Question

A metal ring of initial radius $r$ and cross-sectional area $A$ is fitted onto a wooden disc of radius $R>r$ .if Young's modulus of metal is $Y$ then tension in the ring is:

Answer 1

Step 1. Given data:

Metal ring of radius = $r$

Cross-sectional area = $A$

Wooden ring radius = $R$

Young modulus of metal = $Y$

Step 2. Formula used:

We know, Young's modulus of the metal $\left(Y\right)=\frac{\sigma }{\epsilon }$, where $\sigma =$stress, $\epsilon =$strain.

$\epsilon =$strain.= $\frac{∆l}{L}$, $∆l=$change in length, $L=$original length, stress $\left(\sigma \right)=\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$A$}\right.$, where $T=$tension, $A=$cross-sectional area.

Step 3. Calculations:

$\epsilon =\frac{∆l}{L}=\frac{\left(2\mathrm{\pi R}-2\mathrm{\pi r}\right)}{2\mathrm{\pi r}}=\frac{\left(R-r\right)}{r}$

Thus, $$\begin{gathered} Y=\frac{\sigma }{\epsilon }=\frac{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$A$}\right.}}{\left({\displaystyle \raisebox{1ex}{$R-r$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}\right)} \\ T=\frac{AY\left(R-r\right)}{r} \end{gathered}$$

Hence, the tension in the ring is $\frac{AY\left(R-r\right)}{r}$.