A wire of length 1 text{ m} and radius 2 text | vedclass

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Question

(A) Let $l$ be the length of the wire,$r$ be the radius,$	heta$ be the angle of twist,and $\phi$ be the angle of shear.From the geometry of the twist

Vedclass · Accepted Answer

$0.09$

Vedclass · Answer

(A) Let $l$ be the length of the wire,$r$ be the radius,$	heta$ be the angle of twist,and $\phi$ be the angle of shear.From the geometry of the twisted wire,the arc length $s$ on the circumference is given by $s = r 	heta$.For small angles,the angle of shear $\phi$ is related to the arc length $s$ and the length of the wire $l$ by $\phi = \frac{s}{l}$.Substituting $s = r 	heta$,we get $\phi = \frac{r 	heta}{l}$.Given: $l = 1 	ext{ m}$,$r = 2 	ext{ mm} = 2 	imes 10^{-3} 	ext{ m}$,and $	heta = 45^{\circ}$.Substituting these values: $\phi = \frac{2 	imes 10^{-3} 	ext{ m} 	imes 45^{\circ}}{1 	ext{ m}} = 90 	imes 10^{-3} 	ext{ degrees} = 0.09^{\circ}$.Therefore,the angle of shear is $0.09^{\circ}$.

$A$ wire of length $1 \text{ m}$ and radius $2 \text{ mm}$ is vertically clamped. The lower end is twisted through an angle of $45^{\circ}$. The angle of shear is $.....$ (in $^{\circ}$)

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