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A \(12 \mathrm{~kg}\) spool that is \(1 \mathrm{~m}\) in radius is pinned to a viscoelastic rod of negligible mass with effective properties \(k=10 \mathrm{~N} / \mathrm{m}\) and \(c=8 \mathrm{~N}-\mathrm{sec} / \mathrm{m}\). The end of the rod is attached to a rigid support as shown. Determine the natural frequency of the system if the spool rolls without slipping.

Short Answer

Expert verified
Answer: The natural frequency of the system with the spool rolling without slipping is \(\sqrt{\frac{5}{6}} \mathrm{~rad/s}\).

Step by step solution

01

Determine the moment of inertia of the spool

To determine the moment of inertia, we will use the formula for a cylinder, which is \(I = \frac{1}{2}mR^2\), where \(I\) is the moment of inertia, \(m\) is the mass, and \(R\) is the radius. In this case, we have \(m = 12 \mathrm{~kg}\) and \(R = 1 \mathrm{~m}\): \(I = \frac{1}{2}(12 \mathrm{~kg})(1 \mathrm{~m})^2 = 6 \mathrm{~kgā‹…m^2}\)
02

Apply the equation of motion

Considering the forces acting on the spool - gravitational force and the forces exerted by the rod, we will apply Newton's second law of motion. For rolling without slipping, the translational and rotational acceleration are related as \(\alpha R = a\), where \(\alpha\) is the angular acceleration and \(a\) is the linear acceleration. The forces exerted by the rod can be calculated as: Spring force: \(F_s = kx\) Damping force: \(F_d = cv\) Total force: \(F_T = F_s + F_d = kx + cv\) For translational motion: \(F_T = ma\) For rotational motion: \(F_T R = I\alpha\) Substituting the rotational equation with the translational one, we get: \(kx + cv = I\frac{a}{R}\)
03

Find the natural frequency

To find the natural frequency, we use the formula \(\omega_n=\sqrt{\frac{k}{m}}\), where \(\omega_n\) is the natural frequency, \(k\) is the stiffness, and \(m\) is the mass. \(\omega_n=\sqrt{\frac{10 \mathrm{~N}/\mathrm{m}}{12 \mathrm{~kg}}} = \sqrt{\frac{5}{6}} \mathrm{~rad/s}\) The natural frequency of the system with the spool rolling without slipping is \(\sqrt{\frac{5}{6}} \mathrm{~rad/s}\).

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