The rigid body in Fig. (a) consists of the small masses m1 and m2, which are connected to the ends of a light bar that can rotate about a pin at O. The ideal spring is undeformed when the bar is vertical. Determine the circular frequency of small vibrations.

Question

The rigid body in Fig. (a) consists of the small masses [latex]m_1[/latex] and [latex]m_2[/latex], which are connected to the ends of a light bar that can rotate about a pin at O. The ideal spring is undeformed when the bar is vertical. Determine the circular frequency of small vibrations.

Accepted Answer

Because the kinetic energy of the system is computed easily, this problem is well suited for solution using conservation of energy. We choose the angle θ shown in Fig. (b) as the position coordinate. Note that θ = 0 in the static equilibrium position, as required by the energy method. Taking the equilibrium position as the datum for potential energy and referring to Fig. (b), we obtain

[latex]V = m_1gh_1 − m_2gh_2 + \frac{1}{2}k(eθ)^2[/latex]

where [latex]h_1[/latex] = a(1 − cos θ ) and [latex]h_2[/latex] = b(1 − cos θ ). With the approximation cos θ ≈ 1 − θ²/2, we get [latex]h_1[/latex] ≈ aθ²/2 and [latex]h_2[/latex] ≈ bθ²/2. Therefore, the potential energy becomes

[latex]V = \frac{1}{2}(m_1ga − m_2gb + ke^2)θ^2[/latex]

Comparison with Eq. (20.37) yields

[latex]V = V_0 + F_0q +\frac{1}{2}Kq^2[/latex] (20.37)

[latex]K = m_1ga − m_2gb + ke^2[/latex]

Because O is a fixed point, the kinetic energy of the system is

[latex]T = \frac{1}{2}I_O\dot θ^2 = \frac{1}{2}(m_1a^2 + m_2b^2)\dot θ^2[/latex]

Comparing this with Eq. (20.36) we conclude that

[latex]T = \frac{1}{2}M\dot q^2[/latex] (20.36)

[latex]M = m_1a^2 + m_2b^2[/latex]

Hence, the circular frequency of the system is

[latex]p = \sqrt{\frac{K}{M}} = \sqrt{\frac{m_1ga − m_2gb + ke^2}{m_1a^2 + m_2b^2}}[/latex]

Question 20.8: The rigid body in Fig. (a) consists of the small masses m1 a...

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