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Chapter 14: Problem 30

A spherical aluminum ball of mass \(1.26 \mathrm{kg}\) contains an empty spherical cavity that is concentric with the ball. The ball just barely floats in water. Calculate (a) the outer \(\mathrm{ra}\) dius of the ball and (b) the radius of the cavity.

Short Answer

Expert verified
The outer radius of the ball approximates to \(6.8 \, \mathrm{cm}\) and the radius of the cavity approximates to \(5.7 \, \mathrm{cm}\)

Step by step solution

01

Determine the Total Volume of the Ball

Since the ball just floats in the water, the weight of the water displaced by the ball must be equal to the weight of the ball. This can be determined by dividing the weight of the ball by the density of water (about \(1000 \, \mathrm{kg}/\mathrm{m}^3 \)). Therefore, the volume of the ball (including the cavity volume) is \(v = \frac{1.26 \, \mathrm{kg}}{1000 \, \mathrm{kg}/\mathrm{m}^3 } = 0.00126 \, \mathrm{m}^3\)
02

Calculate the Outer Radius of the Ball

To calculate the outer radius, remember that the volume of a sphere is given by \(\frac{4}{3}\pi r^3\). Rearranging for r gives \(r = \left(\frac{3v}{4\pi}\right)^{1/3}\). Substituting the volume already calculated gives \(r = \left( \frac{3 \times 0.00126 \, \mathrm{m}^3}{4\pi} \right)^{1/3} \approx 0.068 \, \mathrm{m}\) or \(6.8 \, \mathrm{cm}\)
03

Calculate the Volume of the Aluminum

Since the cavity is empty, the volume of aluminum is simply the total volume of the ball less the volume of the cavity. The density of aluminum is about \(2700 \, \mathrm{kg}/\mathrm{m}^3 \). So, the volume of aluminum is the mass of the aluminum (1.26 kg) divided by its density: \(v_{Al} = \frac{1.26 \, \mathrm{kg}}{2700 \, \mathrm{kg}/\mathrm{m}^3 } \approx 0.000467 \, \mathrm{m}^3 \)
04

Calculate the Radius of the Cavity

Since the cavity is also spherical and concentric with the ball, the volume of the cavity is equal to the total volume of the ball less the volume of aluminum. That is \(v_{cavity} = v_{total} - v_{Al} = 0.00126 - 0.000467 \, \mathrm{m}^3 \approx 0.000793 \, \mathrm{m}^3\). Using the formula for the volume of a sphere to find the radius of the cavity gives \(r_{cavity}= \left(\frac{3 \times 0.000793 \, \mathrm{m}^3}{4\pi}\right)^{1/3} \approx 0.057 \, \mathrm{m}\) or approx. \(5.7 \, \mathrm{cm}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Archimedes' Principle
Understanding the buoyant force in physics is greatly aided by a principle formulated over two millennia ago by the Greek mathematician Archimedes. This principle states that when an object is partially or fully submerged in a fluid, it experiences an upward buoyant force equal to the weight of the fluid it displaces.

Let's apply this to our exercise with the aluminum ball. The ball is said to 'just barely float' which implies that the upward buoyant force exactly counteracts the gravitational force pulling it down. According to Archimedes' principle, this means the weight of the displaced water equals the weight of the ball. By determining the weight of the ball and knowing the density of water, we can calculate the volume of water displaced, which indeed is equivalent to the volume of the ball, including its hollow cavity.
Density and Buoyancy
The concepts of density and buoyancy are intricately linked. Density can be defined as the mass per unit volume of a substance. Buoyancy, on the other hand, relates to an object's ability to float in a fluid. The denser an object, the more mass it has in a given volume, and the heavier it is compared to the fluid it's in. An object will float if it is less dense than the fluid it is submerged in.

