Question

Determine what the speed v of the bullet should be such that the rod swings exactly 90 degrees from it's initial equilibrium position. Express v in terms of m, M, g and L.
Picture to come:

Answer 1

Well, lost kinetic energy can be converted to rotational energy, right?

Answer 2

So what I tried was two ways.
1) Assuming it's a perfectly inelastic collision this would imply that kinetic energy is conserved.
so \(KE_{bullet}=\frac{1}{2}mv^2\) and \(KE_{rod}=\frac{1}{2}I\omega^2\) where \(I=\frac{1}{3}ML^2\).
\[KE_{before}=KE_{after}\] \[\frac{1}{2}mv_i^2 + \frac{1}{2}Iw_i^2=\frac{1}{2}mv_f^2 + \frac{1}{2}Iw_f^2\] KE of rod before is 0, and velocity of bullet after is \(\large \frac{v_i}{2}\)
\[\frac{1}{2}mv_i^2=\frac{1}{2}m\frac{v_i^2}{4} + \frac{1}{2}Iw_f^2\]We can take out a factor of 1/2. and rearrange
\[mv_i^2-\frac{1}{4}mv_i^2 = Iw_f^2\]\[mv_i^2(1-\frac{1}{4})= Iw_f^2\]\[\frac{3}{4}mv_i^2 = \frac{1}{3}ML^2w_f^2\]
If we let \(w_f=\sqrt{\frac{g}{L}}\) <- not sure about this part
\[v_i^2=\frac{4MLg}{9m}\]

Answer 3

Do you know the answer?

Answer 4

Nope, which is why I was trying to verify my answer from my last post.

Answer 5

Well, we can use the last post, and just put \(\theta = 90^\circ\)

Answer 6

\[
\frac 12 LMg = \frac 16L^2M\omega ^2
\]

Answer 7

\[
\omega ^2=\frac{3g}{L}
\]

Answer 8

That hasn't verified anything... has it? that just says we need a angular velocity of that much to get to 90.

Answer 9

Wait, how did you get that equation?

Answer 10

which one?

Answer 11

\[\frac 12 LMg = \frac 16L^2M\omega ^2\]

Answer 12

Left side is potential, right side is kinetic rotational

Answer 13

Ah yeah I see. Okay that works.

Answer 14

\[
mv^2 =\frac 14mv^2+I\omega ^2 \implies \frac 34mv^2 = I\omega^2
\]

Answer 15

\[
v^2 = 4\frac{Ig}{mL}
\]

Answer 16

\[
v = \sqrt{\frac{4MLg}{3m}}
\]

Answer 17

A lot of algebra and places that mistakes could have been made.

Answer 18

But you've assumed \(w=\sqrt\frac{g}{L}\) - but shouldn't you be proving your \(w=\frac{3g}{L}\)

Answer 19

The point of the other question was to prove that.

Answer 20

I see what you did. So what I did was wrong in my first post was assuming \(w=\sqrt{\frac{g}{L}}\) when it should be assumed to be the w you need to reach 90 degrees \(\omega^2=\frac{3g}{L}\). And then from conservation of kinetic energy \[\frac{3}{4}mv^2=I\omega^2\] we find the equation for v
\[v^2=\frac{4I\omega^2}{3m}=\frac{4ML^2(3g)}{9m(L)}=\frac{4MLg}{3m}\]
\[v=\sqrt{\frac{4MLg}{3m}}\]