Dynamics Ch8

8/3/2019 Dynamics Ch8

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Chapter 8

Rigid-Body Kinetics

Chapter Abstract. This chapter is the culmination of everything we have done in the pre-

ceding chapters. Newtons Second Law is the foundation of the kinetics of rigid bodies, justas it is for particles and systems of particles. Because a rigid body is the limiting case of an

infinite number of particles for which the mass of a differential-sized particle approaches zero,

we appeal to our analysis of systems of particles to establish the appropriate form of Newtons

Second Law. This involves generalizing the concept of the center of mass for a rigid body.

In examining angular momentum, we discover that we must introduce a new quantity in our

analysis called the inertia tensor.

We consider both two-dimensional motion and three-dimensional motion of rigid bodies.

Then, we turn to the Principle of Work and Energy and the Principle of Impulse and

Momentum. Both principles are easier to apply for a rigid body than for a finite system of

particles because there is no need to include effects of internal forces.

8.1 Equations of MotionIn discussing finite systems of particles, we found that the motion can be conveniently described

in terms of mass-averaged position, velocity and acceleration. Summing over all particles in

a system, the sum of the external forces is equal to the systems mass, m, multiplied by theacceleration of the center of mass, a, viz.,

ni=1

Fi = ma (8.1)

We also found that the rate of change of angular momentum relative to an inertial frame, HO,

is equal to the sum of the moments of the external forces acting on the center of mass and the

sum of the moments relative to the center of mass. Thus, angular momentum is governed by

HO = rni=1

Fi + HCm and HCm =ni=1

MCmi (8.2)

where the MCmi are the external moments relative to the center of mass and HCm is the rate

of change of the bodys angular momentum about its center of mass. While mass averaging

obviates the need to deal with the motion of individual particles in considering linear mo-

mentum, the same is not true for angular momentum. Specifically, evaluation of the angular

251


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252 CHAPTER 8. RIGID-BODY KINETICS

momentum relative to the center of mass, HCm , requires considering the position and velocity

of each particle relative to the center of mass, ri and v

i, respectively, i.e.,

HCm

ni=1

rivimi (8.3)

Since a rigid body is the special limiting case of a system where the number of particles

is extremely large, Equations (8.1), (8.2) and (8.3) govern the bodys motion. Figure 8.1

illustrates this representation of rigid-body motion.

Before we can proceed, we must specify the way in which the center of mass and the

angular-momentum vector are computed. As we will see, the limiting form of Equation (8.3)

for the angular momentum relative to the center of mass can be determined in terms of mass-

averaged properties, albeit with a somewhat complicated calculation.

Cm Cm

F1

F2 F3

F4

F5

HCm

ma

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Figure 8.1: General motion of a rigid body.

8.1.1 Center of Mass

Let mi be the mass of a particle whose position vector is ri. Then, the center of mass fora system of n particles is given by

r =1

m

ni=1

rimi where m =ni=1

mi (8.4)

A rigid body corresponds to the limit n ,mi 0, for which the sum of the differentialmasses is the finite mass of the body. Thus, we replace the summations by integrals, i.e.,

r =1

m

r dm where m =

dm (8.5)

If the body has mass density , then dm = dVwhere V denotes volume. In terms of volumeintegrals, the center of mass is given by

r = 1m

V

r dV where m = V

dV (8.6)

We can use Equation (8.6) to compute the center of mass of any geometry. Obviously, if

the object has uniform density, the center of mass is coincident with its centroid, i.e., its

geometric center. Appendix A includes the center of mass for common two-dimensional

and three-dimensional geometries. The following example shows how the center of mass is

computed for a nonuniform material.


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8.1. EQUATIONS OF MOTION 253

Example 8.1 Determine the location of the center of mass for a cylinder of radius R and heighth, where the lower and upper parts of the cylinder are made of aluminum and lead, respectively.Aluminums density is Al = 2712 kg/m

3and leads density is Pb = 11340 kg/m

3. The

aluminum part of the cylinder is of heighth1.

xy

z

Al

Pb

h

h1

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Solution. Since density is constant in each part of the cylinder, its total mass is

m = AlR2h1 + PbR

2 (h h1) = PbR2h

11 AlPb

h1h

For the given values of Al and Pb, we find

m = PbR2h

1 0.76 h1h

The symmetry of the cylinder tells us that the center of mass lies on the z axis. Thus, we need onlycompute z. The value of z is given by

z =1

m

R2

h10

Alz dz + R2

hh1

Pbz dz

=

R2

2m

Alh

2

1 + Pb

h2 h21

Rearranging terms and substituting the given values of Al and Pb yields

z =PbR

2h2

2m

1 0.76

h1h

2Substituting for m, the center of mass is located at

z =h

2

1 0.76 (h1/h)21 0.76 (h1/h)

The figure below shows the variation of z/h with h1/h. Note that in the limit of a pure leadcylinder (h1 = 0) and a pure aluminum cylinder (h1 = h), the center of mass lies at the center ofthe cylinder, as it must.

0.0 0.2 0.4 0.6 0.8 1.0 h1/h0.50

0.55

0.60

0.65

0.70

z/h

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8.1.2 Angular Momentum and the Inertia Tensor

As shown in Figure 8.2, in terms of a differential element of mass mi, the angular momentumabout the center of mass is

HCm =ni=1

riv

imi =ni=1

ri ( r

i)mi (8.7)

where we use the fact that vi = r

i. As with the center of mass, we focus on the limiting

case n,mi 0, wherefore

HCm =

r ( r) dm (8.8)

O

Cm

mi

x

z

yx

z

y

r

ri

vi

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Figure 8.2: Angular momentum of a rigid body relative to its center of mass, Cm.

