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)`)

Explanation

(1) This situation seems very different from other SHM situations we have analyzed. However, the key idea here is that we can find `omega` if we find an expression for the horizontal acceleration a of the bar as a function of x and then compare it with `a= - omega^(2)x`.
(2) Since vertical forces (as shown in Fig.15.14b), horizontal frictional forces, and torques all act on the bar, we shall apply Newton.s second law both vertically and horizontally, and then in angular form about one of the contact points between bar and roller.
Calculations: The vertical force acting on the bar are the gravitational force `vec(F)_(g)` (with magnitude mg) and supporting forces `vec(F)_(A)` due to roller A and `vec(F)_(B)` due to roller B. Since there is no vertical acceleration of the bar. Newton.s second law for components along a vertical y axis `(F_("net, y") = ma_(y))` gives us
`F_(A) + F_(B) - mg= 0`
The horizontal forces acing on the bar are the kinetic frictional forces `F_(kA) = mu_(k) F_(A)` (toward the right) due to roller A and `f_(kB) = mu_(k) F_(B)` (toward the left) due to roller B. Thus,

Figure 15.14 (a) A bar is in equilibrium on two rotating rollers. A and B, that slip beneath it. (b) The bar is displaced from equilibrium by a distance x and then released.
Newton.s second law for components along a horizontal x axis `(F_("net.x") = ma_(x))` gives us
`mu_(k) F_(A) - mu_(k) F_(B) = ma`
`a= (mu_(k) F_(A) - mu_(k) F_(B))/(m)`
We now consider torques about an axis perpendicular to the plane of Fig. 15.14 and through the contact point between the bar and roller A. The bar experiences no angular acceleration about that axis, so Newton.s second law for torque about an axis `(tau_("net") = I_(a))` gives us
`F_(A) (0) + F_(B) 2L-mg (L+x) + f_(kA) (0) + f_(kB) (0) = 0`
where the forces `vec(F)_(A), vec(F)_(B), m vec(g), vec(f)_(kA), and vec(f)_(kB)` have moment arms about that axis of 0, 2L, L + x, 0 and 0, respectively. Solving Eq.15.36 for `F_(B)` and then solving Eq.15.34 for `F_(A)` we find
`F_(B) = (mg (L+x))/(2L) and F_(A) = (mg (L-x))/(2L)`
Substituting these results into Eq.15.35 yields.
`a= (mu_(k) g)/(L) x`
Thus `omega= sqrt((mu_(k)g)/(L)) = sqrt(((0.40)(9.8 m//s^(2)))/(0.020m))`
= 14 rad/s
Learn: Another important point is that the rollers will be experiencing kinetic friction which cal slow them dwon. There must be some external torque on the rollers to maintain the motion of the rollers.