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A circular disc of mass m and radius R rests flat on a horizontal frictionless surface.A bullet,also of mass m and moving with a velocity v,strikes the disc and gets embedded in it.The angular velocity with which the syatem rotates after the bullet strikes the hoop is:-Ans(v/2R) {I have a conceptual doubt in this question.If external torque is zero then angular momentum is conserved only from axis of centre of mass or from any axis.In this question I tried to solve it by taking center of disc as an axis but I can't solve it.Please clear my doubt with full explanation. please.}

A circular disc of mass m and  radius R rests flat on a horizontal frictionless surface.A bullet,also of mass m and moving with a velocity v,strikes the disc and gets embedded in it.The angular velocity with which the syatem rotates after the bullet strikes the hoop is:-Ans(v/2R)
{I have a conceptual doubt in this question.If external torque is zero then angular momentum is conserved only from axis of centre of mass or from any axis.In this question I tried to solve it by taking center of disc as an axis but I can't solve it.Please clear my doubt with full explanation. please.}

Grade:12th pass

1 Answers

Khimraj
3007 Points
4 years ago
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(r/2)2
I of bullet =m(r/2)2
moment of inertia of the system I=6mR2/4
=3/2mR2
mvR2=3/2mR2w
w=v/3R

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