Question

A wooden wheel of radius $R$ is made of two semicircular parts (see figure). The two parts are held together by a ring made of a metal strip of cross - sectional area $S$ and length $L$. $L$ is slightly less than $2\pi R$. To fit the ring on the wheel, it is heated so that its temperature rises by $\Delta T$ and it just steps over the wheel. As it cools down to surrounding temperature, it presses the semicircular parts together. If the coefficient of linear expansion of the metal is $\alpha$ and its Young's modulus is $Y$, the force that one part of the wheel applies on the other part is-

• A. $2\pi SY\alpha \Delta T$
• B. $SY\alpha \Delta T$
• C. $\pi SY\alpha \Delta T$
• D. $2SY\alpha \Delta T$

Answer 1

The correct option is D $2SY\alpha \Delta T$


$\therefore \frac{\Delta L}{L}=\alpha \Delta T$
The thermal stress developed is
$\frac{T}{S}=Y\frac{\Delta L}{L}=Y\alpha \Delta T$
or $T=SY\alpha \Delta T$
from free body diagram of one point of the wheel
or $F=2T$
where $F$ is the force that one part of the wheel applies on the other part
$\therefore F=2SY\alpha \Delta T$