A homogeneous thin rod of mass m and length l is connected to a vertical shaft S by a smooth hinge bearing at Q. The shaft rotates with a constant angular velocity Ω, as shown in Fig. 10.11. (i) Derive the equation of motion of the rod. (ii) Determine as a function of θ the hinge bearing reaction

Question

A homogeneous thin rod of mass m and length [latex]\mathcal{l}[/latex] is connected to a vertical shaft S by a smooth hinge bearing at [latex]\mathcal{Q}[/latex]. The shaft rotates with a constant angular velocity Ω, as shown in Fig. 10.11. (i) Derive the equation of motion of the rod. (ii) Determine as a function of θ the hinge bearing reaction torque exerted on the rod at [latex]\mathcal{Q}[/latex], for the initial data [latex]\dot{ heta}(0)=0 ext { at } heta(0)= heta_0 [/latex]. (iii) Analyze the infinitesimal stability of the relative equilibrium states of the rod.

Accepted Answer

(i). The central point [latex]\mathcal{Q}[/latex] at the hinge being fixed in the inertial (machine) frame [latex]0=\left\{F ; \mathbf{I} _k ight\}[/latex], the equation of motion for the rod is obtained from Euler's law (10.66) in the principal body frame [latex]2 = \left\{\mathcal{Q ; \mathbf{i}_{k}} ight\} [/latex]. We need [latex]\mathbf{ω} , \mathbf{I}_{\mathcal{Q}}, ext{ and } \mathbf{M}_{\mathcal{Q}}[/latex].

[latex]M_1^Q=I_{11}^Q \dot{\omega}_1+\omega_2 \omega_3\left(I_{33}^Q-I_{22}^Q ight)[/latex],

[latex]M_2^Q=I_{22}^Q \dot{\omega}_2+\omega_3 \omega_1\left(I_{11}^Q-I_{33}^Q ight),[/latex] (10.66)

[latex]M_3^Q=I_{33}^Q \dot{\omega}_3+\omega_1 \omega_2\left(I_{22}^Q-I_{11}^Q ight)[/latex].

Let [latex]\mathbf{k} = \mathbf{i} _3[/latex] denote the hinge axis at [latex]\mathcal{Q}[/latex]. Then the total angular velocity [latex]\omega ≡ \omega_{20}[/latex] of the rod (frame 2) relative to the machine (frame 0) is the sum of the angular velocity [latex]\omega _{21}=\dot{ heta} \mathbf{k}[/latex] of the rod relative to the vertical shaft (frame 1 ) and the angular velocity [latex]\omega _{10}= \Omega =\pmb{\Omega} \mathbf{K}[/latex] of the shaft relative to the machine. Thus, referred to the rod frame 2,

[latex]\omega =-\Omega(\sin heta \mathbf{i} +\cos heta \mathbf{j} )+\dot{ heta} \mathbf{k}[/latex], (10.80a)

and hence the total angular acceleration of the rod is

[latex]\dot{\omega}=-\Omega \dot{ heta}(\cos heta \mathbf{i} -\sin heta \mathbf{j} )+\ddot{ heta} \mathbf{k}.[/latex] (10.80b)

The rod frame at [latex]\mathcal{Q}[/latex] is a principal body frame relative to which

[latex]\mathbf{I} _Q=\frac{1}{3} m \ell^2\left( \mathbf{i} _{11}+ \mathbf{i} _{33} ight)[/latex] (10.80c)

in accordance with (9.46d), with a change of axes in mind. Finally, the moment of the forces and torques acting on the rod about Q is

[latex]\mathbf{I} _H\left( \mathcal{B} _r ight)=\frac{m \ell^2}{3}\left( \mathbf{i}_{11}+ \mathbf{i}_{22} ight)[/latex] (9.46d)

[latex]\mathbf{M} _\mathcal{Q}=\mu_1 \mathbf{i} +\mu_2 \mathbf{j} - m g \frac{\ell}{2} \sin heta \mathbf{k} .[/latex] (10.80d)

Here [latex]\mu_1 ext { and } \mu_2[/latex] are unknown components of the torque exerted on the rod by the smooth hinge bearing for which [latex]\mu_3=0[/latex], and the weight W of the homogeneous rod acts at its center of mass at [latex]\mathbf{x} ^*=\ell / 2 \mathbf{j}[/latex] .

Use of (10.80a) through (10.80d) in Euler's equations (10.66) yields the components of the smooth hinge bearing reaction torque exerted on the rod at Q,

[latex]\mu_1=-\frac{2}{3} m \ell^2 \Omega \dot{ heta} \cos heta, \quad \mu_2=\mu_3=0[/latex] (10.80e)

and the different ial equation of motion of the rod,

[latex]\ddot{ heta}+\left(p^2-\Omega^2 \cos heta ight) \sin heta=0, \quad p^2 \equiv \frac{3 g}{2 \ell} .[/latex] (10.80f)

With the aid of this result, the reader may now determine from Euler' s first law the resultant hinge reaction force R at Q.

When Ω = 0, [latex]\mu_{1} = 0[/latex] and ( 10.80f) reduces to the familiar pendulum equation.
In this case, the circular frequency p for the rod is the same as that of a simple pendulum of length L = 2l/3. Therefore, the exact and approximate solutions for the vibration of the rod when Ω = 0 may be read from those for the simple pendulum. We next consider the case when Ω ≠ 0.

