of the

blocks prior to collision

Example 9.9.2: Engaging Coaxial Disks

As a second example of momentum conservation, consider two disks A and B rotating

freely on a common shaft as depicted in Figure 9.9.2 Let the disks have axial moments

of inertia I A and I B and angular speeds ωA and ωB as shown Let the disks be broughttogether by sliding them toward each other along their common shaft Let the disks engageone another and then rotate together with the common angular velocity ω The objective

Because the disks can rotate freely on their common shaft there is no axial momentumapplied to the disks; thus, there is no angular impulse about the shaft axis This in turnmeans that the angular momentum of the disks about the shaft axis is conserved.Before meshing, the axial angular momentum of the disks is:

(9.9.3)

where n is a unit vector parallel to the shaft axis After meshing, the angular momentum

of the disks is:

Example 9.9.2: Wheel Rolling Over a Step

As a third example of momentum conservation consider a wheel or disk W rolling in a straight line and encountering a small step as in Figure 9.9.3 Let the angular speed of W

before striking the step be ω The objective is to determine the angular speed ω of W justafter impact with the step

Let W have radius r and mass m, let its axial moment of inertia be I, and let the step height be h Then, the angular momentum of W about the step corner O before and after impact is conserved Before impact, the angular momentum of W about O is:

(9.9.6)

where v is the speed of the disk center before impact, and n is an axial unit vector as

shown in Figure 9.9.3 Because W is rolling, v is simply r ω After impact with the step, W

is rotating about O Its angular momentum after impact is thus:

Equating the expressions of Eqs (9.9.6) and (9.9.7), we find to be:

(9.9.8)

9.10 Impact: Coefficient of Restitution

The principles of impulse and momentum are ideally suited to treat problems involvingimpact, where large forces are exerted but only for a short time Although a detailedanalysis of the contact forces, the stresses, and the deformations of colliding bodies requiresextensive analysis beyond our scope here, the principles of impulse and momentum can

be used to obtain a global description of the phenomenon That is, by making a couple

of simplifying assumptions we can develop a direct and manageable procedure for ing impact

study-Our first assumption is that the time interval during which impact occurs is so shortthat the positions and orientations of the colliding bodies do not change significantlyduring the impact time interval, although the velocities may have significant incrementalchanges The second assumption is that after the bodies come together and collide theythen generally rebound and separate again The speed of separation is assumed to be afraction of the speed of approach

To develop these concepts, consider again the principle of linear impulse and

momen-tum Specifically, consider a particle P of mass m initially at rest but free to move along

the X-axis as in Figure 9.10.1 Let P be subjected to a large force F acting for a short time

t* Then, from the linear impulse–momentum principle (Eq (9.5.3)), we have:

Because v* is finite, x* may be made arbitrarily small by making t* small That is, for very

short impact time intervals, the displacement of P remains essentially unchanged Next, consider two particles A and B moving toward each other along a common line

as in Figure 9.10.2 Let the relative speed of approach be v Let the particles collide, rebound,

and separate Let the relative speed of separation be Then, from our assumption aboutthe impact, we have:

(9.10.4)

where e, called the coefficient of restitution, has values between zero and one (0 ≤ e ≤ 1) When e is zero, the separation speed is zero, and the particles stay together after impact Such collisions are said to be plastic When e is one, the separation speed is the same as the approach speed (that is, = v) This is called an elastic collision.

Equation (9.10.4) may be used with the principle of conservation of momentum to study

a variety of collision configurations To illustrate this usage, consider first two colliding

particles A and B, each having the same mass m Let B initially be at rest, and let A approach B with speed v as in Figure 9.10.3 Then, from Eq (9.10.4), the separation speed

of the particles is ev That is, after collision,

Observe in Eq (9.10.7) that if e is zero (plastic collision) and are equal with value

v/2 If e is one, (elastic collision) then is zero and is v In this latter case, the momentum of A is said to be transferred to B.

As a generalization of this example, consider two particles A and B with masses m A and

m B and with each particle moving on the same line before impact Let the pre-impact

speeds be v A and v B Let the post-impact speeds be and (see Figure 9.10.4) As before,

we can determine and using Eq (9.10.4) and the principle of conservation of linearmomentum From Eq 9.10.4, we have:

From the momentum conservation principle (see Eq (9.7.1)), we have:

Consider three special cases: (1) e = 0 , (2) e = 1 , and (3) m A = m B

Case 1: e = 0 (Plastic Collision)

In this case, Eqs (9.10.10) and (9.10.11) simplify to the expressions:

(9.10.13)

and

(9.10.14)

Observe that = , as expected with a plastic collision

Case 2: e = 1 (Elastic Collision)

In this case, Eqs (9.10.10) and (9.10.11) become:

These results are seen to be consistent with those of Eq (9.10.7).

For still another specialization of Eqs (9.10.10) and (9.10.11), let the mass of A greatly exceed that of B, let the collision be elastic, and let B be initially at rest That is,

When colliding particles are moving along the same straight line (as in the foregoing

examples) the collision is called direct impact Generally, however, when particles (or

bodies) collide they are not moving on the same line Instead, they are usually moving

on distinct curves that intersect Such collisions are called oblique impacts It happens that

oblique impacts may be studied using the same principles we used with direct impacts.That is, oblique impact is treated as a direct impact in the direction normal to the plane

of contact of the bodies In directions parallel to the plane of contact, the momenta of theindividual bodies are conserved

To develop the analysis consider the oblique impact depicted in Figure 9.11.1, where A and B are particles moving on intersecting curves Consider also the closer view of the

impact provided in Figure 9.11.2 Let the particles be considered to be “small” bodies.That is, we will neglect rotational or angular momentum effects in the analysis In

Figure 9.11.2, let T represent a plane tangent to the contacting surfaces at the point of

m A>>m B, , e=1 v B=0

ˆv A ˆv B

ˆ

v A=v A and v B=2v A