Airborne -Approach Speed (Revision 2)

Performing a force balance at the onset of liftoff the contact normal force tends to zero, and the net force acting on the body is the weight:

$\begin{aligned} \sum F &= mg \\&= m\frac{v^2}{R} \\ &\Updownarrow \\ v &= \sqrt{gR} \end{aligned}$

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To determine the minimum speed required on the flat to achieve the liftoff speed we can use Conservation of Energy:

$KE + PE = KE' + PE'$

On the LHS ( The Approach Energy ) we have Translational + Rotational (wheels) energy ( taking this to be the datum for potential energy ):

$\frac{1}{2}M v_a^2 + I \omega_a^2 = \frac{1}{2}M v_a^2 + \frac{I}{r^2}v_a^2 = \beta v_a^2$

Where;

$\beta = \frac{1}{2}M + \frac{I}{r^2}$

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On the RHS ( The Liftoff Energy ) we have Translational + Rotational ( wheels ) + Potential energy. Looking at each in turn:

Translational

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$\frac{1}{2}M v^2$

Rotational ( wheels )

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$I \omega^2 = \frac{I}{r^2}v^2$

Potential:

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$mg \left( R - h_{CM} \right)$

Finally, solving for $v_a$

$\beta v_a^2 = mg \left( R-h_{CM} \right) + \beta v^2$

$v_a = \sqrt{ \frac{ mg \left( R-h_{CM} \right) }{ \beta } + v^2 }$

$v_a \approx \SI{7.5}{\frac{m}{s} }$

As always, If I've bungled this analysis ( and problem ) please speak up. It usually takes me several revisions to get things in order.