In component form, the integrand is

r ( r) =

x

y2 + z2

yx

y zxz

i (8.9)

+ xxy + y x2 + z2 zyzj (8.10)+

zx

z yyz + z

x2 + y2

k (8.11)

Because the angular velocity is independent of spatial coordinates within a rigid body, we can

factor it out and rewrite this equation as follows.

r ( r) =

y2 + z2

xy xz

xy

x2 + z2

yz

xz yz

x2 + y2

x

y

z

(8.12)

Integrating the diagonal terms of the matrix yields the moments of inertia defined by1

Ixx = y2 + z2 dm, I yy = x

2 + z2 dm, I zz = x2 + y2 dm (8.13)

Integrating the off-diagonal terms yields the products of inertia defined by

Ixy =

xy dm, I yz =

yz dm, I xz =

xz dm (8.14)

1Strictly speaking, these are the mass moments of inertia. If we replace dm by dV, they are called areamoments of inertia. Some authors prefer the notation Ix = Ixx, Iy = Iyy , Iz = Izz for the moments of inertiaand Jxy = Ixy , Jyz = Iyz , Jxz = Ixz for the products of inertia.


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Therefore, the angular momentum vector can be written as

HCm = [I] (8.15)

where we call the matrix [I], whose elements are the moments and products of inertia, theinertia tensor, i.e.,

[I] =

Ixx Ixy Ixz

Iyx Iyy Iyz

Izx Izy Izz

(8.16)

This matrix is symmetric because Ixy = Iyx , Ixz = Izx and Iyz = Izy . For example,

Ixy =

xy dm =

yx dm = Iyx (8.17)

A special set of coordinate axes called the principal axes of inertia exists in which all

of the off-diagonal terms vanish. The principal axes can be determined through standard

theorems and operations with matrices, a topic that is beyond the scope of this course. Theorientation of the principal axes is usually obvious for simple symmetric geometries. Denoting

these axes by xyz, we have

HCmx = Ixxx, HCmy = Iyyy, HCmz = Izzz (8.18)

The physical meaning of the mass moments of inertia in the principal-axis system is

especially easy to understand. Clearly, each moment is the integral of r2dm, where r is theradial distance from the axis. In this spirit, we define the radii of gyration, Rx,Ry,Rz, by

Ixx = mR2

x, Iyy = mR2

y, Izz = mR2

z (8.19)

The larger the radius of gyration, the more rotational inertia the rigid body has about that

axis. Consider Figure 8.3, which shows a man standing on a platform that rotates about a

vertical axis. He is holding two heavy spherical weights. When he holds the weights with

arms extended, clearly his radius of gyration about the platforms rotation axis is greater than

when he lowers the weights. His angular momentum is HCm = mR2z11k with his arms

extended and HCm = mR2z22k when he moves the weights down to his side. Since there

are no external forces acting, his angular momentum will be the same regardless of where he

holds the weights. Since Rz1 > Rz2, necessarily 2 > 1. Figure skaters who begin a spin

z z

1 2

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Figure 8.3: A man with two spherical weights on a rotating platform.


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with arms outstretched (without the weights) use this effect to dazzle their audience. When a

woman skater pulls in her arms, she reduces her moment of inertia, which causes her to spin

faster.

The products of inertia cannot be related to a single axis but instead involve two separateaxes. Nonzero products of inertia reflect some kind of asymmetry of a geometry. For example,

suppose our rotating man held the weight in his right hand beneath the height of the weight

in his left hand. Referring to Figure 8.4, clearly the change in the mans product of inertia,

Iyz , will be approximately

Iyz m()z mzr = m(z zr) (8.20)

where is the length of the mans arm measured from the center of his body, m is the massof each weight, and the heights of each weight are z and zr as shown. Because his angularmomentum is constant, the man will feel a change in his angular velocity, which will no longer

be aligned with the z axis. This change in direction corresponds to rotations about the x andy axes induced by the change in his product of inertia.

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y

z

zzr

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Figure 8.4: Example with a nonzero product of inertia.

8.1.3 Parallel Axis Theorem

For many applications, the geometry is such that the principal axes are not easily located or

may be an inconvenient choice. For such applications, we can use the Parallel Axis Theorem.

This theorem makes it a simple matter to shift the origin of a coordinate system with no rotation

of the axes.

To prove the theorem, we note first that the location of the center of mass relative to an

inertial coordinate frame is r = x i + yj + z k. So, for a given Point P in a rigid body, wehave

r = r + r = x = x + x, y = y + y, z = z + z (8.21)

where the coordinates xyz and xyz are shown in Figure 8.5. Then, the moment of inertiaabout the x axis is

Ixx =

y2 + z2

dm =

(y + y)

2+ (z + z)

2

dm

=

y2 + z2

dm + 2y

ydm + 2z

zdm +

y2 + z2

dm (8.22)


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O

Cm

P

x

z

yx

z

y

r

rri

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Figure 8.5: Inertial reference frame, xyz , and center-of-mass coordinate frame, xyz.

