Figure 4.1a,b. Oscillation of a pendulum about the vertical axis. The motion takes place in the (ei, e2) vertical plane. The slender rod OA has uniform mass density. A small mass element of the rod is shown in (a). The free-body diagram of the rod showing all the external forces acting on it is given in (b).

The term (d^/dt) e3 is equal to m = a e3 and that [sin2^ + cos2 = 1. After integrating this equation from 0 to L, we obtain:

In the previous chapter we had found for a pendulum of the same geometry but with mass m concentrated at the midway point:

Thus, the assumption that the mass of the rod is localized at the center of the rod rather than being uniformly distributed throughout results in the underestimation of the moment of momentum by about one-third.

As we have seen before, the conservation of momentum for a system of particles dictates that

where Mo denotes the moments of external forces with respect to point O. A free-body diagram of the pendulum is shown in Fig. 4.1b. Because the reaction force exerted by the pin passes through point O, its lever arm with respect to this point is zero and hence the pin force contributes no moment with respect to point O. The only external moment results from the gravitational force acting at the center of the rod. Thus, the conservation of moment of momentum yields the following equation:

(mL2/3) (d20/dt2) e3 = (L/2) (sin 0 ei - cos 0 e2) X (—mg e2)

from which we obtain

In Chapter 2, we have shown that for a pendulum composed of a slender rod and a lump mass m positioned at one end, the corresponding equation was Eqn. 2.17b:

Comparison of Eqns. 4.7a and 4.7b indicates that resistance to angular acceleration decreases when the mass is distributed over the length rather than concentrated at the end of the rod.

When the oscillations of the rod around the vertical axis is small (sin 0 = 0) the solution for Eqn. 4.7a can be approximated by the following equation:

in which A and B are arbitrary constants determined by the intial conditions. If we let

we obtain

This equation shows that the rod with distributed mass swings around the vertical axis with the period of 2^(2L/3g)1/2. For a simple pendulum composed of a slender rod of length L and a bob of mass m, the period of oscillation is equal to 2^(L/g)1/2. Therefore, the rod with uniform mass distribution rotates around point O much like a classical pendulum with effective length equal to 2L/3. Thus, in using the lumped-mass approach, we would have achieved an exact solution if we had placed the lumped mass at a distance 2L/3 from the fixed point O.

Next, let us turn our attention to the forces exerted by the hinge on the rod at point O. The free-body diagram in Fig. 4.1b shows all the external forces acting on the rod. These forces are the gravitational force —mg e2 acting at the center of the rod and the force (Fi ei + F2 e2 ) exerted by the pin at point O. According to the equation of motion of the center of mass, the net resultant force acting on an object must be equal to the mass of the object times the acceleration of the center of mass. The position, velocity, and acceleration of the center of mass are given by the following expressions:

Then, the components of the equation of motion give us

F1 = m (L/2) [(d2$/dt2) cos $ — (d$/dt)2 sin $] (4.10a)

F2 = —mg + m (L/2) [(d2$/dt2) sin $ + (d$/dt)2 cos $] (4.10b)

Once we compute the angle $ as a function of time by using Eqn. 4.8, the reaction forces at the hinge O can be determined with the use of Eqns. 4.10a and 4.10b. Note that when the pendulum is at rest in the vertical position we have F1 = 0 and F2 = — mg.

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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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