The lowest centre of mass while filling a cup of water?

[Aihara]

Homework Statement

Ok, there's a cylindrical mug of mass m. When full of water it has a mass 4m. Assuming the base of the mug has no mass, how full is the water in the mug when the centre of mass is the closest to the base? (as a proportion of the height of the mug I'm guessing)

The Attempt at a Solution

I was trying to set up an equation with the sum of the bits equals the whole thing..
Moment of empty mug: mh/2
Moment of water: I think I get this bit wrong. it takes 3m of water to fill the cup, so 3m=h

But basically I then added the moments of the empty mug and water and equated it to total mass of the system multiplied by the COM.
I tried to differentiate to find a minimum but I had an h and a h^2 which didn't leave me with a nice answer in terms of h.

I rushed this, but if what I have said isn't clear I will make it clear when I come back to it later :)
Thanks

 

[Doc Al]

Show exactly what you did. You'll get a bit of a messy expression for the center of mass as a function of water height. But when you do the differentiation and simplify, you'll end up with a quadratic equation. You'll get a nice answer as a fraction of the cylinder height.

 

[Aihara]

yep ok

Moment of Empty mug = mh/2

When the water has depth h, the mass is 3m.
Take mass of water as km where 0$$\leq$$ k $$\leq$$3
therefore height of water = hk/3
and moment of water is mass x distance from base = km*(hk/3)*1/2 = ((k^2)mh)/6

Moment of the whole system = (m+km)Y (where Y = centre of mass of whole system)


So, mh/2 + (1/6)*(k^2)*m*h = m(1+k)Y
cancel m's
h/2 + (1/6)*(k^2)*h = (1+k)Y

Y= h/(2(1+k)) + ((k^2)*h)/6(1+k)
Y=(h/2)(1/(1+k) + (k^2)/(3(1+k))
so I have centre of mass of the system in turns of k, the mass of the water... (not the water height)

But if I differentiate, and equate to - that should give me the value of k that is a minimum (?)

this gets complicated when I differentiate
dY/dk = (h/2)((-1/(2(k+1)^2)+(k(k+2))/(3(k+1)^2)
So the bit in the brackets equals 0.
Rearranging and cancelling I get k^2 + 2k + 1.5 = 0
Using the formula this gives, (-2+rt(3))/2 (a negative number for k, which is no good)

I'm not sure if what I have put is understandable, but thanks for helping ^^

 

[Doc Al]

yep ok

Moment of Empty mug = mh/2

When the water has depth h, the mass is 3m.
Take mass of water as km where 0$$\leq$$ k $$\leq$$3
therefore height of water = hk/3
and moment of water is mass x distance from base = km*(hk/3)*1/2 = ((k^2)mh)/6

Moment of the whole system = (m+km)Y (where Y = centre of mass of whole system)


So, mh/2 + (1/6)*(k^2)*m*h = m(1+k)Y
cancel m's
h/2 + (1/6)*(k^2)*h = (1+k)Y

Y= h/(2(1+k)) + ((k^2)*h)/6(1+k)
Y=(h/2)(1/(1+k) + (k^2)/(3(1+k))
so I have centre of mass of the system in turns of k, the mass of the water... (not the water height)

But if I differentiate, and equate to - that should give me the value of k that is a minimum (?)

Looks good.

this gets complicated when I differentiate
dY/dk = (h/2)((-1/(2(k+1)^2)+(k(k+2))/(3(k+1)^2)

Redo that derivative.