Physics stress/strain problem?

dY = T*L/(A*E) where T = tension in rod = weight supported by rod, and E is Young's modulus for steel (200E9 Pa)

dY = 1800*16.5/(7.80E-4*200E9) = 0.00019 m = .019 cm

When rotating, the problem is to find T and substitute into the above equation:
a = centripetal acceleration
w = angular velocity, rad/sec = 8.0 rpm*2π/60 = .83776
g = gravitational acceleration = 9.8 m/s²
r = radius of revolution
L = length of rod = 16.5 m
Θ = angle of rod wrt the axis of rotation
The upper end of the rod is assumed to be attached at the axis of rotation.

a = r*w².....r = L*sinΘ.....→
1) a = w²*L*sinΘ
A vector diagram will show that

2) T = m*√[g² + w^4*L²*sin²Θ]
also,
m*g = T*cosΘ
substituting from (2) and squaring,

g² = cos²Θ*[g² + w^4L²sin²Θ], which can be manipulated to

3) sinΘ = √[1 - g²/(w^4L²)]

Using the value for Θ in (2), T can be found.

I got T = 2138.78 N
and
dY = 0.000226 m = 0.026 cm


The case where the upper end of the rod is held at some distance Ro from the axis gives a more interesting requirement for getting Θ; email me if you're interested.......

Good problem......I'll record the results for future reference.