Solid Circular Rod

Consider a solid circular rod with a polar moment of inertia of 0.1 m^4 experiences a 100 Pa torsional stress. If the applied torque was 20 N∙m, what is the rod’s diameter in meters?

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
Solve for $$r$$ , then multiply by 2 to get the diameter.
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 radius:
$$$100 N/m^2 =\frac{(20N\cdot m)r}{0.1m^4}$$$
$$$r=\frac{100 N(0.1m^4)}{20N\cdot m^3}=0.5\:m$$$
To solve for the rod’s diameter:
$$$D=(2)r=(2)0.5m=1\:m$$$
1 m
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