Question

A thin ring of volume mass density $\rho$, cross-sectional area A, radius R and young's modulus of elasticity E, is rotated about an axis passing through its centre with angular velocity $\omega$. If increases in cirumference (assumed small) is $\frac{18\pi R^{3}\rho\omega^{2}}{aE}$. Find a.

Answer 1

9

Answer 2

1. Tension in the rotating ring: Consider a small element of the ring of mass $dm$ at radius $R$. The centrifugal force acting on this element is $dF = dm \cdot R\omega^2$. If the volume mass density is $\rho$ and the cross-sectional area is $A$, the mass of an infinitesimal arc element $ds = R d\theta$ is $dm = \rho \cdot A \cdot ds = \rho A R d\theta$. The centrifugal force on this element is $dF = (\rho A R d\theta) R\omega^2 = \rho A R^2 \omega^2 d\theta$. This outward force is balanced by the inward radial component of the tension $T$ in the ring. For a small angle $d\theta$, the inward radial component of tension is $T d\theta$. Equating these, $T d\theta = \rho A R^2 \omega^2 d\theta$, which gives the tension $T = \rho A R^2 \omega^2$.

2. Strain and increase in circumference: Young's modulus $E$ is defined as stress/strain. The stress in the ring material is $\sigma = T/A$. The strain is $\epsilon = \sigma/E = (T/A)/E = T/(AE)$. The increase in the circumference $\Delta C$ is given by $\Delta C = \epsilon \cdot C$, where $C = 2\pi R$ is the original circumference. Substituting the expression for strain: $\Delta C = \frac{T}{AE} \cdot (2\pi R)$ Substituting the expression for tension $T = \rho A R^2 \omega^2$: $\Delta C = \frac{\rho A R^2 \omega^2}{AE} \cdot (2\pi R) = \frac{2\pi R^3 \rho \omega^2}{E}$

3. Finding 'a': The problem states that the increase in circumference is $\frac{18\pi R^{3}\rho\omega^{2}}{aE}$. Equating our derived expression for $\Delta C$ with the given expression: $\frac{2\pi R^3 \rho \omega^2}{E} = \frac{18\pi R^{3}\rho\omega^{2}}{aE}$ Cancelling the common terms ($\pi$, $R^3$, $\rho$, $\omega^2$, $E$) from both sides, we get: $2 = \frac{18}{a}$ Solving for $a$: $a = \frac{18}{2} = 9$