Linear momentum

Classical Mechanics Level 4

A uniform rod X Y XY of length L L and mass M M can freely rotate around the pivot at one end X . X. A bullet of mass m m flying at a velocity of v v hits and gets stuck in the rod at the height h ( < L ) h (< L) from end X . X. Find the linear momentum of the system right after the hit.

ω \omega denotes the angular velocity right after the hit.

ω ( m h + M L / 2 ) \omega (mh+ML/2) ω ( m h + M L / 3 ) \omega (mh+ML/3) ω ( m h + M L / 6 ) \omega (mh+ML/6) ω ( m h + M L ) \omega (mh+ML)

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Srinivasa Satti by Srinivasa Satti

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1 solution

Rohit Ner
Jan 26, 2016

Center of mass of the system with respect to X r c m = M ( L 2 ) + m h M + m Linear velocity v c m = r c m × ω Linear momentum = ( M + m ) × v c m = ω ( m h + M L 2 ) \text{Center of mass of the system with respect to} X\\ \begin{aligned}{r}_{cm}&=\frac{M\left(\frac{L}{2}\right)+mh}{M+m}\\ \text{Linear velocity} {v}_{cm}&= {r}_{cm}\times \omega \\ \text{Linear momentum }&=(M+m)\times {v}_{cm}\\&\huge\color{#3D99F6}{=\boxed{\omega\left(mh+\frac{ML}{2}\right)}}\end{aligned}

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