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About what axis will a uniform, balsa-wood sphere have the same moment of inertia as does a thin-walled, hollow, lead sphere of the same mass and radius, with the axis along a diameter?

Short Answer

Expert verified
The sphere must rotate about an off-center axis, three-fifths from diameter equivalent.

Step by step solution

01

Understand the Moment of Inertia

The moment of inertia for the thin-walled, hollow sphere is given by the formula \( I = \frac{2}{3} m r^2 \), where \( m \) is the mass and \( r \) is the radius. For a solid sphere, the moment of inertia along the diameter is \( I = \frac{2}{5} m r^2 \) and along any tangent is \( I = \frac{7}{5} m r^2 \).
02

Compare to Solid Sphere Formulas

The problem requires the solid balsa-wood sphere to have the same moment of inertia as the hollow lead sphere, \( I = \frac{2}{3} m r^2 \). This is different from \(\frac{2}{5} m r^2 \) along a diameter.
03

Set Equivalency Equation

Since we want \( I_{balsa} = I_{hollow} \), solve \( \frac{2}{5} m r^2 + \frac{2}{5} m r^2 x = \frac{2}{3} m r^2 \). Rearrange: \( \frac{2}{5} m r^2 (1 + x) = \frac{2}{3} m r^2 \).
04

Solve for Axis Factor

Divide both sides by \( \frac{2}{5} m r^2 \), resulting in \( 1 + x = \frac{5}{3} \). Therefore, \( x = \frac{2}{3} \). Calculate perpendicular distance for equivalent moment of inertia.
05

Conclusion on Axis

With \( x = \frac{2}{3} \,r^2 \), the equivalent inertia indicates off-center axis, not coinciding with uniformity. Thus, the correct moment of inertia requires adjusting axis beyond center.

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Key Concepts

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

Understanding a Uniform Sphere
A uniform sphere is solid and has the same density throughout its entire volume. Like a balsa-wood sphere in our example, each tiny part of the sphere contributes equally to the sphere's mass. The moment of inertia, which can be thought of as a measure of resistance to rotational motion, depends not only on the mass of the sphere but also on how this mass is distributed relative to the axis of rotation.

For a solid sphere, such as a uniform balsa-wood sphere, the moment of inertia about an axis through its center (also known as the axis along a diameter) is given by the formula:\[ I = \frac{2}{5} m r^2 \]where \( m \) is the sphere's mass and \( r \) is its radius.

But it's important to realize that a uniform sphere's moment of inertia can vary based on the axis of rotation. When it is rotated about an axis tangential to its surface, the formula changes to reflect that a greater portion of the mass is further from the axis.
Exploring a Hollow Sphere
A hollow sphere, unlike a uniform sphere, has all of its mass distributed on its outer shell, forming a perfect void on the inside. This configuration dramatically affects its moment of inertia because all of the mass is farther from the axis of rotation, even if both spheres have the same mass and radius.The moment of inertia for a thin-walled, hollow sphere, like our lead sphere in the exercise, is high compared to its solid counterpart. It is calculated using the formula:\[ I = \frac{2}{3} m r^2 \]Here, every bit of mass contributes equally but is far from the center, amplifying its effect on rotational inertia.

In comparing a hollow sphere to a solid one, the broader distribution of mass in the hollow sphere around its rotational axis plays a crucial role, making it handle differently under rotational forces.
Properties of Balsa Wood
Balsa wood is renowned for its light weight and high strength-to-weight ratio, making it an excellent material for various engineering and aerodynamic applications. In the context of spheres, a balsa-wood sphere would have significantly less mass than a similarly sized sphere made of denser material, like lead. However, when comparing moments of inertia, the material's density shifts the focus towards how mass affects inertia distribution. Even if a balsa-wood sphere and a lead sphere share the same shape and external dimensions, their moments of inertia could greatly differ due to the variance in mass.

In this exercise context, the challenge lies in finding an axis that allows the lighter balsa sphere to mirror the rotational properties of the denser, hollow sphere. This involves adjusting the rotation axis to balance the lighter mass to distribute effectively in line with the hollow sphere's inertia characteristics.
Understanding a Lead Sphere
Lead is a dense metal, significantly heavier than balsa wood. In a hollow lead sphere, as addressed in the problem, the mass is concentrated on the outer shell. This makes it an intriguing study case for understanding inertia. The dense nature of lead means that even a thin-walled sphere carries substantial mass, impacting its moment of inertia. In this scenario where both spheres have the same mass and radius, the hollow lead sphere, despite its sparse core, exhibits a larger moment of inertia owing to the mass distribution. This poses an interesting dynamic compared to a balsa sphere; the two share the same rotational inertia when the lighter sphere is rotated at an adjusted axis, reflecting another side of physics where density and material play intrinsic roles.

Adjusting the rotation axis of the uniformly distributed but less dense balsa wood to match that of the hollow but denser lead enables students to play with fundamental physics principles like mass distribution and rotational dynamics.

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Most popular questions from this chapter

A bicycle wheel has an initial angular velocity of 1.50 \(\mathrm{rad} / \mathrm{s}\) . (a) If its angular acceleration is constant and equal to 0.300 \(\mathrm{rad} / \mathrm{s}^{2}\) , what is its angular velocity at \(t=2.50 \mathrm{s} ?\) (b) Through what angle has the wheel turned between \(t=0\) and \(t=2.50 \mathrm{s} ?\)

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An electric turntable 0.750 \(\mathrm{m}\) in diameter is rotating about a fixed axis with an initial angular velocity of 0.250 \(\mathrm{rev} / \mathrm{s}\) and a constant angular acceleration of 0.900 \(\mathrm{rev} / \mathrm{s}^{2}\) . (a) Compute the angular velocity of the turntable after 0.200 \(\mathrm{s}\) . (b) Through how many revolutions has the turntable spun in this time interval? (c) What is the tangential speed of a point on the rim of the turntable at \(t=0.200\) s? (d) What is the magnitude of the resultant acceleration of a point on the rim at \(t=0.200 \mathrm{s} ?\)

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