Finding radii of aluminum ball with cavity that barely floats

[Honda47]

I'm doing some practice problems because I have a final tomorrow. I ran across this problem and I haven't seen an equation or example like it in our textbook. If you could give me a little direction I'd appreciate it I probably won't have one like it on the test but I'd still like to figure it out.

A spherical aluminum ball of mass 1.26 kg 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.

Thanks

 

[Astronuc]

You can calculate the radius of the sphere from the volume of the sphere, which you get by dividing the density of water by the mass of the ball. Why?

Because in order to barely float, the smear density must be the same as water (1 gm/cc).

Then since you know the real density of all (~2.7 g/cc), you can find the actual volume of Al

Then the volume of the cavity is just the difference in volumes.

Then find the radius of the cavity.

 

[wais]

for reaching out! I'm glad you're practicing for your final tomorrow. Fluid mechanics can definitely be tricky, but with some practice and understanding of the concepts, you'll do great on the test.

To solve this problem, we can use the concept of buoyancy and Archimedes' principle. The buoyant force on the ball must equal its weight in order for it to float in water. This means that the weight of the displaced water by the ball must equal its weight.

We can set up an equation using this concept:

Weight of ball = Weight of displaced water

mg = ρVg

Where m is the mass of the ball, ρ is the density of water (1000 kg/m³), V is the volume of the ball, and g is the acceleration due to gravity (9.8 m/s²).

Since the ball contains a cavity, we can calculate the volume using the equation for the volume of a sphere:

V = (4/3)πr³

Where r is the radius of the ball.

Now, we can plug in the values and solve for r:

1.26 kg * 9.8 m/s² = 1000 kg/m³ * (4/3)πr³ * 9.8 m/s²

r = (1.26 kg * 9.8 m/s²) / (1000 kg/m³ * (4/3)π * 9.8 m/s²)

r = 0.051 m

Therefore, the outer radius of the ball is 0.051 m.

To calculate the radius of the cavity, we can use the same concept as before, but with a different volume. The volume of the cavity can be calculated by subtracting the volume of the ball from the volume of the ball with the cavity:

Vcavity = (4/3)πrb³ - (4/3)πr³

Where rb is the radius of the cavity.

Now, we can plug in the values and solve for rb:

1.26 kg * 9.8 m/s² = 1000 kg/m³ * [(4/3)πrb³ - (4/3)πr³] * 9.8 m/s²

rb = [(1.26 kg * 9.8 m/s²) / (1000 kg/m³ * 9.8 m/s²)] + r³