Moment of Momentum of a Tree of Rigid Bodies

In the analysis of some movements, the human body can be considered as a tree of rigid body segments. What is the angular momentum of a tree of rigid objects about the center of mass of the system?

We can address this question by computing the angular momentum H of each rigid body Bi in the system about the center of mass of the system by using Eqn. 9.13a:

iHc = mi rCi/C X EvCi + Hci where mi and Ci represent the mass and the center of mass of body Bi. Summing over all bodies in the multibody system, we obtain

The first term on the right-hand side, H*c, is the moment of momentum about point C of the system of lumped masses where the mass of each body in the system is lumped in the center of mass of the body. The second term is the sum of angular momentums of each rigid body in the system about its own center of mass.

If the velocity of center of mass of each of these bodies are the same, then the first term will be equal to zero because %mi rCl/C defines position of the center of mass of the system. In this case the moment of momentum will be the sum of angular momentum of each body about its own center of mass.

In Chapter 3, we have seen that the laws of motion for a system of particles can be written as follows:

where 2F is the resultant external force acting on the system, ac is the acceleration of the center of mass, and 2Mc is the resultant external moment with respect to the center of mass of the system. These laws apply for a rigid body because it is composed of a large number of particles. When the resultant external moment acting on a system is zero, then Hc must be equal to a constant. This is the case for airborne heavy bodies where the only significant external force is the force of gravity and it passes through the center of mass of the body.

If a point within the body is fixed in the inertial reference system, Eqn. 9.20a can be replaced with the following:

where Moext is the resultant external moment with respect to the fixed point O. In Eqns. 9.20 we have specified the frame in which a time derivative must be taken. If the frame is not specified, it is understood to be the inertial reference frame E, that is:

Let the angular momentum of a body B expressed in the following form: Hc = Hcidi + Hc2 d2 + Hc3 d3

where dj represents the unit vectors in a reference frame D.

Let ft = 01 d1 + 02 d2 + 03 d3 be the angular velocity of D in E. Note that if D were a reference frame embedded in body B then the angular velocity ft would be equal to the angular velocity m of body B in reference frame E.

Using Eqn. 9.7, the time derivative of Hc in the inertial reference frame E can be expressed as follows:

Substituting Eqns. 9.14 into this equation, we derive three differential equations governing the time rate of change of angular velocity of B. These equations are called the equations of angular momentum:

M1 = Ic11 (d«x/dt) + Ic12 (d«2/dt) + Ic13 (d«3/dt)

- O3 (Ic21«1 + Ic22«2 + Ic23 «3) + ^2 (I°31«1 + I°32 «2 + I°33 «3)

M2 = Ic21 (d«1/dt) + Ic22 (d«2/dt) + Ic23 (d«3/dt) +

+ O3 (Ic12 «2 + Ic11 «1 + Ic13 «3) - (Ic31«1 + Ic32 «2 + Ic33 «3)

M3 = Ic31 (d«1/dt) + Ic32 (d«2/dt) + Ic33 (d«3/dt) -

- O2 (Ic11«1 + Ic12 «2 + Ic13 «3) + (Ic21«1 + Ic22«2 + Ic23 «3)

Taken together with the primary equation for angular velocity:

Equation 9.22 determines both the angular velocity and the orientation of the body in space, provided that the orientation and the angular velocity are known at an initial time and that the resultant external moment acting on the body is specified as a function of time. If an object rotates about a fixed point O, the equations of angular momentum with respect to point O are identical to that of Eqns. 9.22 with Ici;- replaced by I°j.

Integration of Euler's equations is not trivial even for simple-shaped bodies. Often, however, the angular velocity of a body may be a function of only one or two variables. In other cases, as in the study of airborne motion, the rotational velocity is found by setting the angular momentum about the center of mass equal to a constant. Additionally, commercially available programs such as Working Model 3D can be used to determine both the motion of the center of mass and the rotation of the body around the center of mass for a rigid body or a tree of rigid bodies. Yet another option is to quantify the motion using photogrammetric means, and then use the laws of motion to evaluate the resultant force and the resultant moment acting on a body segment.

