Conversation started Jan 1, 2023 at 11:52.
Wolgwang
Wolgwang
Jan 1, 2023 11:52
@JohnRennie Can you take a look here when you get time?
 
4 hours later…
John Rennie
John Rennie
Jan 1, 2023 15:31
@Wolgwang I drew a diagram to show what happens:
user image
The period is T = 2𝜋√𝓁/g so it depends only on 𝓁
And we will assume 𝓁 is the distance from the pivot to the COM of the bob. So we are assuming:
1. the string mass is negligible
2. the radius of the bob isn't big enough that we have to calculate its moment of inertia i.e. we treat it as a point mass a distance 𝓁 from the pivot
When full of mercury and when empty the COM of the bob is at the centre of the sphere, so 𝓁 is the same.
But when half full of mercury the COM has moved downwards by ... erm ... wherever the COM of a hemisphere is. So when half full the distance 𝓁 is longer then when completely full or empty.
So as the mercury empties 𝓁 increases then decreases again, so the period increases then decreases again.
@Wolgwang Does this make sense?
 
Conversation ended Jan 1, 2023 at 15:36.