A thin uniform copper rod of length l and mass m rotates uniformly with an angular velocity `omega` in a horizontal plane about a vertical axis passing through one of its ends. Determine the tension in the rod as a function of the distance r from the rotation axis. Find the elongation of the rod.

Answer

`T = ( m omega^(2))/(wl) (l^(2) - x ^(2))`

Explanation

Choose a small element of width dx at a distance x from one end of the rod . The mass of the element = `dm(m)/(l)` dx . Let T is the tension in the rod at a distance x . By Newton's second law for the motion of element of mass dm we have
`T - (T + dT) = (dm)omega^(2) x`
or `" " -dT = ((m)/(l) dx) omega^(2)x`
Integrating above equation , we get
`-int_(T)^(0)dT = (momega^(2))/(l) int_(x)^(l)x dx`
or `" " -|T|_(T)^(0)= (momega^(2))/(2l)|x^(2)|_(x)^(l)` lt brgt or `" " -(0-T) = (m omega^(2))/(2l)(l^(2) - x^(2))`
or `" " T = ( momega^(2))/(2l) ( l^(2) - x^(2))`
At x = 0 , T = `T_("max") = (m omega^(2)l)/(2)`
At x = `(l)/(2) , T = (momega^(2)l)/(8)`