A rod of mass m and length hinged at its centre is placed on a horizontal surface. A bullet of mass m moving with velocity v strikes the end A of the rod and gets embedded in it. The angular velocity with which the system rotates about its centre of mass after the bullet strikes the rod
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We know that momentum must be conserved .
Thus the momentum of that part of the rod is
mu .
And Angular momentum = rP
= (L/2)mu
= ωI = ω(∫( 2M/L)r² dr + m(L/2)²)
= ω((2M/L)(L/2)³/3 + m(L/2)²)
Solve the equation between the first and last line , and you will have your answer .
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