The armature slender rod AB has a mass of 0 22 kg and can pi

The armature (slender rod) AB has a mass of 0 22 kg and can pivot about the pin at A Movement is controlled by the electromagnet E Which exerts a horizontal attractive force on the armature at B of F_b = (0.25(10^3) l^-2) N where I in meters is the gap between the armature and the magnet at any instant If the armature lies in the horizontal plane and is originally at rest determine the speed of the contact at B the instant l = 0.01 in Orignally l = 0.02 m Express your answer to three significant figures and include the appropriate units.

Solution

>> As, Force, F = - m*a

where, m = Mass

and, a = acceleration = vdv/dL

As, F = 0.25*10^-3*L^-2

=> - 0.25*10^-3*L^-2 = v dv/dL

=> - 0.25*10^-3*L^-2 dL = v dv

>> Now, Integrating both sides from L = 0.02 m, v = 0 to L = 0.01 m to v = ?

=> +0.25*10^-3*| L^-1 |_0.02^0.01  = (1/2) | v² |₀^v

Simplifying,

v = 0.158 m/s .......ANSWER.......