Centre of Mass of an empty and filled goblet

[Taniaz]

1. A goblet consists of a uniform thin hemispherical cup of radius r, a circular base of the same material thickness and radius as the cup, and an intervening stem of length r and whose weight one quarter of that of the cup.
(a) Show that the height of the centre of gravity above the base is (13/14)r.
(b) If the mass of the goblet is M and that of the amount of liquid that fills it is M' . Show that filling it raises the centre of gravity through a distance (39/65) r (M'/M+M')2. X = x1M1 + x2M2+...xnMn / M1+M2+...Mn
where x is the centre of mass of each body that makes up the goblet and M is their respective mass3. I've managed to do part a where I've considered the surface densities of the hemispherical shell, the rectangular stem and since it's from the base, I considered the mass off the circle but the distance to the base would be 0.

So C.O.M of the hemispherical shell = r/2 + r (from the start till the base)
C.O.M of the stem is r/2 to the base
C.O.M circle = 0 since it's to the base itself

Then Mass = surface density x area ; surface density is represented by sigma
Area of hemisphere = 2 pi r^2 so m= 2 pi r^2 (sigma) Area of circle = pi r^2 so m = pi r^2 (sigma)
And m of stem = 1/4 (mass of hemisphere) = pi r ^2 /2

Plug it into the equation and you get 13/14 r

What I don't get is part b since they haven't mentioned whether how full the cup is? How do we find its centre of mass?

 

[Doc Al]

What I don't get is part b since they haven't mentioned whether how full the cup is?

They say the liquid fills the cup.

 

[Taniaz]

They say the liquid fills the cup.

So wouldn't the centre of mass of the liquid be the same as that of the cup? I tried that, fit it into my equation but it still doesn't give me the required answer.
M( 13/14)r +M' (x) / (M + M') where x is the centre of mass of the liquid which I don't know.
Am I right in using surface densities? I used this because it said a thin cup and it's hollow as well. Or should volume densities be used?

 

[Doc Al]

So wouldn't the centre of mass of the liquid be the same as that of the cup?

Are the centers of mass of the solid and hollow hemispheres the same?

 

[Taniaz]

The centre of mass of a hemispherical shell is r/2 from the flat side and for a solid hemisphere it is 3r/8 from the flat side.

 

[Doc Al]

The centre of mass of a hemispherical shell is r/2 from the flat side and for a solid hemisphere it is 3r/8 from the flat side.

Good. Now make use of that fact.

 

[Taniaz]

I've tried but it's not worked. I have just been using surface densities and not volume densities to find the mass.

So for part a this is what I did: My diagram included a horizontal goblet with the x-axis as it's line of symmetry.
C.O.M of the hemispherical shell = r/2 + r (from the start till the base)
C.O.M of the stem is r/2 to the base
C.O.M circle = 0 since it's to the base itself

Then Mass = surface density x area ; surface density is represented by sigma
Area of hemisphere = 2 pi r^2 so m= 2 pi r^2 (sigma) Area of circle = pi r^2 so m = pi r^2 (sigma)
And m of stem = 1/4 (mass of hemisphere) = pi r ^2 /2 (sigma)

Then my equation for an empty goblet became:

(r/2 + r) ( 2 pi r^2 sigma) + (r/2)( pi r^2 /2 sigma) + 0( pi r^2 sigma) / ( 2 pi r^2 sigma + pi r^2 /2 sigma +( pi r^2 sigma)

the pi r^2 sigma terms cancel out leaving me with:

(r/2 + r) + (r/2)(1/2) / 2 + (1/2) + (1) = (13/4) r / (7/2) = (13/14)r

For part b, when they put in the liquid, what am I supposed to take, the surface density or the volume density? I'm not sure how to go about it. Even if I take it as the completely filled and I take the surface density, all I get is:

M(13/14)r + M' (r/2 +r) / (M+M') and simplifying this does not give (39/65) r (M'/M+M')

 

[Doc Al]

You have the center of mass of the empty goblet and of the filled liquid. Now find the center of mass of the system. (No need for any densities.)

 

[Taniaz]

For the centre of mass of the liquid, should I take it till the base if the goblet or just it's centre of mass alone?
So for the latter case my equation would become:

M (13/14)r +M'(r/2) / (M+M')
but how do I simplify this equation? I can't add the variables in the numerator?

 

[Taniaz]

Got it!
Basically we assume the liquid to be a solid hemispherical shell and not hollow so the c.o.m of the shell from the curved surface is 5/8 r but since it's to the base it will be 5/8 r + r where r is the length of the stem. Then because it's asking for the difference we basically have to subtract it from that of the empty goblet.
So the equation is
M'(5/8 r + r) + M(13/14)r / M+M'
- M (13/14)r/M
= 39/56 r (M'/M+M')

 

[Doc Al]

Good!

 

[Taniaz]

Thank you.

Doc Al would you mind taking a look at the new question I posted please?