Example 9.5. Precession of a Football. When a top or an American football is set to spinning about the vertical axis, the long axis of the top (football) may initially remain vertical (Fig. 9.8). As friction reduces the spin rate, the spin axis begins to lean over and rotate about the vertical axis. This phase of the top's motion approximates steady precession. Thus it is reasonable to assume the rate of rotation to be constant. Determine an equation that relates the spin rates to the mass moment of inertia and the angle of inclination of the top with the vertical axis.

Solution: Let ei, e2, and e3 denote the unit vectors of reference frame E. We choose origin O of E to coincide with the point of the top and the e3 vertical upward. The reference frame D has also its origins at O.

Let d3 denote the unit vector along the axis of the top as shown in Fig. 9.8. The d1 axis is taken to be in the plane of e3 and d3. The unit vector d2 is equal to the vector product of d1 and d3. The nutation angle between d3 and e3 is denoted by the symbol 0. This angle remains constant during the motion. The top spins about the d3 axis with a constant spin rate equal to (d0/dt). The center of mass of the top traverses a circle about the vertical axis e3 at a constant precession rate equal to (d¥/dt).

The unit vector e3 remains constant in both reference frames D and E during the spinning of football, indicating that the motion is planar. The angular velocity of D in E is then given by the following expression:

emd = ft = (d^/dt) e3 = (d^/dt) (-sin 0 d1 + cos 0 d3)

O1 = -(dW/dt) sin 0, O2 = 0 , O3 = (d^/dt) cos 0 (9.23a)

The angular velocity of the top with respect to D is dmb = (d0/dt) d3

where B denotes a reference frame that is fixed in B. Angular velocity of the top with respect to E is obtained using the relation:

emb = m = emd + dmb = (d^/dt) (-sin 0 d1 + cos 0 d3) + (d0/dt) d3 -(d^/dt) sin 0 d1 + [(d^/dt) cos 0 + (d0/dt)] d3

«1 = -(d^/dt) sin 0, «2 = 0, «3 = (d^/dt) cos 0 + (d0/dt) (9.23b)

Because 0 remains constant during motion and the angles ^ and 0 vary with time at constant rates, angular acceleration is equal to zero:

As the top is symmetric in the reference frame D, the mass moment of inertia in D can be written as

The only external moment about point O is the moment of the force of gravity. Let h be the distance between the center of mass and the point O. The moment M is then given by the equation:

M = h d3 X (-mg) e3 = h d3 X (-mg) (-sin 0 d1 + cos 0 d3)

Now we can use Euler's equations for rotation about the stationary point O. Substituting Eqns. 9.23a-e into Eqns. 9.22 and replacing Icj by Ioij, we find:

M2 = mgh sin 0 = Io21 (d«1/di) + Io22 (d«2/di) + Io23 (d«3/di)

+ O3 (Io12 «2 + Io11 «1 + Io13 «3) — O1 (Io31«1 + Io32 «2 + Io33 «3)

mgh sin 0 = —(d^/di)2 cos 0 sin 0 Io + (d^/di)2 sin 0 cos 0 I3

mgh = (d^/di)2 cos 0 (I3 — Io) + (d^/di) (d^/di) I3 (9.24a)

The components of Euler's equations in d1 and d3 directions are identically satisfied. If we know the spin rate (dfi/di) and the nutation angle 0, we can solve Eqn. 9.24a for the top's precession rate (d^/di). Note that the top must spin fast for this motion to occur. This is because (I3 — Io) < 0, and thus Eqn. 9.24a can be put in the form:

(d4>/di) = mgh/[(d^/di)l3] + (d¥/di) cos 0 (Io — WI3 (9.24b)

A spinning top seems to defy gravity. The solution presented indicates that a top could remain inclined to the vertical axis with only the tip touching a horizontal plane so long as it spins fast enough about its own long axis. As the spin rate decreases, the top leans and the center of mass moves downward toward the horizontal plane. If the body under consideration is slender (I3 = 0), then the motion described is not possible. The example of a spinning top has implications for human movement and motion. If an aging ballet dancer spins faster on stage in between leaps than he did years before, it is because it is easier to stay in certain postures at higher spin rates.

Getting Started With Dumbbells

Getting Started With Dumbbells

The use of dumbbells gives you a much more comprehensive strengthening effect because the workout engages your stabilizer muscles, in addition to the muscle you may be pin-pointing. Without all of the belts and artificial stabilizers of a machine, you also engage your core muscles, which are your body's natural stabilizers.

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