By definition, we know that

Ixx =

y2 + z2

dm and

dm = m (8.23)

Also,

y dm and

z dm are the y and z components of the center of mass relative to thecenter of mass, respectively. Thus, both of these integrals are zero. Therefore,

Ixx = Ixx + m

y2 + z2

(8.24)

A similar computation shows that

Iyy = Iyy + m

x2 + z2

, Izz = Izz + m

x2 + y2

(8.25)

Turning to the products of inertia, consider Ixy. We have

Ixy =

xydm =

(x + x) (y + y) dm

=

xy dm x

ydm y

xdm xy

dm (8.26)

Therefore, the product of inertia Ixy is

Ixy = Ixy mxy (8.27)

Similarly, the other two products of inertia are

Iyz = Iyz myz and Ixz = Ixz mxz (8.28)

Therefore, in matrix form, the Parallel Axis Theorem is

Ixx Ixy Ixz

Iyx Iyy Iyz

Izx Izy Izz

=

Ixx Ixy Ixz

Iyx Iyy Iyz

Izx Izy Izz

+

m

y2 + z2

mxy mxzmxy m

x2 + z2

myz

mxz myz m

x2 + y2

(8.29)


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Example 8.2 Compute the inertia tensor relative to Origin O of coordinate frame xyz. The rodsare very slender and made of the same material. The mass of the horizontal rod is m.

O

Cm1Cm2

z

y

x

2

correspond to zxy for Rod 1 correspond to yzx for Rod 2

Mass = MLength = LCm

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Solution. Since the mass of the horizontal rod is m, necessarily the mass of the vertical rod is 2m.Appealing to Appendix A, we find that for a rod of mass M and length L, the principal momentsof inertia are

I = I =1

12ML2, I 0

The center of mass of each rod is at its center. We construct the following table.

Rod x y z I xx Iyy Izz Mass

1 0 0 23

m2 23

m2 0 2m

2 0 12

112

m2 0 112

m2 m

Thus, the moment of inertia about the x axis is

Ixx =

2i=1

Ixixi

+ mi y2i + z2i = 23 m2+2m 02 + 2+ 112 m2+m14 2 + 2 = 4m2Similarly, a little algebra shows that

Iyy =11

3m2 and Izz =

1

3m2

Turning to the off-diagonal terms,

Iyz =

2i=1

Iyizimiyizi

= 2m(0 )m

1

2

= 1

2m2

As can be easily demonstrated,

Ixy = Ixz = 0

Thus, the inertia tensor is

[I] =

4m2 0 0

0 113

m2 12

m2

0 12

m2 13

m2

= 1

6m2

24 0 0

0 22 30 3 2


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8.2. MOTION OF A RIGID BODY IN TWO DIMENSIONS 259

8.1.4 Equations of Motion Summary

In this section, we have developed the equations of motion for rigid bodies in terms of mass-

averaged quantities. As with finite systems of particles, we have demonstrated that Newtons

Second Law yields the following equations for linear and angular momentum [see Equa-tions (8.1) and (8.2)].

ni=1

Fi = ma and HCm =ni=1

MCmi (8.30)

We also established the relation between angular momentum and angular velocity, viz.,

H = [I] (8.31)

where [I] is the inertia tensor. Although we could conceptually develop the inertia tensor fora system of particles, it would not be terribly useful in general applications because it would

change every time the relative distance between particles changes. By contrast, the distance

between particles in a rigid body is constant so that the inertia tensor is well defined. In

words, these equations tell us the following

The resultant force,m

i=1 Fi, goes through the center of mass and causes it to accelerate.

The sum of the external moments about the center of mass,m

i=1 MCmi, causes the

spin angular momentum, HCm , to change.

At first glance, this would suggest that we can consider translation and rotation separately,

similar to what we have done for the kinematical description. Although this is sometimes

possible, the translational and rotational motions of a rigid body often cannot be separated

because they are inherently coupled.

8.2 Motion of a Rigid Body in Two Dimensions

For rigid-body motion in two dimensions, we consider a plane with x and y axes that lies atz = 0. In this plane, the velocity is

v = vx i + vy j (8.32)

If the body rotates in the xy plane, its angular-velocity vector must be normal to the plane sothat2

= k (8.33)

The inertia tensor simplifies also. Since z = 0, the moments and products of inertia are

Ixx =

y2dm, I yy =

x2dm, I zz =

x2 + y2

dm (8.34)

Ixy = Iyx =

xy dm, I xz = Izx = Iyz = Izy = 0 (8.35)

2Rotation about the x or y axis would involve motion out of the plane.


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Therefore, in matrix form, the inertia tensor is

[I] = Ixx Ixy 0

Ixy

Iyy

0

0 0 Izz (8.36)

Consequently, the angular momentum relative to a Point O is given by

HO =

Ixx Ixy 0

Ixy Iyy 0

0 0 Izz

0

0

= Izz (8.37)

Thus, the angular-momentum vector is always aligned with the angular-velocity vector. This is

a major simplification in the equations of motion for a rigid body. For notational convenience,

we drop the subscript zz so that HO is

HO = I (8.38)

Also, the Parallel Axis Theorem simplifies to

IO = I + m

x2 + y2

(8.39)

where IO is the moment of inertia relative to Point O and I is the moment of inertia relative

to the center of mass of the rigid body. No other moments or products of inertia are relevant.

Therefore, the equations governing planar motion of a rigid body are

ni=1

Fi = ma andni=1

Mi k = Id

dtk (8.40)

In component form, we thus arrive at the following three equations of motion.

ni=1

Fxi = max (8.41)

ni=1

Fyi = may (8.42)

ni=1

Mi = I (8.43)

These equations underscore one of the most important differences between motion of an

individual particle and motion of a rigid body. Because of a rigid bodys finite size, we can

discern rotation about its center of mass. By contrast, a particle, which we idealize as a point

mass, has no finite size and rotation about its center of mass is ill defined.

