Let's say the block has mass $m$, and the rod has mass $M$ and length $L$. The block slides down the ramp and collides with the rod, sticking to it.
At the bottom of the ramp, the block has $v = \sqrt{2gh}$.
I am not sure whether I should use conservation of linear momentum or angular momentum to determine the velocity of the block-rod system after the block collides with the rod.
When I use linear momentum, I have:
$mv_{block} = (m+M)v_{f}$
$v_{f}=\frac{m\sqrt{2gh}}{m+M}$
When I use angular momentum, I have:
$mv_{block}L=I\omega_{f}$
$\omega_{f}=\frac{m\sqrt{2gh}L}{L^2(\frac{1}{3}M+m)}=\frac{m\sqrt{2gh}}{L(\frac{1}{3}M+m)}$
If I use the equation: $v_{f}=L\omega_{f}$, I get:
$v_{f}=\frac{m\sqrt{2gh}}{\frac{1}{3}M+m}$
, which does not match the $v_{f}$ found in linear momentum.
I probably used the linear momentum incorrectly in this situation, or $v_{f}=L\omega_{f}$ incorrectly, but I do not know the reason why. Could this be explained to me?
Furthermore, after I determine the velocity/angular velocity of the block-rod system, how do I find the height at which the center of mass of the block-rod will go to?
Should I use:
$\frac{1}{2}I_{rod}\omega^2 + \frac{1}{2}I_{block}\omega^2=(m+M)gh_{CM}$
or:
$\frac{1}{2}mv^2 + \frac{1}{2}I_{rod}\omega^2=(m+M)gh_{CM}$
or do both work? And why?
As for using this equation: $\frac{1}{2}mv^2 + \frac{1}{2}I_{rod}\omega^2=(m+M)gh_{CM}$
would I use $v$ from $v_{f}$ (in the linear momentum equation), and $\omega$ from $\omega_{f}$ (in the angular momentum equation)?
In summary, I am having trouble knowing when I should use linear vs. angular quantities, or when both are acceptable. Also when $v=r\omega$ can be used (this is still vague to me), as seen in the linear and angular momentum question I had above where $v$ and $\omega$ do not seem to be related by the radius.
Best Answer
(a) Your linear momentum conservation equation is wrong on two counts:
(1) Momentum is not conserved because the rod has a force on it from the hinge at O.
(2) Only the bottom of the rod acquires velocity $v_f$. The closer parts of the rod are to O the slower they will move!
(b) But I think your angular momentum equation is correct. Angular momentum about O is conserved at the collision because there is no external torque about O on rod or block during the collision.
(c) "or do both [energy equations] work? And why?" The first is alright, but the second isn't, because again you seem to be assuming that the whole of the rod has the same speed. The right hand sides of the equations are fine.