Question

A circular loop of mass m and R rests flat on a horizontal frictionless surface.A bullet also of mass m, and moving with a velocity v, strikes the loop and gets embedded in it. The thickness of the hoop is much smaller than R. The anguler velocity with system rotates after the bullet strikes the hoop is

Answer 1

centre of mass of the loop - bullet system is R/2 from centre of loop.

By momentum conservation :

$mv=2mVcm$

$Vcm=v/2$

Applying angular momentum conservation about the centre of mass.

$MVR/2=Iw$

I for ring $=mr2+m\left(r/2\right)2$

I of bullet$=m\left(r/2\right)2$

moment of inertia of the system I=6mR2/4

$=3/2mR2$

$mvR2=3/2mR2w$

$w=v/3R$