8.2.1 Rolling Motion

One particularly interesting example of two-dimensional rigid-body motion is that of an ax-

isymmetric body rolling on a surface. This type of motion involves both translation of and

rotation about the objects center of mass. For present purposes, we assume the object has

uniform mass density so that its center of mass is coincident with its centroid. This type of

application is aptly described as centroidal rotation.


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Consider a round object of radius R rolling on a planar surface as shown in Figure 8.6.The forces acting on the object are its weight, mg, a reaction force from the surface, N, adriving force, F, and a rolling-friction force, f. Our goal is to compute the rolling-friction

force as a function ofF = |

F|

and/orN = |

N|.

y

x

F

f

N

m g

a2R

Cm

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Figure 8.6: Round object rolling on a horizontal surface.

This problem includes an important kinematical constraint. Specifically, if the object rolls

without sliding, there is a direct relation between the linear acceleration of its center of mass,a, and its angular acceleration, . That is, if the object rotates clockwise through an angle

about its center of mass, Cm, the distance traveled by its center of mass in the x directionis R, where R is the objects radius. Note that clockwise rotation corresponds to negativevalues of. Differentiating twice with respect to time tells us that x = R. But, a = x and = . Thus, we conclude that for an object that rolls without sliding,

a = R (Rolling without sliding) (8.44)

As discussed in Section 3.3.2, there are three different regimes for rolling friction, viz.,

rolling without sliding, rolling with imminent sliding and rolling with sliding. If the object

rolls without sliding, the coefficient of proportionality is the coefficient of rolling friction,

r. When the object is on the brink of sliding as it rolls, the coefficient of proportionality is

the coefficient of static friction, s. Finally, when the object rolls and slides, the coefficientof kinetic friction, k, applies. Summarizing, we have the following.

f =

rN (a = R), Rolling without slidingsN (a = R), Rolling with imminent slidingkN (a independent of R), Rolling with sliding

(8.45)

The most interesting case is that of rolling without sliding. Newtons Second Law tells us

that i

Fi = F i fi + Njmgj = ma i (8.46)

Therefore, we have

F f = ma and N = mg (8.47)

If we assume that the driving force, like the gravitational and reaction forces, acts through the

objects center of mass, then only the rolling-friction force contributes to the moment about

the center of mass. Therefore,i

MCmi = (Rj) (fi) = Rfk= Ik (8.48)


12/42


where I is the objects moment of inertia. Using the constraint of Equation (8.44) in Equa-tion (8.48), there follows

f = I

mR2ma (8.49)

Also, substituting this value for f into the first of Equations (8.47) yields the relation betweenthe driving force and the acceleration, viz.,

F =

1 +

I

mR2

ma (8.50)

There are three interesting observations we can make based on Equations (8.49) and (8.50).

Combining Equations (8.49) and (8.50) shows that f and F are related by

f =F

1 + mR2/I(8.51)

so that the rolling-friction force is independent of the reaction force, N. This holdsup to the point where F is equal to the minimum value for sliding to begin, i.e., the

point at which f = sN, where s is the static-friction coefficient. For larger appliedforces, the object will roll and slide. When this happens, f = kN, where k is thekinetic-friction coefficient. Thus, as with a block sliding on a horizontal surface, the

rolling-friction force behaves as shown in the following graph.

fk

fs

f

Fs F

fs =Fs

1 + mR2/I

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The point at which sliding is imminent is often of interest in rolling-object applications.It is simply the point at which the rolling-friction force equals the static-friction force.

Letting fs denote this critical friction-force value, we conclude that

fs = smg =Fs

1 + mR2/I= Fs = s

1 +

mR2

I

mg (8.52)

The rolling-friction coefficient is

r =f

N

= I

mR2

a

g

(8.53)

If the object is not accelerating, the rolling-friction coefficient is zero. This does not

contradict what we discussed in Subsection 3.3.2. The analysis here is for a rigid body,

which by definition does not deform. The rolling-friction coefficients listed in Table 3.2

are for real materials, which experience some degree of deformation. The values for

commercial materials are in fact much smaller than kinetic-friction coefficients, and a

rigid body is simply a special limiting case for which r 0.


13/42


Example 8.3 Compute the rolling-friction coefficient, r, for a cylinder of mass m and radius R,which rolls without sliding down an incline that makes an angle with the horizontal. Compare

your value for = 30

o

to that of an oak disk rolling on a horizontal oak surface.

y

x

f

N

mga

2R

Cm

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Solution. The most convenient way to solve is to align our coordinate system with the incline.

Then, the force and acceleration vectors of interest are

N = Nj, f = f i, mg = mg sin img cos j, a = a i, = k

The force and moment equations simplify as follows.i

Fx = mg sin f = ma,i

Fy = Nmg cos = 0

i

MCm = (Rj) (f i) = Rfk= I k

Because the cylinder is rolling without sliding, we know that

a = R

Substituting into the moment equation and solving for a shows that

a =R2

If

Hence, the x component of the force equation becomes

f = mg sin m R2

If = f = mg sin

1 + mR2/I

Making use of the y component of the force equation, the rolling-friction coefficient is

r =f

N=

tan

1 + mR2/I

Reference to Appendix A shows that the moment of inertia for a cylinder is I =

1

2mR

2

, so that(1 + mR2/I) = 3. Consequently,

r =1

3tan

For a 30o incline, the numerical value of r is 0.192. From Table 3.2, the rolling-friction coefficientfor an oak disk rolling on a horizontal oak surface is 0.072, which is 3/8 of 0.192.


