A steel ring of radius r and cross-section ar | vedclass

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Question

(B) The initial length (circumference) of the ring is $L = 2\pi r$.The final length (circumference) of the ring after fitting it onto the disc is $L'

Vedclass · Accepted Answer

$AE\left( \frac{R - r}{r} \right)$

Vedclass · Answer

(B) The initial length (circumference) of the ring is $L = 2\pi r$.The final length (circumference) of the ring after fitting it onto the disc is $L' = 2\pi R$.The change in length is $\Delta L = 2\pi R - 2\pi r = 2\pi(R - r)$.The longitudinal strain is defined as $	ext{strain} = \frac{\Delta L}{L} = \frac{2\pi(R - r)}{2\pi r} = \frac{R - r}{r}$.According to the definition of Young's modulus $E = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\Delta L/L}$.Substituting the values,we get $E = \frac{F/A}{(R - r)/r}$.Rearranging for force $F$,we get $F = AE\left( \frac{R - r}{r} ight)$.

$A$ steel ring of radius $r$ and cross-section area $A$ is fitted onto a wooden disc of radius $R$ $(R > r)$. If Young's modulus is $E$,then the force with which the steel ring is expanded is:

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