Can someone check my answer to a rotational problem?

[toesockshoe]

Good, except for the v² on the left.
Fix that, then use the above to determine $v_{fs}$.

uhhh, that gets really messy cause there's a quadratic doesn't it?
anyway, after I find $v_{fs}$ ... I find $v_{fp}$ using the vfs + wb formula right?

also, my friend did the priblem this way:

$L_i = L_f$
$m_pv_{pi}b = i\omega$
$m_pv_{pi}b = (\frac{m_sL^2}{12} + m_pb^2) w_f$ because the angular velocity are the same for both masses right?

isnt this method much easier?

 

[haruspex]

that gets really messy cause there's a quadratic doesn't it?

But it shouldn't be quadratic. I said that was an error in your equation:

Good, except for the v² on the left.

Correct that equation and try again.

isnt this method much easier?

Oh yes, that's much easier. Unfortunately it's wrong. It's the same error you made in your initial post on this thread.

 

[toesockshoe]

Oh yes, that's much easier. Unfortunately it's wrong. It's the same error you made in your initial post on this thread.

you can't do it because the point is not rotating on its around its own mass center right?

it shouldn't be quadratic. I said that was an error in your equation.

I did try solving it after I fixed it... I got $m_p v_{pi}^2 = m_p(v_{fs}+\omega b)^2 + m_sv_{fs}^2$... and that gets messy because you need to square the $v_{fs} + wb$ and that makes it difficult to factor out $v_{fs}$ becasue there is two v_fs terms to the 2nd power and one v_fs term to the first power. where am i going wrong?

 

[haruspex]

you can't do it because the point is not rotating on its around its own mass center right?

Not exactly. It's wrong because the particle's angular momentum about the point in space where the mass centre of the stick had been is not just $m_pb^2\omega_f$. There is also the $m_pv_{fs}b$ term. Look at your relative velocity equation.

I did try solving it after I fixed it... I got $m_p v_{pi}^2 = m_p(v_{fs}+\omega b)^2 + m_sv_{fs}^2$

I'll try once more... IT SHOULD NOT BE QUADRATIC! You appear to be adding up energies, not angular momenta.

 

[toesockshoe]

Not exactly. It's wrong because the particle's angular momentum about the point in space where the mass centre of the stick had been is not just $m_pb^2\omega_f$. There is also the $m_pv_{fs}b$ term. Look at your relative velocity equation.

I'll try once more... IT SHOULD NOT BE QUADRATIC! You appear to be adding up energies, not angular momenta.

OHH STUPID ME. yes sorry. i added up energies.

 

[toesockshoe]

Not exactly. It's wrong because the particle's angular momentum about the point in space where the mass centre of the stick had been is not just $m_pb^2\omega_f$. There is also the $m_pv_{fs}b$ term. Look at your relative velocity equation.

I'll try once more... IT SHOULD NOT BE QUADRATIC! You appear to be adding up energies, not angular momenta.

ok so $v_{fs} = \frac { m_pv_{fs}- m_p\omega b}{m_p + m_s}$

 

[toesockshoe]

Not exactly. It's wrong because the particle's angular momentum about the point in space where the mass centre of the stick had been is not just $m_pb^2\omega_f$. There is also the $m_pv_{fs}b$ term. Look at your relative velocity equation.

I'll try once more... IT SHOULD NOT BE QUADRATIC! You appear to be adding up energies, not angular momenta.

ok, so let's say that the mass hit the stick at an angle... thus sending the stick in a somewhat diagonal direction. in this case, you would need to add the L=r x p to the stick as well right? (in addition to doing hte similar thing to the point mass_

 

[haruspex]

ok so $v_{fs} = \frac { m_pv_{fs}- m_p\omega b}{m_p + m_s}$

Not quite. The $v_{fs}$ on the right-hand side should be a different variable.

lets say that the mass hit the stick at an angle... thus sending the stick in a somewhat diagonal direction. in this case, you would need to add the L=r x p to the stick as well right? (in addition to doing hte similar thing to the point mass

Your question is not precise enough for me to answer.
As I say, there are several approaches that work - the trick is to use one consistently and not count any momentum contributions twice over or not at all.
If you care to post the algebra for such an oblique impact I'm happy to check it.

 

[toesockshoe]

Not quite. The $v_{fs}$ on the right-hand side should be a different variable.

$v_{fs} = \frac { m_pv_{pi}- m_p\omega b}{m_p + m_s}$

Not quite. The $v_{fs}$ on the right-hand side should be a different variable.
Your question is not precise enough for me to answer.
As I say, there are several approaches that work - the trick is to use one consistently and not count any momentum contributions twice over or not at all.
If you care to post the algebra for such an oblique impact I'm happy to check it.

sorry i copied it wrong to latex... this is what i have:

Disregard me other question... basically we did r x p on the mass and v happened to be v_stick + wb... would you say that in problems where there is translational speed its a good idea to always use r x p for a point mass?

 

[toesockshoe]

Not quite. The $v_{fs}$ on the right-hand side should be a different variable.

Your question is not precise enough for me to answer.
As I say, there are several approaches that work - the trick is to use one consistently and not count any momentum contributions twice over or not at all.
If you care to post the algebra for such an oblique impact I'm happy to check it.

also can you explain to me how to find that $v_{pf} = v_{sf} + \omega b$ ? to be i just used it cause it made sense conceptually, but how would you show your work to get that naswer?

 

[haruspex]

... basically we did r x p on the mass and v happened to be v_stick + wb... would you say that in problems where there is translational speed its a good idea to always use r x p for a point mass?

Yes.

also can you explain to me how to find that $v_{pf} = v_{sf} + \omega b$ ? to be i just used it cause it made sense conceptually, but how would you show your work to get that naswer?

It's just relative velocity. This only applies here at the instant after collision, note. In the direction of the particle's original motion, the stick gets a velocity $v_{fs}$, and the particle's velocity relative to that is $\omega b$, in the same direction, so the two add to give the particle's total velocity. Later, of course, as the stick rotates, the relative velocity will be in a different direction from the system's mass centre, but we don't need to worry about that.