(ii).

To determine the hinge bearing reaction torque (10.80e) as a function of θ, we need [latex] \dot{ heta}( heta)[/latex], the first integral of (10.80f). With [latex]\ddot{ heta}=d\left(\frac{1}{2} \dot{ heta}^2 ight) / d heta[/latex],

separation of the variables and use of the initial data [latex]\dot{ heta}(0) = 0 ext{ at } heta(0) = heta_{0}[/latex] yields

[latex]\dot{ heta}= \pm \sqrt{\Omega^2\left(\cos ^2 heta_0-\cos ^2 heta ight)+3 \frac{g}{\ell}\left(\cos heta-\cos heta_0 ight)}[/latex] (10.80g)

Substitution of this result into (10.80e) delivers as a function of 0 the hinge bearing
reaction torque exerted on the rod at Q:

[latex]\mu_1=\mp \frac{2}{3} m \ell^2 \Omega \cos heta \sqrt{\Omega^2\left(\cos ^2 heta_0-\cos ^2 heta ight)+3 \frac{g}{\ell}\left(\cos heta-\cos heta_0 ight)}[/latex]. (10.80h)

Notice that the reaction torque [latex]\mu_{1}( heta)[/latex] vanishes when [latex] heta= heta_0[/latex] or nπ/2 (n odd) and, of course, also when Ω =o.

(iii). The relative equilibrium states [latex] heta_{s}[/latex] of the rod, by (10.80f), are given by

[latex]\left(p^2-\Omega^2 \cos heta_s ight) \sin heta_s=0[/latex], (10.80i)

which yields three distinct states

[latex] heta_s=0, \pi, \quad heta_s=\cos ^{-1}\left(\frac{p^2}{\Omega^2} ight) .[/latex] (10.80j)

To examine the infinitesimal stability of these states , introduce [latex] heta = heta_{s} + δ[/latex], where δ is an infinitesimal angular disturbance from [latex] heta_{s}[/latex] Then recalling (10.80i) and retaining only terms of the first order in δ, we obtain from (10.80f),

[latex]\ddot{\delta}+\left\{\left[p^2-\Omega^2 \cos heta_s ight] \cos heta_s+\Omega^2 \sin ^2 heta_s ight\} \delta=0[/latex]. (10.80k)

Therefore, the disturbance is bounded and simple harmonic, and hence infinitesimally stable, if and only if the coefficient

[latex]\omega^2 \equiv\left[p^2-\Omega^2 \cos heta_s ight] \cos heta_s+\Omega^2 \sin ^2 heta_s>0 .[/latex] (10.80l)

We now recall (10.80j). First consider [latex] heta_{s} = 0[/latex] Then (10.80l) requires [latex]\omega^{2} = p^2-\Omega^2>0[/latex]. Hence, the relative equilibrium state [latex] heta_{s} = 0[/latex] is infinitesimally stable if and only if [latex]\Omega

[latex]\Omega_c \equiv p=\sqrt{\frac{3 g}{2 \ell}}[/latex], (10.80m)

which is independent of the mass of the rod. In this case, the small amplitude circular frequency of the rod oscillations about the vertical state [latex] heta_s=0[/latex] is [latex] \omega_v \equiv \left(p^2-\Omega^2 ight)^{1 / 2} ext {. }[/latex]

Of course, the inverted vertical configuration of the rod [latex] heta_{s} = \pi[/latex] is impractic al in respect of the suggested design in Fig. 10.11. Even so, (10.80l) fails for [latex] heta_{s} = \pi[/latex] , and hence the inverted relative equilibrium state of the rod is inherently unstable for all angular speeds of the vertical shaft.

Notice that when [latex]\Omega=\Omega_c=p, \omega^2 \leq 0[/latex] for all positions (10.80j ). Hence, no infinitesimally stable relative equilibrium states of the rod exist at the critical angular speed [latex]\Omega=\Omega_c[/latex].

Finally, consider the case [latex]\cos heta_s=p^2 / \Omega^2[/latex] This requires p/ Ω < l ; hence (10.80l) is satisfied, and the relative equilibrium state [latex] heta_{s} = cos^{- 1}(p^2/ \Omega^2)[/latex], regardless of the mass of the rod, is infinitesimally stable if and only if [latex]\Omega>p=\Omega_c[/latex]. The small amplitude vibrational frequency of the rod about this displaced relative equilibrium state is [latex]\omega_d \equiv \Omega\left(1-p^4 / \Omega^4 ight)^{1 / 2}[/latex].

In sum, if [latex]\Omega<\Omega_c[/latex], the vertical relative equilibrium position of the rod is its only infinitesimally stable relative equilibrium state. When the vertical shaft attains its critical angular speed, [latex]\Omega =\Omega_c[/latex] , however, no infinitesimally stable relative equilibrium positions of the rod exist. Afterwards, when [latex]\Omega >\Omega_c[/latex], the displaced configuration of the rod at [latex] heta_s=\cos ^{-1}\left(p^2 / \Omega^2 ight)[/latex] is its only infinitesimally stable relative equilibrium position.

Question 10.10: A homogeneous thin rod of mass m and length l is connected t...

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