A wooden wheel of radius R is made of two sem | vedclass

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(A) The ring is heated to fit over the wheel. When it cools down,it exerts a tension force $F$ on the wheel.The thermal strain produced by the tempera

Vedclass · Accepted Answer

$SY \alpha \Delta T$

Vedclass · Answer

(A) The ring is heated to fit over the wheel. When it cools down,it exerts a tension force $F$ on the wheel.The thermal strain produced by the temperature change $\Delta T$ is given by $\epsilon = \alpha \Delta T$.From the definition of Young's modulus,$Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/S}{\Delta L/L}$.Here,the strain $\frac{\Delta L}{L}$ is equal to $\alpha \Delta T$.Therefore,the tension force $F$ in the ring is $F = Y S \alpha \Delta T$.Since the ring is circular and exerts force on two semicircular parts,consider a small element of the ring subtending an angle $d	heta$ at the center. The radial force component is $dF_r = F d	heta$. Integrating this over a semicircle (from $0$ to $\pi$),the total force exerted by one half of the ring on the other half is $2F$.However,the force exerted by one part of the wheel on the other part is balanced by the tension in the ring. The force $F$ calculated above is the tension in the ring. The force exerted by one semicircular part on the other is $2F_{radial} = 2F / \pi$ (for a full ring),but for this specific setup,the force exerted by one part of the wheel on the other is simply the tension $F$ acting at the two contact points. Thus,the force is $F = SY \alpha \Delta T$.

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