Question

A rigid structure consists of a thin ring of radius $R=\dfrac{75}{290}\,m$and a thin radial rod of equal mass and length 2R. The structure is pivoted around a horizontal axis xx' in the plane of the ring, passing through its centre. The structure is released from rest and it rotates around the axis xx' from the initial upright orientation. Find its angular speed in rad/s about the axis when it is inverted. Assume no frictional losses.

Answer 1

When the structure is inverted, there will be no change or a decrease in the potential energy of the ring. Thus, we will equate the decrease in the potential energy of the rod to the gain in the rotational kinetic energy of the structure. This rotational kinetic energy will be in the form of rotational inertia.
Formula used:

& PE=mgh \\\ & KE=\dfrac{1}{2}I{{\omega }^{2}} \\\ \end{aligned}$$ **Complete answer:** From the given information, we have the data as follows. When the structure is inverted, there will be no change or a decrease in the potential energy of the ring. Thus, we will equate the decrease in the potential energy of the rod to the gain in the rotational kinetic energy of the structure. $$M\times g\times 4R=\dfrac{1}{2}I{{\omega }^{2}}$$ Now we will find the expression for the moment of inertia. In this case, the moment of inertia represents the sum of the moment of inertia of a disk rotating around its centre, the moment of inertia of a uniform rod of length L rotating around its centre and the moment of inertia of a simple point mass around an axis. The mathematical representation of the same is, $$\begin{aligned} & I=\dfrac{M{{R}^{2}}}{2}+\left[ \dfrac{M\times 4{{R}^{2}}}{12}+M\times 4{{R}^{2}} \right] \\\ & \Rightarrow I=\dfrac{M{{R}^{2}}}{2}+13\times \dfrac{M\times 4{{R}^{2}}}{12} \\\ & \Rightarrow I=\dfrac{M{{R}^{2}}}{2}+\dfrac{13M{{R}^{2}}}{3} \\\ & \therefore I=\dfrac{29M{{R}^{2}}}{6} \\\ \end{aligned}$$ Now represent the equation in terms of the angular frequency . $$\begin{aligned} & {{\omega }^{2}}=\dfrac{2\times M\times g\times 4R}{I} \\\ & \therefore \omega =\sqrt{\dfrac{8MgR}{I}} \\\ \end{aligned}$$ Now substitute this expression of the moment of inertia in the above equation. $$\begin{aligned} & \omega =\sqrt{\dfrac{8MgR}{{}^{29M{{R}^{2}}}/{}_{6}}} \\\ & \therefore \omega =\sqrt{\dfrac{49g}{29R}} \\\ \end{aligned}$$ Substitute the values of the radius of the ring and the acceleration due to gravity. $$\begin{aligned} & \omega =\sqrt{\dfrac{49\times 10}{29\times {}^{75}/{}_{290}}} \\\ & \Rightarrow \omega =\sqrt{64} \\\ & \therefore \omega =8\,{rad}/{s}\; \\\ \end{aligned}$$ $$\therefore $$The angular speed of a rigid structure consisting of a thin ring in rad/s about the axis when it is inverted is $$8\,{rad}/{s}\;$$. **Note:** Usually, we consider the kinetic energy as linear, so, we compute it as the product of mass and the velocity (the result will be in metres), but, in the case of the rotational kinetic energy, it should be computed as the product of inertia and the angular frequency (the result will be in radians).