14/42


8.2.2 General Plane Motion

As an example of general plane motion, consider the small cart shown in Figure 8.7. The cart

rides on horizontal tracks and has a slender rod of length 2 and mass m attached that can

swing about a frictionless pivot. The cart begins accelerating from rest with a = a i at timet = 0. We would like to determine the angular acceleration and the angular velocity of therod as a function of the angle .

Cm

O

2

g = gj

a = a i

x

y

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v v f fq q

q

q

Figure 8.7: Accelerating cart with a swinging slender rod.

Kinematics. As indicated in the figure, we select a cart-fixed coordinate system. To solve for

the rods motion about the pivot, we will need the position and acceleration of the rods center

of mass, r and a, respectively. Also, because we will use the moment equation to determine

the rods angular acceleration, we will need the moment of inertia, I, relative to the rodscenter of mass. Reference to Appendix A shows that the slender rods center of mass is at

the midpoint of the rod so that

r =

sin i

cosj (8.54)The rods moment of inertia relative to its center of mass is

I =1

12m(2)2 =

1

3m2 (8.55)

Finally, we appeal to Equation (7.39) to determine the acceleration of the center of mass, i.e.,

a = a + k r 2r

= a i + k ( sin i cosj) 2( sin i cosj)

= a i ( sinj cos i) + 2( sin i + cosj)

=

a + cos + 2 sin

i + sin + 2 cos

j (8.56)

Therefore, in component form, the acceleration of the center of mass isax = a + cos +

2 sin and ay = sin + 2 cos (8.57)

Kinetics. Since our objective is to solve for the rods angular acceleration, we combine

Equations (8.1) and (8.2) to arrive at the following form of the moment equation.i

MOi = mra +i

MCmi (8.58)


15/42


Because this is two-dimensional motion, the net moment about the center of mass is simply

iMCmi = Ik=

1

3m2 k (8.59)

where k is a unit vector pointing out of the page. Turning to the cross product ra, substi-

tuting for r from Equation (8.54), we have

ra =

i j k

sin cos 0

ax ay 0

= (ax cos ay sin) k (8.60)Now, we substitute the components of a from Equation (8.57) to obtain

ra =

a + cos + 2 sin

cos sin + 2 cos

sin

k (8.61)

Regrouping terms and simplifying, we conclude that

ra = (a cos + ) k (8.62)

Finally, the external forces acting on the rod are gravity, which acts through its center of mass,

and a reaction force at the pivot. So, taking moments about the pivot, only gravity makes a

contribution, viz.,i

MOi = r (mg) = ( sin i cosj) (mgj) = mg sink (8.63)

Substituting Equations (8.59), (8.62) and (8.63) into moment-balance Equation (8.58) gives

mg sin = m(a cos+)+1

3m2 =

4

3m2 = m(a cosg sin) (8.64)

Therefore, the angular acceleration of the rod is

= 3

4

g

a

gcos sin

(8.65)

Equation (8.65) shows that is negative for < tan1(a/g). This is consistent with thefact that must have a negative value for small , which corresponds to clockwise rotation.

Finally, since we now know the rods angular acceleration, we can solve for its angular velocity,

. First, we note that the angle is related to the standard cylindrical coordinate angle by

=

2= = (8.66)

By definition, = and = . Therefore,

= ddt = ddt dd = dd =

34 gag cos sin (8.67)

Integrating over and using the fact that the rod starts from rest, there follows

=

3

2

g

a

gsin + cos 1

(8.68)


16/42


Example 8.4 A slender rod of length and mass m starts from rest and slides along frictionlesssurfaces as shown. Determine the reaction forces on the rod at Points A and B.

vf

vfx

y

sCm

A

B

Cc

Length =

NA

NB

g = gj

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Solution. To solve, we must first take account of all relevant kinematical considerations. Then, we

proceed to kinetics. We assume the rod is always in contact with both surfaces.

Kinematics. For a slender rod, reference to Appendix A tells us that the center of mass lies at the

rods geometric center. It also gives the moment of inertia, I, about the rods center of mass, viz.,

r =1

2(cos i + sin j) and I =

1

12m2

We can relate the acceleration of the center of mass, a, to its angular acceleration, , by locating

the instantaneous center of rotation, Point C. Since Points A and B are constrained to movevertically and horizontally, respectively, we draw dashed lines normal to each surface in order to

locate Point C. Their intersection lies at

rC = (cos i + sin j)

The acceleration of the center of mass is thus

a = rCm/C = k1

2(cos i + sin j)

=

1

2(sin i cos j)

Kinetics. From Newtons Second Law, we havei

Fi = NAi + NBjmgj = 12

m(sin i cos j)

Thus, the reaction forces, NA and NB are

NA =1

2m sin and NB = mg 1

2m cos

The moment equation isi

MCmik =1

2( cos i + sin j) (NAi) + 1

2(cos i sin j) (NBj)

=1

2 (NB cos

NA sin ) k=

1

12m2 k

Therefore, we have a third equation relating the three unknowns, , NA and NB , viz.,

NB cos NA sin = 16

m

We now have sufficient information to complete the solution.


