Question

A thin uniform bar of length l and mass 8 m lies on a smooth horizontal table. Two point masses m and 2 m are moving in the same horizontal plane from opposite sides of the bar with speeds 2v and v respectively. The masses stick to the bar after collision

Answer 1

Moment of inertia= I = Ml^2 / 12

As per angular momentum's conservation law:

τ = dL / dt

Torque = 0

dL / dt = zero

L = constant

Ang. momentum would be conserved if torque is 0

Center of masses of system from zero

= 8m x 0 + m (L/3) - 2m ( L/6 )

Center of masses = 0

As angular momentum is conserved:

Li=Lf

Li = m (2v) x ( L/3) + 2mv x ( L/6 )

Lf = [ (8m) L^2 / 12 + m (L/3)^2 + m (L/6)^2 ω

= { 2/3 mL^2 + mL^2 / 9 + mL^2 / 18 } ω

= 5/6 mL^2 ω