/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 About what axis will a uniform, ... [FREE SOLUTION] | 91Ó°ÊÓ

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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
Rotate the balsa-wood sphere about an axis \(\frac{2}{\sqrt{15}}R\) from its center.

Step by step solution

01

Understanding the Problem

We have two spheres with the same mass and radius: a uniform balsa-wood sphere and a thin-walled hollow lead sphere. We need to find the axis about which the moment of inertia of the balsa-wood sphere equals that of the lead sphere, where the lead sphere's moment of inertia is calculated about its diameter.
02

Moment of Inertia for Thin-Walled Lead Sphere

The moment of inertia for a thin-walled, hollow sphere about a diameter is given by \[ I_{ ext{lead}} = \frac{2}{3}MR^2 \] where \( M \) is the mass and \( R \) is the radius of the sphere.
03

Moment of Inertia for Uniform Balsa-Wood Sphere

The moment of inertia for a uniform solid sphere about a diameter is \[ I_{ ext{balsa}} = \frac{2}{5}MR^2 \], but this does not equal \( I_{lead} \), so we need to align the axis differently to match the values.
04

Setup the Equation for Rotational Axis

Since the axis isn't about the diameter, we'll use the parallel axis theorem: \[ I = I_{cm} + Md^2 \] where \( I_{cm} = \frac{2}{5}MR^2 \) is the moment of inertia about its center of mass (diameter), and \( d \) is the distance from the new axis to the center of mass.
05

Finding the Required Axis Distance

Set the moment of inertia equation for the balsa sphere equal to the lead sphere: \[ \frac{2}{5}MR^2 + Md^2 = \frac{2}{3}MR^2 \] Solving this, we find:\[ Md^2 = \frac{2}{3}MR^2 - \frac{2}{5}MR^2 = \frac{10}{15}MR^2 - \frac{6}{15}MR^2 = \frac{4}{15}MR^2 \]\[ d^2 = \frac{4}{15}R^2 \]\[ d = \frac{2}{\sqrt{15}}R \]
06

Conclusion: Axis about Which Moment of Inertia are Equal

The balsa-wood sphere should be rotated about an axis \[ \frac{2}{\sqrt{15}}R \] from its center to have the same moment of inertia as the lead sphere rotated about its diameter.

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Key Concepts

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

Balsa-Wood Sphere
A balsa-wood sphere is a perfect example of a uniform solid sphere. Balsa wood is known for its lightweight properties, which makes it an interesting material for physics problems. When we examine the moment of inertia of a balsa-wood sphere, it's essential to recall that it is generally calculated about its diameter. Moment of inertia depends heavily on both mass distribution and geometry.

For a uniform solid sphere, like our balsa-wood sphere, the formula to calculate the moment of inertia about a diameter is:
  • \( I_{\text{balsa}} = \frac{2}{5}MR^2 \)
where \( M \) is the mass and \( R \) is the radius.
These spheres have mass uniformly spread throughout their volume, which makes the moment of inertia calculation more straightforward compared to objects with less uniform mass distribution, such as a hollow sphere.

To find an axis where this balsa-wood sphere shares the same moment of inertia as another object, like a hollow lead sphere, we often need to use advanced techniques such as the parallel axis theorem. This theorem allows us to find the moment of inertia about different axes, that are not through the center of the sphere.
Hollow Lead Sphere
A hollow lead sphere differs significantly from a solid sphere in terms of mass distribution. Lead is much denser, and a hollow sphere only has mass on its outer shell, which changes how its moment of inertia is calculated.

For a thin-walled, hollow sphere, the mass is situated away from the axis of rotation in the center. This unique distribution gives the hollow sphere a distinct formula for its moment of inertia:
  • \( I_{\text{lead}} = \frac{2}{3}MR^2 \)
where \( M \) is the mass of the sphere and \( R \) is the radius.
The moment of inertia is generally larger for hollow objects compared to uniform ones of the same mass and radius because more mass is situated further from the rotational axis.

In practical terms, understanding this difference is crucial for comparing the rotational properties of different objects, such as the uniform balsa-wood sphere and the hollow lead sphere. This understanding allows us to solve problems like identifying an axis for the solid sphere where its moment of inertia matches that of the hollow sphere.
Parallel Axis Theorem
The "Parallel Axis Theorem" is a helpful tool in physics that allows us to calculate the moment of inertia of an object about an axis parallel to an axis through its center of mass. It tells us how to adjust the moment of inertia when the rotation isn't about the axis at the object's center of mass.

The theorem formula is:
  • \( I = I_{\text{cm}} + Md^2 \)
where \( I_{\text{cm}} \) is the moment of inertia at the center of mass, \( M \) is the mass, and \( d \) is the distance between the two axes.
When using this theorem in practice, we calculate the moment of inertia about the center of mass and then add \( Md^2 \) to account for the new axis.

This theorem is particularly useful in exercises like the one at hand, where the goal is to find an equivalent moment of inertia for different objects with distinct axes. For instance, the balsa-wood sphere, which normally has its moment of inertia calculated over its diameter, needs adjustment to match the moment of inertia of the hollow lead sphere, which was given for the diameter.
Using the parallel axis theorem allows us to find the specific axis for the balsa sphere that matches the moment of inertia of the lead sphere, thus solving the problem through mathematical insight and understanding of physical properties.

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Most popular questions from this chapter

A flywheel with a radius of 0.300 m starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s\(^2\). Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim (a) at the start; (b) after it has turned through 60.0\(^\circ\); (c) after it has turned through 120.0\(^\circ\).

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