Question

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 $c=8\mathrm{N}-\sec /\mathrm{m}$. 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

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

Step-by-step solution

Determine the moment of inertia of the spool

To determine the moment of inertia, we will use the formula for a cylinder, which is $I=\frac{1}{2}m{R}^2$, 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 $R=1\mathrm{m}$: $I=\frac{1}{2}(12\mathrm{kg})(1\mathrm{m}{)}^2=6\mathrm{kg}\cdot {\mathrm{m}}^2$

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: $kx+cv=I\frac{a}{R}$

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. ${\omega }_n=\sqrt{\frac{10\mathrm{N}/\mathrm{m}}{12\mathrm{kg}}}=\sqrt{\frac{5}{6}}\mathrm{rad}/\mathrm{s}$ The natural frequency of the system with the spool rolling without slipping is $\sqrt{\frac{5}{6}}\mathrm{rad}/\mathrm{s}$.