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

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.

• I don't know what the discriminant rule is, but when I solve for $k$ directly I get $k = r\omega + \sqrt{\frac{2E - I\omega^2}{m}}$
– d_b
Jan 7, 2020 at 21:27
• I'm confused what you think you're doing. You derive a quadratic equation for $k$ but don't solve it correctly. You are trying to maximise $k$ but never differentiate. Have you heard of Lagrange multipliers? Jan 7, 2020 at 21:43
• @d_b If you put that in a comment I'll mark the answer as solved. Jan 7, 2020 at 21:45
• @jacob1729 I'm a highschool student whose just using the tools accessible to me. I likened it too much to a maths problem about adjusting the coeffecients of a line for the line to become a tangent to a circle. Jan 7, 2020 at 21:48
• @Krish I didn't mean that to sound condescending. I am genuinely confused where you get the equation for $k$ from. The mention of disciminants in particular is interesting. Lagrange multipliers aren't that hard and do I think make the algebra easier to organise in this case - you might want to take this as an opportunity to learn about them. Jan 7, 2020 at 21:53

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.