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(A) The ring is heated by $\Delta T$ to increase its length by $\Delta L = L\alpha\Delta T$ so that it fits the wheel of circumference $2\pi R$. When

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$2SY\alpha\Delta T$

Vedclass · Answer

(A) The ring is heated by $\Delta T$ to increase its length by $\Delta L = L\alpha\Delta T$ so that it fits the wheel of circumference $2\pi R$. When it cools down,it exerts a tension $F$ in the ring. The stress in the ring is $\sigma = F/S$. The strain is $\epsilon = \Delta L/L = \alpha\Delta T$. Using Young's modulus $Y = \sigma / \epsilon$,we have $Y = \frac{F/S}{\alpha\Delta T}$,which gives the tension in the ring as $F = SY\alpha\Delta T$.Consider one semicircular part of the wheel. The ring exerts a force $F$ at each end of the semicircle in the tangential direction. The total force pressing the two semicircular parts together is the sum of the forces exerted by the ring at both contact points. Since the ring exerts a force $F$ at each end,the force pressing one part against the other is $2F$. Therefore,the net force is $F_{net} = 2F = 2SY\alpha\Delta T$.

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