Maximise $v + r \omega$ in $E = \frac{1}{2}m v^2 + \frac{1}{2} I \omega^2 $ [closed]

Question

I'm working out the maximum 'speed' that an object can cut a light gate, but the object can both rotate and move linearly.

The maximum speed measured by the light gate can be approximated as $k = r\omega + v$, with $\omega r$ being the angular velocity of the bar. The total energy in the system limits the angular and linear velocity of the bar, $E = \frac{1}{2}m v^2 + \frac{1}{2} I \omega^2 $.

Combining both equations leads to $0 = mk^2 - 2 m r \omega k + ((I + mr^2 )\omega^2 - 2E)$ because:

$$ 2E = mv^2 + I \omega^2$$ $$ 0 = m(k-r\omega)^2 + I \omega^2 - 2E$$

which can be simplified by discriminant rule to find that $$k = \frac{\sqrt{m((I+m)\omega^2 - 2E)}}{m r\omega}$$

This is confusing to me because if you increase the energy in the system, $((I + mr^2)\omega^2 - 2E)$ decreases, decreasing $k$. Any help surrounding this would be greatly appreciated.

Answer 1

I don't know what the discriminant rule is (I guess you mean the quadratic formula), but from \begin{align} 2E = m\left(k - r\omega\right)^2 + I \omega^2 \end{align} we can solve for $k$ directly: \begin{align} \left(k - r\omega\right)^2 = \frac{2E - I \omega^2}{m}\\ \rightarrow k = r\omega + \sqrt{\frac{2E - I \omega^2}{m}} \end{align} No quadratic formula needed.