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Three-dimensional kinematics of rigid bodies

Hiroki Okubo

1 Introduction

Although a large percentage of dynamics problemsin engineering can be solved by the principles of plane motion, modern developments have focusedincreasing attention on problems which call for theanalysis of motion in three dimensions. The intro-duction of a third dimension adds the possibility of two additional components for vectors represent-ing angular quantities including moments of forces,

angular velocity, angular acceleration, and angularmomentum.

2 Translation

Any two points in the body, such as  A  and  B , willmove along parallel straight lines if the motion isone of   rectilinear translation  or  curvilinear transla-tion . In either case, every line in the body, such asAB, remains parallel to its original position.

The position vectors and their first and secondtime derivatives are

rA   =   rB + rA/B   (1)

vA   =   vB   (2)

aA   =   aB   (3)

where  rA/B  remains constant.

3 Fixed-axis rotation

We consider the   rotation   of a rigid body about afixed axis n-n   in space with an angular velocity  ω.For fixed-axis rotation,  ω  does not change its direc-

tion. We choose the origin O of the fixed coordinatesystem on the rotation axis. Any point such as  Awhich is not on the axis moves in a circular arc inthe plane normal to the axis and has a velocity

v =  ω × r   (4)

The acceleration of  A is given by the time derivativeEq. (4).

a =  ω̇ × r + ω × (ω × r) (5)

4 Rotation about a fixed point

When a body rotates about a fixed point, theangular-velocity vector no longer remains fixed in

direction.

4.1 Rotation and proper vectors

We consider  Infinitesimal  rotations dθ1  and  dθ2  of a rigid body about the respective axes through thefixed point  O. As a result of  dθ1, point  A  has adisplacement   dθ1  × r, and   dθ2   causes a displace-ment   dθ2   ×  r   of point   A. Either order of ad-dition of these infinitesimal displacements clearlyproduces the same resultant displacement, which isdθ1  × r +  dθ2  × r   = (dθ1  +  dθ2) × r. Thus, the

two rotations are equivalent to the single rotationdθ =  dθ1 + dθ2. It follows that the angular veloc-ities  ω1   =  θ̇1   and  ω2   =  θ̇2  may be added to giveω =  θ̇ =  ω1 + ω2.

4.2 Angular acceleration

The angular acceleration α of a rigid body in three-dimensional motion is the time derivative of its an-gular velocity,  α =  ω̇.

When the magnitude of  ω  remains constant, theangular acceleration  α  is normal to  ω. If we let  Ωstand for the angular velocity with which the vectorω itself rotates (precesses ), the angular accelerationmay be written

α =  Ω × ω   (6)

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 y

 z

 N A

Oβ  =30

Figure 1: Rotating arm

5 Example

The 0.8-m arm  OA   is pivoted about the horizon-

tal  x-axis, and the entire assembly rotates aboutthe  z -axis with a constant speed  N  = 60 rev/min.Simultaneously, the arm is being raised at the con-stant rate  β̇   = 4 rad/s. For the position whereβ  = 30◦, determine (a) the angular velocity of  OA,(b) the angular acceleration of  OA, (c) the velocityof point  A, and (d) the acceleration of point  A.

(a) Since the arm  OA   is rotating about both thex- and the z-axes, it has the components  ωx =β̇  = 4 rad/s and  ωz   = 2πN/60 = 6.28 rad/s.The angular velocity is

ω =  ωx + ωz  = 4i + 6.28k rad/s (7)

(b) The angular acceleration of  OA  is

α =  ω̇x +  ω̇z   (8)

Since  ωz   is not changing in magnitude or di-rection,  ω̇z  =  0.   ωx   is changing direction andhas a derivative which is

ω̇x =  ωz × ωx = 25.1 j  rad/s2 (9)

Therefore,

α = 25.1 j rad/s

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(10)

(c) With the position vector of  A  given by   r   =0.693 j + 0.4k  m, the velocity of  A  becomes

v =  ω×r = −4.35i−1.60 j+ 2.77k m/s (11)

(d) The acceleration of  A  is

a   =   α× r + ω × (ω × r) (12)

= 20.1i− 38.4 j − 6.40k m/s2 (13)

References[1] J. L. Meriam and L. G. Kraige, (2001), Engi-

neering Mechanics, Volume 2, Dynamics, 5thedition, Wiley

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