A uniform bar with mass m lies symmetrically across two rapidly rotating fixed rollers, A and B with distance L=2.0 cm between the bar's centre of mass and each roller. The rollers, whose directions of rotation are shown in figures slip against the bar with coefficient of kinetic friction `mu_(k)=0.40`. suppose the bar is displaced horizontally by the distance x as shown in figure and then released. the angular frequency `omega` of the resulting horizontal simple harmonic motion of the bar is (in `rad" "s^(-1)`)

Options

A. 14
B. 15
C. 16
D. 17

Answer

A

Explanation

Consider horizontal forces
`mu_(k)N_(A)-mu_(k)N_(B)=ma`
or `a=(mu_(k)N_(A)-mu_(k)N_(B))/(m)` . . . (i)

Taking torque about an axis perpendicular to the plane and through the contact point between bar and roller A. the bar experiences no angular acceleration about that axis.
`N_(A)(0)+N_(B)(2L)-Mg(L+x)f_(KA)(0)+k_(KB)(0)=0` . . . (ii)
By balancing vertical forces, we have
`N_(A)+N_(B)-mg=0` . . . (iii)
solving eqns. (i), (ii) and (iii)
`N_(B)=(mg(L+X))/(2L),N_(A)=(mg(L-X))/(2L),a=-(mu_(k)g)/(L)X`
Comparing it with `a=-omega^(2)X`
we get, `omega=sqrt((mu_(k)g)/(L))=sqrt((0.40xx10)/(2xx10^(-2)))=14" rad "s^(-1)`