Circular Solid

Consider a solid circular rod with with a 2 inch diameter experiences a 10 psi torsional stress. If the applied torque was 5 lb∙inch, what is the rod’s polar moment of inertia (in^4)?

Expand Hint
Torsional stress in circular solid or thick-walled (t<0.1r) shafts:
$$$\tau =\frac{Tr}{J}$$$
where $$\tau$$ is torsion, $$T$$ is the torque at the section of interest, $$r$$ is the radius to the point of interest, and $$J$$ is the section's polar moment of inertia.
Hint 2
Divide the rod’s diameter by two to get $$r$$ .
Torsional stress in circular solid or thick-walled (t<0.1r) shafts:
$$$\tau =\frac{Tr}{J}$$$
where $$\tau$$ is torsion, $$T$$ is the torque at the section of interest, $$r$$ is the radius to the point of interest, and $$J$$ is the section's polar moment of inertia.

Solving for polar moment of inertia:
$$$10\frac{lb}{in^2}=\frac{(5lb\cdot in)(\frac{2in}{2})}{J}$$$
$$$J=\frac{5lb\cdot in^2}{10\frac{lb}{in^2}}=0.5\:in^4$$$
$$$0.5\:in^4$$$
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