Question

A circular ring of radius $R$ and mass $m$ made of a uniform wire of cross-sectional area $A$ is rotated about a stationary vertical axis passing through its center and perpendicular to the plane of the ring. The breaking stress of the material of the ring is ${\sigma }_b$. Determine the maximum angular speed ${\omega }_{max}$ at which the ring may be rotated without failure.

• A. $\sqrt{\frac{\pi A{\sigma }_b}{mR}}$
• B. $\sqrt{\frac{\pi A{\sigma }_b}{2mR}}$
• C. $\sqrt{\frac{2\pi A{\sigma }_b}{mR}}$
• D. $\sqrt{\frac{4\pi A{\sigma }_b}{mR}}$

Answer 1

The correct option is C $\sqrt{\frac{2\pi A{\sigma }_b}{mR}}$

Let's consider a small portion of the ring subtending an angle $d\theta$.

The ring, as a result of rotational motion, experiences a tension $F$ along its circumference.

The net tangential component of this tension will be $0$, since tensions of $F\cos \left(\frac{d\theta }{2}\right)$ will act in opposite directions and cancel each other out, as shown in the figure.


Therefore, the tension on the ring comes entirely from the radial components of the tension, given by

$T_{net}=F\sin \left(\frac{d\theta }{2}\right)+F\sin \left(\frac{d\theta }{2}\right)$

$\Rightarrow T_{net}=2F\sin \left(\frac{d\theta }{2}\right)\approx 2F\left(\frac{d\theta }{2}\right)$

since $d\theta$ is very small.

$\Rightarrow T_{net}=Fd\theta$

Therefore, the mass of the ring over this small component will be: $dm=m\times \frac{d\theta }{2\pi }.$

The centripetal force experienced by this component of the ring on rotating it about its vertical axis with the maximum angular velocity ${\omega }_{max}$ will be:

$F_{centripetal}=\left(dm\right){\omega }_{max}^{2}R$

$=(m\times \frac{d\theta }{2\pi }){\omega }_{max}^{2}R$

Also, given that the breaking stress of the wire is ${\sigma }_b$.

$\Rightarrow {\sigma }_b=\frac{F}{A}\Rightarrow F={\sigma }_{b}A$

$\Rightarrow T_{net}=Fd\theta =\left({\sigma }_{b}A\right)d\theta$

Now, the centripetal force is nothing but the radial tension of the ring:

$\Rightarrow F_{centripetal}=T_{net}$

$\Rightarrow (m\times \frac{d\theta }{2\pi }){\omega }_{max}^{2}R=\left({\sigma }_{b}A\right)d\theta$

$\Rightarrow \frac{m{\omega }_{max}^{2}R}{2\pi }={\sigma }_{b}A$

$\Rightarrow {\omega }_{max}^2=\frac{2\pi {\sigma }_{b}A}{mR}$

$\Rightarrow {\omega }_{max}=\sqrt{\frac{2\pi {\sigma }_{b}A}{mR}}$

Hence, option $\left(\text{C}\right)$ is correct.

$\text{Key concepts:}$ Breaking stress is the maximum force that can be applied on a cross-sectional area of a material in such a way that the material is unable to withstand any additional amount of stress before breaking. Breaking stress may also be known as ultimate tensile stress or breaking strength.