A uniform ring of mass `M` of outside radius `r_(2)` is fitted tightly with a shaft of radius `r_(1)`. If the shaft is rotated with a constant angular acceleration. About it's axis, the moment of the elastic force in the ring about the axes of rotation is

Options

A. `(M(r_(2)^(4) - r_(1)^(4))alpha)/(2(r_(2)^(2) - r_(1)^(2)))`
B. `(M(r_(2)^(4) + r_(1)^(4))alpha)/(2(r_(2)^(2) + r_(1)^(2)))`
C. `(M(r_(2)^(4) - r_(1)^(4))alpha)/(2(r_(2)^(2) + r_(1)^(2)))`
D. `(M(r_(2)^(4) + r_(1)^(4))alpha)/(2(r_(2)^(2) - r_(1)^(2)))`

Answer

A

Explanation

`sigma = (M)/(pi (r_(2)^(2) - r_(1)^(2))), dT = (dm)alpha x`
`d tau = (dm) alpha x^(2), d tau = sigma (2pi x) dx alpha x^(2)`
`tau = int d tau = (M2pi alpha)/(pi(r_(2)^(2)-r_(1)^(2))) ((x^(4))/(4))_(r_(1))^(r_(2)), tau = (M(r_(2)^(4)-r_(1)^(4)))/(2(r_(2)^(2)-r_(1)^(2))) `