17/42

8.3. MOTION OF A RIGID BODY IN THREE DIMENSIONS 267

Example 8.4 (Continued) First, we can solve for by substituting for NA and NB from above,which yields

mg cos

1

2m cos2

1

2m sin2 =

1

6m =

mg cos =

2

3m

Solving for , we find

=3

2

g

cos

Thus, the reaction forces are

NA =1

2m

3

2

g

cos

sin =

3

4mg sin cos

NB = mg 12

m

3

2

g

cos

cos = mg

1 3

4cos2

8.3 Motion of a Rigid Body in Three DimensionsAs with two-dimensional rigid-body motion, the equations of motion are

ni=1

Fi = ma andni=1

MCmi = HCm (8.69)

Since these are vector equations and we are considering full three-dimensional motion, we

have a total of six scalar equations to solve. The equation relating the sum of the external

forces and the acceleration of the center of mass involves a minor adjustment relative to

two-dimensional applications, viz., we simply add a third component. However, the number

of active components in the moment equation increases from one to three. Furthermore, the

moment equation now involves the entire inertia tensor as opposed to the single diagonal

moment, Izz .Before we can proceed, we must find a convenient way to compute HCm . Some choices of

coordinate axes with Cm at the origin yield extremely complicated equations. To understandwhy, consider the rate of change of HCm = [I], which is

HCm =d

dt([I]) =

d[I]

dt + [I]

d

dt(8.70)

Unless we choose axes for which [I] does not change with time, the moment equation caneasily prove to be intractable for even relatively simple motion. Thus, we seek a coordinate

frame centered at Cm for which [I] is independent of time.It is worthwhile to pause at this point and review the basis of Equation (8.69). This

equation reflects our kinematics formulation in which we make use of Chasles Theorem (see

Section 7.3). That is, we have found that motion of a rigid body can be represented as a

combination of translations and rotations, which is valid in both two- and three-dimensional

motion. We have also found that it is especially convenient to focus on the center of mass.

We thus handle the translational part of the motion by observing the way in which the center

of mass moves relative to an inertial frame, viz., the first equality stated in Equation (8.69).

To handle the rotational part of the motion, we use a coordinate system whose origin, Cm, isfixed on the center of mass. Because this coordinate frame translates, any rotation of the rigid

body must be about an axis passing through Cm.


18/42


O

Cm

x

z

yx

z

y

r

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Figure 8.8: Inertial reference frame, xyz, and a center-of-mass coordinate frame, xyz, thatrotates about its z axis.

We can often accomplish the objective of locating a coordinate frame in which the inertia

tensor does not change in time by using a rotating coordinate system as shown in Figure 8.8.

For certain applications such as gyroscopes, it is convenient to select an axis parallel to anangular-velocity vector, , that differs from . Then, using the Coriolis Theorem,

dHCmdt

=d HCm

dt+HCm (8.71)

where, in component form, we have

d HCmdt

= HCmx i + HCmy j

+ HCmz k (8.72)

and HCmx = Ixxx + Ixyy + Ixzz , etc. This equation assumes an especially elegantform if xyz coincide with the principal axes of the rigid bodys inertia tensor. When this istrue, the angular-momentum vector is

HCm = Ixxx i + Iyyy j

+ Izzz k (8.73)

So, if we select = , then

HCm = Ixxx i + Iyyyj

+ Izz zk +

i

j

k

x y z

Ixxx Iyyy Izzz

(8.74)Therefore, after a little algebra, we find

HCm = [Ixx x (Iyy Izz)yz] i

+ [Iyyy (Izz Ixx)zx]j

+ [Izzz (Ixx Iyy)xy] k (8.75)

Although no products of inertia appear in this equation, we cannot avoid terms proportional

to products of the angular-velocity components such as yz . This is a ramification of

rotational-mode coupling that is inherent to three-dimensional motion. These nonlinear terms

arise because of the difference between the rate of change of seen by non-rotating and

rotating observers, which are related through the Coriolis Theorem. Nevertheless, this vector


19/42


equation is a dramatically simplified equivalent of the full three-dimensional vector equation

with all of the moments and products of inertia appearing. In component form, we arrive at

the following six equations that govern three-dimensional motion of rigid bodies. They are

known as Eulers Equations.

ni=1

Fxi = max (8.76)

ni=1

Fyi = may (8.77)

ni=1

Fzi = maz (8.78)

ni=1

Mxi = Ixxx (Iyy Izz)yz (8.79)

ni=1

Myi = Iyyy (Izz Ixx)zx (8.80)

ni=1

Mzi = Izz z (Ixx Iyy)xy (8.81)

To see how we approach a problem involving three-dimensional motion of a rigid body,

consider a disk of radius r and mass m that is mounted on Axle OA of length L = 2r andmass 9

16m. As shown in Figure 8.9, the axis pivots about Point O, and the disk rolls without

sliding on a horizontal surface counterclockwise with angular velocity a. We will proceed

step by step and ultimately we will compute the reaction forces on the disk and at Point O.

ss qO

A

x

z

yr

L

a

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Figure 8.9: Rolling disk rotating about a central axis.

Kinematics. We begin by determining the absolute angular-velocity vector, . To do this, we

observe that the disk moves in a circle of perimeter P = 2L. If Axle OA rotates clockwisewith angular velocity about the z axis, then one revolution takes t = 2/ to complete.

