A uniform rod is placed on two spinning wheels as shown in figure. The axes of the wheels are separated by a distance `l=20 cm`, the coefficient of friction between that in this case the rod performs harmonic oscillations. Find the period of these oscillations.

Answer

`1.5`

Explanation

Each roller exerts on the board forces of friction equal of `f_(1) = KN_(1) and f_(2) = KN_(2) , "where" N_(1) , N_(2)` are normals of the board on the rollers . There forces `f_(1),f_(2)` are shown in the figure

Suppose the centre of gravity of the board is slightly displaced. Let it be x from the middle . Then the forces `f_(1), f_(2)` will not be equal because `N_(1) and N_(2)` will different
`N_(1) = (1 + x)/( l) W `
When W is the weight of board and the distance between the wheel axes is l
`N_(2) = (l - x)/( l)`
`:. f_(1) = (K (l + w))/(l) W `
and `f_(2) = (K ( l - x))/( l) W`
`:.` Resultant force ` = (K)/(l) W [ l + x - l + x] = ( 2 KW)/(l) x `
`:.` Acceleration of rod ` = ( 2 K W x)/( l (W//g))` , since mass of rad ` = (W)/( g)`
`:.` Acceleration `= (2 K gx)/(l)` So, acceleration is proportional to displacement . So, it executes SHM
`:.` Period ` T = 2 pi sqrt((l)/( 2 Kg)) = 2 pi sqrt((20)/(2 xx 100 xx 0.18 xx 9.8)) = 2 pi sqrt((1)/( 2 xx 49 xx 0.18)) = 1.5 ` sec