In our exercise, the buoyancy allows the aluminum ball to float because the combined density of the aluminum and the air in the cavity is less than the density of the water. By understanding the relationship between mass, volume, and density, we were able to deduce the volume of aluminum and the cavity within the ball to ensure that it matched up with the initial conditions for flotation.
Spherical Volume Calculations
Spherical volume calculations are vital in solving problems involving round objects. For a sphere, the volume is calculated using the formula \( V = \frac{4}{3}\pi r^3 \), where \( V \) is the volume and \( r \) is the radius. This formula arises from integrating the area of circular slices of the sphere across its diameter.

In the steps given for the exercise solution, we first find the total volume of the ball (including the cavity) to calculate the outer radius. Then, we calculate the volume of just the aluminum to find the volume of the cavity alone. Knowing the volume of these individual parts enables us to use the volume formula of a sphere to solve for their respective radii, giving us the size of the ball's outer surface and the size of the inner cavity.

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Most popular questions from this chapter

The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkably high speed, creating low pressure. The higher atmospheric pressure in the still air inside the buildings can cause windows to pop out. As originally constructed, the John Hancock building in Boston popped window panes, which fell many stories to the sidewalk below. (a) Suppose that a horizontal wind blows in streamline flow with a speed of \(11.2 \mathrm{m} / \mathrm{s}\) outside a large pane of plate glass with dimensions \(4.00 \mathrm{m} \times 1.50 \mathrm{m}\) Assume the density of the air to be uniform at \(1.30 \mathrm{kg} / \mathrm{m}^{3} .\) The air inside the building is at atmospheric pressure. What is the total force exerted by air on the window pane? (b) What If? If a second skyscraper is built nearby, the air speed can be especially high where wind passes through the narrow separation between the buildings. Solve part (a) again if the wind speed is \(22.4 \mathrm{m} / \mathrm{s}\), twice as high.

A cube of wood having an edge dimension of \(20.0 \mathrm{cm}\) and a density of \(650 \mathrm{kg} / \mathrm{m}^{3}\) floats on water. (a) What is the distance from the horizontal top surface of the cube to the water level? (b) How much lead weight has to be placed on top of the cube so that its top is just level with the water?

A Pilot tube can be used to determine the velocity of air flow by measuring the difference between the total pressure and the static pressure (Fig. \(\mathrm{P} 14.49) .\) If the fluid in the tube is mercury, density \(\rho_{\mathrm{Hg}}=13600 \mathrm{kg} / \mathrm{m}^{3},\) and \(\Delta h=5.00 \mathrm{cm},\) find the speed of air flow. (Assume that the air is stagnant at point \(A,\) and take \(\rho_{\mathrm{air}}=1.25 \mathrm{kg} / \mathrm{m}^{3} .\) )

For the cellar of a new house, a hole is dug in the ground, with vertical sides going down \(2.40 \mathrm{m} .\) A concrete foundation wall is built all the way across the \(9.60-\mathrm{m}\) width of the excavation. This foundation wall is \(0.183 \mathrm{m}\) away from the front of the cellar hole. During a rainstorm, drainage from the street fills up the space in front of the concrete wall, but not the cellar behind the wall. The water does not soak into the clay soil. Find the force the water causes on the foundation wall. For comparison, the weight of the water is given by \(2.40 \mathrm{m} \times 9.60 \mathrm{m} \times 0.183 \mathrm{m} \times 1000 \mathrm{kg} / \mathrm{m}^{3} \times\) \(9.80 \mathrm{m} / \mathrm{s}^{2}=41.3 \mathrm{kN}\).

In about 1657 Otto von Guericke, inventor of the air pump, evacuated a sphere made of two brass hemi. spheres. Two teams of eight horses each could pull the hemispheres apart only on some trials, and then "with greatest difficulty," with the resulting sound likened to a cannon firing (Fig. P14.62). (a) Show that the force Frequired to pull the evacuated hemispheres apart is \(\pi R^{2}\left(P_{0}-P\right),\) where \(R\) is the radius of the hemispheres and \(P\) is the pressure inside the hemispheres, which is much less than \(P_{0}\). (b) Determine the force if \(P=0.100 P_{0}\) and \(R=0.300 \mathrm{m}\)
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