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songoku
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- Homework Statement
- A rod with mass ##M## and length ##L## meter is pivoted at its center and put in vertical position, then a bullet of mass ##m## with initial speed ##u## is directed at a point ##x## meter from the top of the rod (##x## < ##0.5~L##) and the bullet will penetrates through the rod with final speed ##v##. Considering no energy lost and ignoring mass of rod lost when the bullet penetrates, find the angular speed of the rod
- Relevant Equations
- Conservation of linear momentum
Conservation of angular momentum
Circular motion
1) Applying conservation of linear momentum:
$$m.u = M.V + m.v$$
where ##V## is final linear speed of the rod
$$V=\frac{m.u-m.v}{M}$$2) Applying formula of circular motion:
$$V=\omega . r$$
$$\omega = \frac{\left(\frac{mu-mv}{M} \right)}{\frac{1}{2}L-x}$$
Is this correct?And can this be solved by using conservation of angular momentum? This is what I tried:
Applying conservation of angular momentum:
$$m.u.(\frac{1}{2}L-x)=m.v.(\frac{1}{2}L-x)+I \omega$$
$$(\frac{1}{2}L-x)(m.u-m.v)=\frac{1}{12} ML^2 \omega$$
$$\omega = \frac{(\frac{1}{2}L-x)(m.u-m.v)}{\frac{1}{12} ML^2}$$I am really not sure about my working. Thanks
$$m.u = M.V + m.v$$
where ##V## is final linear speed of the rod
$$V=\frac{m.u-m.v}{M}$$2) Applying formula of circular motion:
$$V=\omega . r$$
$$\omega = \frac{\left(\frac{mu-mv}{M} \right)}{\frac{1}{2}L-x}$$
Is this correct?And can this be solved by using conservation of angular momentum? This is what I tried:
Applying conservation of angular momentum:
$$m.u.(\frac{1}{2}L-x)=m.v.(\frac{1}{2}L-x)+I \omega$$
$$(\frac{1}{2}L-x)(m.u-m.v)=\frac{1}{12} ML^2 \omega$$
$$\omega = \frac{(\frac{1}{2}L-x)(m.u-m.v)}{\frac{1}{12} ML^2}$$I am really not sure about my working. Thanks
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