wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> medium >> Tumbler
        
(Message started by: THUDandBLUNDER on Jan 12^th, 2004, 1:09pm)

Title: Tumbler
Post by THUDandBLUNDER on Jan 12^th, 2004, 1:09pm

A perfectly cylindrical tumbler is 12 cm long, has a radius of 4 cm, and is of uniform density and thickness.
As water is poured into the tumbler the centre of mass of the water and tumbler changes.
This centre of mass is at its lowest when the water is 4.5 cm deep.

What is the mass of the tumbler?

Title: Re: Tumbler
Post by Eigenray on Jan 12^th, 2004, 11:06pm

The centre of mass of the empty tumbler is
x = [hide](2pi R H)(H/2) / ( 2pi R H + pi R² ) = H²/(2H+R) = 36/7[/hide].
The main observation is that [hide]with 4.5 cm of water in the tumbler, the combined centre of mass must be at 4.5 cm. Otherwise, it could be lowered by removing water if it were lower, or adding water if it were higher[/hide]. Let y = 4.5, M = mass of tumbler, m = mass of water = pi y R² ~ 226 gm; then:[hide]
(Mx + my/2) / (M + m) = y
M = my/[2(x-y)] ~ 792 gm[/hide].