Hence, the speed of any point on the perimeter of the disk is

v =P

t=

2L

2/= L (8.82)

But, the disk rotates at angular velocity a about the y axis so that its velocity is also givenby v = ra. Thus, we conclude that

L = ra = a =L

r (8.83)


20/42


However, we are given L = 2r, so that a = 2. Therefore,

= 2j1

2k (8.84)

The following table includes kinematical properties of the axle and the disk that we need in

order to compute the position, velocity and acceleration of the center of mass. It also includes

inertia-tensor information for each object.

Table 8.1: Kinematical Properties of the Axle and Rod Assembly

Object x y z I xx Iyy Izz Mass

Disk 0 2r 0 14

mr2 12

mr2 14

mr2 m

Axle 0 r 0 112

ML2 0 112

ML2 M = 916

m

Table 8.1 shows that the center of mass lies at x = z = 0 and

y = 9

16 mr + 2mr9

16m + m

= 4125

r (8.85)

Therefore, the center of mass is located at

r =41

25rj (8.86)

The center of mass is constrained to circular motion. Specifically, it moves on a circle of

radius 4125

r about the z axis with angular velocity = k. Thus, its velocity is

v = r = (k)

41

25rj

=

41

25r i (8.87)

Also, the acceleration of the center of mass is purely centripetal, which means

a = ( r) = (k)

(k)

41

25rj

= ( k)

41

25r i

=

41

25

2rj (8.88)

Of greatest importance, Ixx, Iyy and Izz are constant in the coordinate system xyz rotatingwith = k. For the given values of M and L, note that

1

12ML2 =

1

12

9

16m(2r)2 =

3

16mr2 (8.89)

Thus, using the Parallel Axis Theorem, we determine Ixx according to

Ixx = 14

mr2 + m

y2d + z2

d

Disk

+ 316

mr2 + 916

m

y2r + z2

r

Rod

(8.90)

where subscripts r and d correspond to the rod and the disk, respectively. Appealing toTable 8.1, yd = 2r, zd = 0, yr = r and zr = 0. A short algebraic exercise shows that

Ixx = 5mr2 (8.91)


21/42


For Iyy , we have

Iyy =1

2mr2 + m x

2

d + z2

d Disk

+ 0 +9

16m x

2

r + z2

r Rod

(8.92)

Reference to Table 8.1 shows that xd = zd = xr = zr = 0, which yields

Iyy =1

2mr2 (8.93)

Finally, for Izz , we have

Izz =1

4mr2 + m

x2d + y

2

d

Disk

+3

16mr2 +

9

16m

x2r + y2

r

Rod

(8.94)

Again referring to Table 8.1, xd = 0, yd = 2r, xr = 0 and yr = r. Consequently, weconclude that

Izz = 5mr2 (8.95)

Similar computations quickly reveal the fact that all of the products of inertia, i.e., the off-

diagonal terms are zero. Thus, the inertia tensor is as follows.

[I] =

5mr2 0 0

0 12

mr2 0

0 0 5mr2

(8.96)

We can now compute the inertia tensor relative to the center of mass, [I]. To determine [I],we must again use the Parallel Axis Theorem, and we find

Ixx = Ixx 25

16 m y2 + z2 = 5mr2 2516 m41252

r2

=

319

400 mr2

(8.97)

Iyy = Iyy 25

16m

x2 + z2

=1

2mr2 0 =

1

2mr2 (8.98)

Izz = Izz 25

16m

x2 + y2

= 5mr2 25

16m

41

25

2r2 =

319

400mr2 (8.99)

Therefore, the inertia tensor relative to the center of mass is

[I] =

319

400mr2 0 0

0 12

mr2 0

0 0 319400

mr2

(8.100)

Kinetics. Equations (8.84) and (8.100) provide sufficient information to compute the angular-momentum vector relative to the center of mass. We simply multiply the inertia tensor [I] bythe absolute angular-velocity vector, viz.,

HCm =

319

400mr2 0 0

0 12

mr2 0

0 0 319400

mr2

0

2

=

0

mr2

319

400mr2

(8.101)


22/42


Thus, in conventional vector form, the angular-momentum vector relative to Cm is

HCm = mr2

j319

400k (8.102)

We use the Coriolis Theorem to compute the rate of change of HCm . Since HCm is constant

in the rotating coordinate frame, we have

dHCmdt

=d HCm

dt+HCm = (k)mr

2

j

319

400k

(8.103)

Evaluating the cross product, we conclude that

HCm = m2r2 i (8.104)

In the final part of our computation, we will determine the reaction force from the surface

on the disk, N, and the reaction force at the top of the pivot rod, R. Figure 8.10 shows a

side view of the configuration, including N, R and the gravitational force that acts throughthe center of mass, 25

16mg k, where g is gravitational acceleration.

O Cm A

z

y

41

25r

9

25r

R 2516

mg k

N

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Figure 8.10: Forces acting on the pivot-rod/axle/disk assembly.

The force equation in vector form is

i

Fi =25

16ma = R+ Nk

25

16mg k=

25

16m

41

25

2rj

(8.105)

where we have used Equation (8.88) for the acceleration of the center of mass. Solving forR

yields

R = 41

16m2rj +

25

16mg N

k (8.106)

The angular-momentum equation is

i

MCmi = HCm =

9

25rj

(Nk) +

41

25rj

R = m2r2 i (8.107)

Substituting for R from Equation (8.106), we have

9

25rN i

41

25r

25

16mg N

i = m2r2 i (8.108)


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8.4. PRINCIPLE OF WORK AND ENERGY 273

Solving for the magnitude of the reaction force on the disk from the surface, N, yields

N =1

2m2r +

41

32mg (8.109)

Combining Equations (8.106) and (8.109) shows that the forces N and R are as follows.

