# Calculating the Radius of a Floating Aluminum Ball | Fluids Question

• mirandasatterley
In summary, the spherical aluminum ball has an outer radius of 1.26cm and a radius of 0.79cm from the center of the sphere to the surface of the ball.
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)

mirandasatterley said:
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.

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?

mirandasatterley said:
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.

Doc Al said:
Yes, that will give you the outer radius.

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

Doc Al said:
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.

mirandasatterley said:
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.)

mirandasatterley said:
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.

Doc Al said:
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?

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?

mirandasatterley said:
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

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.

All good.

mirandasatterley said:
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.

Thank you so much for all your help

## 1. How do you determine the weight of the ball?

The weight of the ball can be determined by using a scale to measure its mass. This can be done by placing the ball on the scale and recording the reading.

## 2. What is the density of the fluid?

The density of the fluid can be found in a reference table or by using a density meter to measure the mass and volume of the fluid. Density is typically measured in units of grams per cubic centimeter (g/cm3).

## 3. What is the buoyant force acting on the ball?

The buoyant force acting on the ball can be calculated using the Archimedes' principle, which states that the buoyant force is equal to the weight of the fluid displaced by the object. This can be calculated using the density of the fluid and the volume of the ball.

## 4. How do you calculate the volume of the ball?

The volume of the ball can be calculated using the formula for the volume of a sphere: V = (4/3)Πr3, where r is the radius of the ball.

## 5. What is the formula for calculating the radius of the ball?

The formula for calculating the radius of the ball is r = (3M/(4Πρ))1/3, where M is the mass of the ball, ρ is the density of the fluid, and Π is the mathematical constant pi (approximately equal to 3.14).

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