Question

A 80 g , 37-cm-long rod hangs vertically on a frictionless, horizontal axle passing through its center. A 14 g ball of clay traveling horizontally at 2.1 m/s hits and sticks to the very bottom tip of the rod. Part A To what maximum angle, measured from vertical, does the rod (with the attached ball of clay) rotate?

Answer 1

angular mometum L= R*p

where p is the momentum

so , L= m*v*r        [ here r is the distance from the pivot]

so , L = 0.014*2.1* 0.37 = 0.0108 kgm/s

now conservation of monetum =

0.108 kgm/s= omega*[M_rod*L²/3 + M_ball*L² ]=omega*[(M_rod/3 + M_ball)*L²]

=>omega= 1.939

now initial energy = potential energy at top

=> 0.5* (I)*omega²= (1 - cosθ)*( M_ball*L + M_rodL/2)

=>0.010465= (1 - cosθ)*( 0.01998)

=> cos(theta) = 0.533

=> theta= 58.41 degrees