
Determine the relationship between the torque N and the torsion angle \[\phi \] for the tube whose wall thickness \[\Delta r\] is considerably less than the tube radius.
A. \[N = \dfrac{{2\pi {r^3}\Delta r\phi }}{{3l}}G\]
B. \[N = \dfrac{{3\pi {r^3}\Delta r\phi }}{l}G\]
C. \[N = \dfrac{{2\pi {r^3}\Delta r\phi }}{l}G\]
D. None of the above
Answer
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Hint: To solve this question, we have to know about torque. We know that, A torque angle, otherwise called a dihedral angle, is framed by three successive bonds in an atom and characterized by the point made between the two external bonds. The foundation of a protein has three distinctive twist points.
Complete step by step answer:
We know that, keeping the lower end of the hollow tube fixed, its upper end is twisted by angle \[\phi \] by applying a force F. Due to the twist, a shear stress is generated between the lower end and upper end of the tube. Thus, we can say, the point A is displaced to A’ due to the force such that, \[AA' = dx\]
Now. From sector, AOA’, \[AA' = r\phi \]
Also we can say, from the sector,\[ABA'\], \[AA' = 1\theta \]
Therefore, \[\theta = \dfrac{{r\phi }}{l}\]
Tangential stress equal to, force upon area. Which is equal to, \[\dfrac{F}{{dx\Delta r}}\]
Therefore, shear modulus,
$G = \dfrac{{stress}}{\theta } = \dfrac{{\dfrac{F}{{\Delta rdx}}}}{{\dfrac{{r\phi }}{l}}} \\
\Rightarrow F = \dfrac{{G\phi r}}{l}\Delta rdx \\$
Moment of force,
\[dM = Fr = \dfrac{{G\phi {r^2}}}{l}\Delta rdx\]
So, we can say, the total restoring torque on the annual surface,
$N = \int {dM} \\
\Rightarrow N= \dfrac{{G{r^2}\phi }}{l}\Delta r\int {dx} \\
\therefore N = \dfrac{{2\pi G{r^3}\phi }}{l}\Delta r \\$
Hence,option C is correct.
Note:We also have to know that, S.I unit of torque is Newton- meter. We have to keep that in our mind. We have calculated here the torsion angle which is denoted by\[\phi \]. We know that, a power that produces or will in general create turn or twist a vehicle motor conveys force to the drive shaft likewise: a proportion of the adequacy of such a power that comprises of the result of the power and the opposite separation from the line of activity of the power to the pivot of revolution.
Complete step by step answer:
We know that, keeping the lower end of the hollow tube fixed, its upper end is twisted by angle \[\phi \] by applying a force F. Due to the twist, a shear stress is generated between the lower end and upper end of the tube. Thus, we can say, the point A is displaced to A’ due to the force such that, \[AA' = dx\]
Now. From sector, AOA’, \[AA' = r\phi \]
Also we can say, from the sector,\[ABA'\], \[AA' = 1\theta \]
Therefore, \[\theta = \dfrac{{r\phi }}{l}\]
Tangential stress equal to, force upon area. Which is equal to, \[\dfrac{F}{{dx\Delta r}}\]
Therefore, shear modulus,
$G = \dfrac{{stress}}{\theta } = \dfrac{{\dfrac{F}{{\Delta rdx}}}}{{\dfrac{{r\phi }}{l}}} \\
\Rightarrow F = \dfrac{{G\phi r}}{l}\Delta rdx \\$
Moment of force,
\[dM = Fr = \dfrac{{G\phi {r^2}}}{l}\Delta rdx\]
So, we can say, the total restoring torque on the annual surface,
$N = \int {dM} \\
\Rightarrow N= \dfrac{{G{r^2}\phi }}{l}\Delta r\int {dx} \\
\therefore N = \dfrac{{2\pi G{r^3}\phi }}{l}\Delta r \\$
Hence,option C is correct.
Note:We also have to know that, S.I unit of torque is Newton- meter. We have to keep that in our mind. We have calculated here the torsion angle which is denoted by\[\phi \]. We know that, a power that produces or will in general create turn or twist a vehicle motor conveys force to the drive shaft likewise: a proportion of the adequacy of such a power that comprises of the result of the power and the opposite separation from the line of activity of the power to the pivot of revolution.
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