Calculating Speed and Angular Momentum | Physics Forums Discussion

[bobie]

Homework Statement

this is not homework, I am discussing angular momentum here: https://www.physicsforums.com/threa...d-angular-momentum.776258/page-2#post-4884631, we have 2 equations$$\frac{i \omega _0^2}{2}+\frac{m v_i^2}{2}=\frac{I \omega _1^2}{2}+\frac{m
v_f^2}{2}$$
$I \omega _0+m r v_0=~I \omega _1+m r v_1$
$~M=10,~m=1,~l=1, ~I=\frac{[l^2] M = 10}{12},~r=\frac{l}{2} = 0.5,~v_i=22,~\omega _0=0, ~\omega_1 = x$

Homework Equations

$$\frac{m v_i^2}{2}=\frac{I \omega ^2}{2}+\frac{m
v_f^2}{2} \rightarrow v_f^2 = (v_i^2= ~22^2) = 484 - \frac{5}{6}\omega^2 $$
$m r v_i= ~I \omega +m r v_f \rightarrow v_f = 22 - \frac{10}{6}\omega$

The Attempt at a Solution

The problem is simple, but I get a funny result. I tried hundreds of times with different approaches to no avail; can you tell me where I go wrong, or if the problem has no solution?
$v_f = 22 - \frac{10}{6}\omega → v_f^2 = 22^2 + \frac{10^2}{6^2} \omega^2 + \frac{2*22*-10}{6}\omega$
$v_f^2 = 484 + \frac{100}{36} \omega^2 - \frac{440}{6}\omega$
plugging in the first equation:

$(v_f^2 = ) ~484 - \frac{5}{6}x^2 = 484 + \frac{100}{36} x^2 - \frac{440}{6}x \rightarrow - \frac{30}{36}x^2 = \frac{100}{36}x^2 - \frac{6*440}{36}x = 130x^2 = 2640 x$
$x= 2640/130 = 20.3$

 

[haruspex]

$$ v_f^2 = (v_i^2= ~22^2) = 484 - \frac{5}{6}\omega^2 $$

Check that last term.