Delay of Moon passing a defined meridian each day

On the website Lunar Synodic Curve a nice image of an Moon analemma is shown together with an explantion on the reason why for the different phases of the Moon the images have to be taken with 24h 51min apart. It is explained that this difference is due to the longer synodic month of 29days 12hrs 44mins (compared to the sidereal month of 27days 7h 43mins). Whereas the explanation that due to the additional rotation time for the moon to reach the same angular relation to the sun an additional angle of 29.1 degs has to be travelled, the last step to come to the additional 51mins each day is not clear to me. Any clues on how to calculate this from the sidereal and synodic month duration or the associated mean motions of moon and sun would be very welcome.

$$\omega = \frac{2*\pi}{T_{stars}} - \frac{2*\pi}{T_M} = \frac{2*\pi}{T}$$
where $$T_M$$ and $$T_{stars}$$ are the sidereal period of the Moon and the sidereal day length. Substituting the values above, we get that the Moon takes about 51 minute over 24h to go around the celestial sphere, crossing the same meridian. Note that this is a rough approximation because the Moon's orbit is not circular, but it explains the value you mentioned.