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3 E l A uniform straight rigid bar of mass m and length b is placed in a horizontal position across the top of two identical cylindncal rollers. Aster; of the two rollers are 2d apart. It'li'is the frictional coefcient between the cylinder surface and the bar. show that if' the bar is displaced a distance x from its central position then the net horizontal force on the bar is F = ngw'd, and the bar will execute simple harmonic motion with a pc- l7 nod of 2:: we r jg . Solution: 7 . Fig. 1.? "l is a force diagram for the systern. 0 Jr. C \"(unmask/x\" L K Displacement of the center of mass of the bar is " ' m a\" it. positive to the right, measured from the mid- ' way point between the rollers. Equations of mo- tion of the bar, in the component forms, are mi?\" 2}?" 21%.\"ng mi' : 0 1 Z a :N1 -- N1 * me where. as indicated in Fig. 2.20, N and N are normal forces and F and F are the friction forces on the bar. N and N can be determined by using the rotation equilibrium condition: 2 from (total' a torque) : 0 Taking torques about. 0 ,we obtain 1 N3 (Ed) ; mglfet + d). or N: - 3mg\". +- x d) . (3) Then ti-om Eq.( 2) we obtain 1 N1: mg L \"x = 5mm x: d) (4) The friction forces can now be determined: It F. = m "is. - fmg. d Subslimting this into Equation (1) we obtain .. f -, .t' + LEE;- = it i eight\" :: 0, (no: :2 f3 d C which describes simple harmonic motion at angular frequency to. The period of oscillation 15 P x 2:: m : batik.972