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 2 v and v respectively. The masses stick to the bar after collision at a distance L 3 and L 6 respectively from the centre of the bar. If the bar starts rotating about its center of mass as a result of collision, the angular speed of the bar will be :

Answer 1

solution ;

moment of inertia uniform rod of lemgth (I)

I=M t^2/12

about axis passing through the center and perpendicular to its length .

law of conservation angular moment

r=dL/dT

it net torque is zero

i.e dL/dT=0

L =constant

angular moment conserved only when torque  is zero

when central of mass from system from 0

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

so center of mass at zero

from conservation angular moment Li=Lf

Li=m(.2v)*(L/3)+2mv*(L/3)=mvL

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

 = [2/3mL^2+L^2/9+mL^2/18]ω=(12+2+1/18)m L^2 ω   =5/6 m L^2 ω

 =5/6 m L^2 ω = mvL∵ω =6v/5L

hope it is helpful