A thin circular hoop is suspended from a knife edge. Its rotational inertia about the rotation axis (along the knife) is \(I=2 m r^{2} .\) Show that it oscillates with the same frequency as a simple pendulum of length equal to the diameter of the hoop.

The hoop is suspended from a knife-edge and has a moment of inertia \(I=2mr^2\), where \(m\) is the mass of the hoop and \(r\) is the radius. We will use small angle approximation, as in the simple pendulum, and consider the restoring torque caused by the gravitational force on the hoop when it is rotated by a small angle \(\theta\). The gravitational force acting on the hoop can be considered to act on the center of mass of the hoop, located at the center of the circle. The torque due to this force is \(\tau = mgr\sin(\theta) \approx mgr\theta\). Now, from Newton's second law for rotation, we have \(\tau = -I\alpha\), where \(\alpha\) is the angular acceleration. Therefore: \(I\alpha = -mgr\theta\)

For a simple pendulum of length \(L\) and mass \(M\), when it is displaced by a small angle \(\phi\), the equation of motion can be found as: \(MgL\sin(\phi) = -Ml\ddot{\phi}\) Using small angle approximation \(\sin(\phi) \approx \phi\), we obtain: \(MgL\phi = -Ml\ddot{\phi}\)

Now, we want to relate the hoop's equation of motion to that of the simple pendulum. We have: \(2mr^2\ddot{\theta} = -mgr\theta\) This equation looks similar to the equation of motion of a simple pendulum: \(Ml\ddot{\phi} = -MgL\phi\) Comparing the two equations, we can see that they have the same form. If we find that the ratios of the angular accelerations to the angular displacements are equal, then both systems will oscillate with the same frequency.

For the simple pendulum, we have: \(\frac{\ddot{\phi}}{\phi} = -\frac{g}{l}\) For the hoop, we have: \(\frac{\ddot{\theta}}{\theta} = -\frac{g}{2r}\) Now, we want to show that the frequency \(\omega\) of the simple pendulum is equal to the frequency \(\omega'\) of the hoop: \(\omega^2 = \frac{g}{l}\) \(\omega'^2 = \frac{g}{2r}\) Since we are asked to show that the simple pendulum has a length equal to the diameter of the hoop, we set \(l = 2r\): \(\omega^2 = \frac{g}{2r}\) Comparing the two equations, we can see that \(\omega^2 = \omega'^2\), which implies \(\omega = \omega'\). Hence, the hoop and the simple pendulum oscillate with the same frequency when the length of the pendulum is equal to the diameter of the hoop.