Figure (a) shows a homogeneous semicylinder that rocks back and forth without slipping on the horizontal plane. The angular position of the semicylinder is defined by the angle θ. Determine the natural circular frequency of the oscillations for small amplitudes, assuming the motion to be simple

Question

Figure (a) shows a homogeneous semicylinder that rocks back and forth without slipping on the horizontal plane. The angular position of the semicylinder is defined by the angle θ. Determine the natural circular frequency of the oscillations for small amplitudes, assuming the motion to be simple harmonic. The mass center G is located at the distance 4R/3π = 0.4244R from point O.

Accepted Answer

This problem is not well suited for solution using conservation of energy, because the kinetic energy cannot be easily computed in an arbitrary position. The source of the difficulty is the displacement of the contact point between the semicylinder and the horizontal surface, which gives rise to geometric complications. We can overcome this problem by using Rayleigh’s principle.
The position of maximum potential energy is shown in Fig. (b). Choosing the horizontal surface as the datum, the potential energy is

[latex]V_{max} = mg(R − 0.4244R cos θ_{max}) ≈ mg ( R − 0.4244R ( 1 - \frac{θ_{max}^2}{2}))[/latex]
= 0.5756 mg R + 0.2122 mg R[latex]θ_{max}^2[/latex]

Note that [latex]V_0[/latex] = 0.5756mg R is the potential energy in the equilibrium (θ = 0) position.

The maximum kinetic energy of the semicylinder occurs in the position θ = 0, which is shown in Fig. (c). Because the contact point C is the instant center for velocity, the kinetic energy is [latex]T_{max} = (1/2)I_C\dot θ_{max}^2[/latex]. By the parallel-axis theorem

[latex]\bar I = I_O − m(\overline{OG})^2[/latex] = 0.5 m R² − m(0.4244 R)² = 0.3199 m R²

[latex]I_C = \bar I + m(\overline{GC})^2[/latex] = 0.3199 m R² + m(0.5756 R)² = 0.6512 m R²

so that

[latex]T_{max} = \frac{1}{2}(0.6512 m R^2)\dot θ_{max}^2 = 0.3256 m R^2 \dot θ_{max}^2[/latex]

[latex]= 0.3256 m R^2 p^2 θ_{max}^2[/latex]

In the last step, we substituted [latex]\dot θ_{max}^2 = p^2 θ_{max}^2[/latex], thereby assuming the motion to be harmonic.

Rayleigh’s principle [latex]T_{max} = V_{max} − V_0[/latex] now becomes

[latex]0.3256 m R^2 p^2θ_{max}^2 = 0.2122 mg Rθ_{max}^2[/latex]

yielding for the circular frequency

[latex]p = 0.807 \sqrt{\frac{g}{R}}[/latex]

Question 20.9: Figure (a) shows a homogeneous semicylinder that rocks back ...

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