Calculating the Radius of a Floating Aluminum Ball | Fluids Question

[mirandasatterley]

My Question is " A spherical aluminum ball of mass 1.26kg contains an empty spherical cavity that is concentric with the ball. The ball just barely floats in water. Calculate (a) the outer radius of the ball and (b) the radius of the cavity."
I'm notsure if it's right, but so far I have:
the buoyancy force = the gravitational force.
B = (density(fluid))(g)(V(object))
Fg = m(object)g = (density(object))(V(object))(g)
Setting them equal,
d(fluid)gV(object) = d(object)V(object)g
I was then thinking that V(object) = 4/3(pi)(r^3), sothat i can solve for the radius of the object.

I'm unsure of how to solve for the radius because the V of water is unknown, and I don't know any other equations that i can use to solve it. I'm also unsure of how to incorperate the radius of the cavity in part b. Would it just be r(found in part a)-r(cavity)

 

[Doc Al]

B = (density(fluid))(g)(V(object))

OK.

Fg = m(object)g = (density(object))(V(object))(g)

No need to go beyond the second expression, since you are given the mass of the object. Set those equal to solve for V. Then use your volume formula to solve for the radius.

 

[mirandasatterley]

So,

d(fluid)gV(object) = m(object)g
d(fluid) (4/3(pi)(r^3) = m(object)

Is my idea for part b right at all?

 

[Doc Al]

So,

d(fluid)gV(object) = m(object)g
d(fluid) (4/3(pi)(r^3) = m(object)

Yes, that will give you the outer radius.

Is my idea for part b right at all?

What idea?

Hint for part b: The density of aluminum will come in handy.

 

[mirandasatterley]

Yes, that will give you the outer radius.

Does this mean the radius from the surface to the center (including the cavity)?

 

[mirandasatterley]

Hint for part b: The density of aluminum will come in handy.

This is what confused me before.
I have the equation;
d(fluid)(4/3(pi)(r^3))g = d(object)(4/3(pi)(r^3))g
d(fluid)(4/3(pi)(r^3)) = d(object)(4/3(pi)(r^3))

And when I am this far, the term;
(4/3(pi)(r^3), will cancel.

 

[Doc Al]

Does this mean the radius from the surface to the center (including the cavity)?

Yes. (Radius is always from the center of the sphere to some point.)

 

[Doc Al]

This is what confused me before.
I have the equation;
d(fluid)(4/3(pi)(r^3))g = d(object)(4/3(pi)(r^3))g
d(fluid)(4/3(pi)(r^3)) = d(object)(4/3(pi)(r^3))

And when I am this far, the term;
(4/3(pi)(r^3), will cancel.

You are assuming here that the aluminum is in the shape of a solid ball, but it's really a shell--a hollow ball. (Thus, the radius in each expression would be different, so they would not cancel.) But don't go there.

When you go from Fg = m(object)g to d(object)V(object)g, realize that V(object) is just the volume of the material, not really the volume of the object. You should use m(ball) = d(aluminum)V(aluminum); V(aluminum) is the volume of the aluminum (the shell), not the entire ball.

 

[mirandasatterley]

When you go from Fg = m(object)g to d(object)V(object)g, realize that V(object) is just the volume of the material, not really the volume of the object. You should use m(ball) = d(aluminum)V(aluminum); V(aluminum) is the volume of the aluminum (the shell), not the entire ball.

Okay, that makes more sense,

d(fluid)gV(object) = d(aluminum)gV(aluminum)
d(fluid)(4/3(pi)(r^3)) = d(aluminum)(m(aluminum)/d(alumimun))
d(fluid)(4/3(pi)(r^3)) = (m(aluminum))

So, in this case, when I solve for r, this will give me the radius of only the aluminum and I subtract that from the total radius i have already found in part a to find the radius of the cavity?

 

[mirandasatterley]

Wait, that's not right, It gives me the same r, I'm missing something does the equation V=4/3 (pi)(r^3) have to be modified to represent only the aluminum as well?

 

[Doc Al]

Okay, that makes more sense,

d(fluid)gV(object) = d(aluminum)gV(aluminum)
d(fluid)(4/3(pi)(r^3)) = d(aluminum)(m(aluminum)/d(alumimun))
d(fluid)(4/3(pi)(r^3)) = (m(aluminum))

You're going in circles. That last equation is what you should have started with:

d(fluid)gV(object) = m(object)
d(fluid)g(4/3(pi)(r^3)) = m(object)​

The radius here is the radius of the object, the sphere of aluminum. (The volume of the object equals the volume of displaced fluid.) m(object) is given; use it to solve for part (a).

So, in this case, when I solve for r, this will give me the radius of only the aluminum and I subtract that from the total radius i have already found in part a to find the radius of the cavity?

No. Do this:
(1) Solve for the volume of the object
(2) Solve for the volume of the aluminum (as discussed earlier)
(3) Subtract one from the other to find the volume of the space
(4) Since the space is spherical, find its radius

 

[mirandasatterley]

I think I have it this time;
d(fluid)gV(object) = m(aluminum)g
d(fluid)(4/3(pi)(r^3)) = m(aluminum)
I solve this for r, and that is the answer for part a.

Next,
V(object) = 4/3(pi)(r^3)
i sub in the radius i found in part a to find V(object)

Next,
V(aluminum) = m(aluminum)/ d(aluminum)
and solve for v aluminum

I then subtract,
V(cavity) = V(object) - V(aluminum)

Finally,
4/3(pi)(r^3) = V(cavity),
and solve for r, this is the answer for b.

 

[Doc Al]

All good.

I think I have it this time;
d(fluid)gV(object) = m(aluminum)g
d(fluid)(4/3(pi)(r^3)) = m(aluminum)
I solve this for r, and that is the answer for part a.

Next,
V(object) = 4/3(pi)(r^3)
i sub in the radius i found in part a to find V(object)

You can also just use your first equation and get V(object) directly.

 

[mirandasatterley]

Thank you so much for all your help