A thin uniform rod of length `l` and masses `m` rotates uniformly with an angularly velocity `omega` in a horizontal plane about a verticle axis passing through one of its ends determine the tension in the rot as a funtion of the distance x from the rotation axis

Options

A. `(1)/(2) ML omega^(2)`
B. `(1)/(4) ML omega^(2)`
C. `(1)/(8) ML omega^(2)`
D. `(3)/(8) ML omega^(2)`

Answer

D

Explanation

Let `m` be the mass per unit length of the rod. Then `M = mL`
Consider a small element of length `dx` located at `C` at a distance `x` from `A`
The mass of element of length `dx = mdx`
The centripetal force at `C` is
`dF = (mdx)x omega^(2)`
`F =m int_(x = x)^(x = L) (mdx) xx omega^(2) = (1)/(2) m omega^(2) (L^(2) - x^(2))`
Now, `m = (M)/(L), F = (1)/(2) ML omega^(2) (1 - (x^(2))/(L^(2)))`
`F = (1)/(2) pi r^(2) rho L^(2) omega^(2) (1 - (x^(2))/(L^(2)))`
Tension in the middle put `x = L//2`