N =

1

2m2r +

41

32mg

k (8.110)

R = 41

16m2rj

1

2m2r

9

32mg

k (8.111)

Example 8.5 Taking moments about Point O for the rolling-disk problem depicted in Figure 8.9, show that Equation (8.109) follows.

Solution. The first thing we must do is determine the angular-momentum vector relative to Point O.Using Equations (8.84) and (8.96),

HO =

5mr2 0 0

0 12

mr2 0

0 0 5mr2

0

2

=

0

mr2

5mr2

= mr2 (j 5k)

The inertia tensor does not change in time for a rotating observer, regardless of his position on the

y axis. Thus, the Coriolis Theorem tells us that

dHOdt

=d HO

dt+HO = (k)mr2 (j 5k) = m2r2 i

Taking moments about Point O, we have

i MOi = 0R+ 41

25rj

25

16mg k+ (2rj) (Nk) = m

2r2 i

Evaluating the cross products gives

4116

mgr i + 2Nr i = m2r2 i = N = 12

m2r +41

32mg

which is identical to Equation (8.109).

8.4 Principle of Work and Energy

As with single particles and systems of particles, the work done by external forces moving

from State 1 to State 2, U12, is equal to the change in kinetic energy, T2 T1, viz.,

U12 = T2 T1 (8.112)

Furthermore, if our application involves several rigid bodies, we can say

mi=1

(U12)i =mi=1

(T2)i mi=1

(T1)i (8.113)


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Following our formulation for systems of particles, we generalize Equation (6.50) by replacing

the summation with an integral. Since the velocity relative to the center of mass is v = r,the total kinetic energy for a rigid body is thus given by

T = 12

mv v + 12

( r) (r) dm (8.114)

With a bit of algebra, we find

T =1

2mv2 +

1

2

Ixx

2

x +Iyy2

y +Izz2

z +2Ixyxy +2Iyzyz +2Ixzxz

(8.115)

In terms of principal axes, we have

T =1

2mv2 +

1

2

Ixx

2

x + Iyy2

y + Izz2

z

(8.116)

For two-dimensional motion, since x = y = 0, Equation (8.116) simplifies to

T =1

2mv2 +

1

2I2 (Two-dimensional motion) (8.117)

where we replace Izz with I for consistency with the notation used in Section 8.2.

Example 8.6 An axisymmetric rigid body of mass m , radius R and moment of inertia about its symmetry axis, I , is released from rest on an incline. Assuming the body rolls without sliding,determine the speed of the body after it has rolled a distance corresponding to z = h.

z

x

f

N

mga

2R

Cm

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Solution: Because the body rolls without sliding, a = R. Integrating once over time and usingthe fact that the body is initially at rest, we conclude that = v/R. The initial kinetic energy isT1 = 0 and, at the lower elevation, it is

T2 =1

2mv2 +

1

2I2 =

1

2mv2 +

1

2

I

R2v2 =

1

2

m +

I

R2

v2

The point of contact between the body and the surface is the instantaneous center of rotation, so

that the bodys velocity relative to the surface is zero. Consequently, the rolling-friction force does

no work as there is no relative displacement between the point of contact and the surface. Also,

the reaction force, N, is normal to the direction of motion and thus does no work. The work doneby gravity is

U12 = mgh

Hence, the Principle of Work and Energy yields

mgh =1

2

m +

I

R2

v2


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8.5. PRINCIPLE OF IMPULSE AND MOMENTUM 275

Example 8.6 (Continued) Solving for v, we obtain

v = 2gh

1 + I/ (mR2)

For a block sliding down a frictionless incline, we know that v =

2gh. To show the effect ofrolling friction, the following table lists the ratio of v to

2gh for three different objects. Note

that the values of I are from Appendix A.

Object I/

mR2

v/

2gh

Sphere 0.40 0.85

Cylinder 0.50 0.82

Hoop 1.00 0.71

8.5 Principle of Impulse and Momentum

As with systems of particles, we must take account of both linear and angular momenta. It

is often most convenient to compute moments about the rigid bodys center of mass, and

integrating the equations governing the motion from time t1 to time t2 yields

ni=1

t2t1

Fi dt = mv2 mv1 (8.118)

ni=1

t2t1

rFi dt = HCm2 HCm1 (8.119)

For two-dimensional applications, the angular-momentum vector is given by HCm = I k.Thus, for two-dimensional motion of a rigid body, Equation (8.119) simplifies to

ni=1

t2t1

rFi dt = I(2 1) k (Two-dimensional motion) (8.120)

We refer to the sum of the external forces integrated from time t1 to t2 as the linearimpulse acting on a rigid body. We denote it by Imp. Similarly, the sum of the external

moments integrated over the same time interval is the angular impulse, and we denote it by

Impa. Thus, by definition we have

Imp

ni=1

t2t1

Fi dt and Impa ni=1

t2t1

rFi dt (8.121)

In terms of the linear and angular impulses, the Principle of Impulse and Momentum for arigid body is

Imp = mv2 mv1 and Impa = HCm2 HCm1 (8.122)

In this section, we will apply the Principle of Impulse and Momentum to a rigid body

for a given impulsive force. This illustrates how we must account for both translation and

rotation attending the post-impulse motion. The astute reader might pose a question about

why impacts between rigid bodies have been excluded from this section. The reason we dont


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generalize our analysis from single particles and systems of particles is quite clear. Classical

imp

Documents

Dynamics Ch8