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Chapter 9: Problem 55

A thin, rectangular sheet of metal has mass \(M\) and sides of length \(a\) and \(b .\) Use the parallel-axis theorem to calculate the moment of inertia of the sheet for an axis that is perpendicular to the plane of the sheet and that passes through one corner of the sheet.

Short Answer

Expert verified
The moment of inertia of the sheet through one corner is \( \frac{1}{3}M(a^2 + b^2) \).

Step by step solution

01

Identify the Moment of Inertia About the Center

The moment of inertia of a rectangular sheet about an axis perpendicular to the plane and passing through its center is given by \( I_{center} = \frac{1}{12}M(a^2 + b^2) \). This value will form the basis for applying the parallel-axis theorem.
02

Use the Parallel-Axis Theorem

The parallel-axis theorem states that if the moment of inertia about any axis parallel to and a distance \(d\) from an axis through the center of mass is given by \(I_{parallel} = I_{center} + Md^2\). Here, \(d\) is the distance from the center of mass to the corner of the sheet.
03

Calculate the Distance to the Corner

The distance \(d\) from the center to a corner of the sheet can be calculated using the Pythagorean theorem. It is given by \(d = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2} = \frac{1}{2}\sqrt{a^2 + b^2}\).
04

Apply the Parallel-Axis Theorem

Substitute \(I_{center}\) and \(d\) into the parallel-axis theorem: \(I_{corner} = \frac{1}{12}M(a^2 + b^2) + M\left(\frac{1}{2}\sqrt{a^2 + b^2}\right)^2\).
05

Simplify the Expression

Simplify the expression: \[I_{corner} = \frac{1}{12}M(a^2 + b^2) + \frac{1}{4}M(a^2 + b^2)\]Combine the terms: \[I_{corner} = \left(\frac{1}{12} + \frac{1}{4}\right)M(a^2 + b^2)\].
06

Simplify Further to Find Final Solution

Combine the fractions: \[\frac{1}{12} + \frac{1}{4} = \frac{1}{12} + \frac{3}{12} = \frac{4}{12} = \frac{1}{3}\].The final moment of inertia is therefore: \[I_{corner} = \frac{1}{3}M(a^2 + b^2)\].

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Key Concepts

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

parallel-axis theorem
The parallel-axis theorem is a fundamental principle in physics used to determine an object's moment of inertia when the axis of rotation is not located at the center of mass. In essence, it allows you to calculate the moment of inertia about any axis parallel to an axis through the mass's center. If you know the moment of inertia about the center, the parallel-axis theorem lets you adjust for different rotational axes located a distance away. The theorem is mathematically expressed as:
  • Iparallel = Icenter + Md2
where:
  • Iparallel is the moment of inertia about the new axis.
  • Icenter is the moment of inertia about the center of mass.
  • M is the mass of the object.
  • d is the perpendicular distance between the center of mass and the new axis.
This theorem is crucial for problems where the rotational axis is not conveniently passing through the center of mass but at some distance from it. By applying the parallel-axis theorem, you can calculate the necessary adjustments efficiently.
rectangular sheet
A rectangular sheet can be thought of as a flat two-dimensional object with a specific length and width. For our problem, this sheet is a thin piece of metal with a uniform surface and mass distributed evenly across its plane.The dimensions are a significant factor when calculating the moment of inertia as they determine the distances involved in rotation. The longer the sides, the further the mass is distributed from the center, and thus affects how much rotational inertia the sheet has.For an axis that passes perpendicularly through the center of a uniform rectangular sheet, the moment of inertia is given by:
  • Icenter = \( \frac{1}{12}M(a^2 + b^2) \)
where:
  • M is the mass of the sheet.
  • a and b are the lengths of the sides of the sheet.
Understanding this baseline is key before employing additional theorems like the parallel-axis theorem for offset calculations.
mass distribution
Mass distribution refers to how mass is spread across an object. This is essential in physics as it influences how easily an object can be rotated or how it moves in response to forces. In terms of moment of inertia, an evenly distributed mass means the calculation adheres to straightforward mathematical expressions because each infinitesimal piece of the object contributes equally to the overall inertia. For a rectangular sheet, which is considered thin and uniform, the mass is evenly spread out over its area. This consistent mass distribution enables simplification in calculations, as assumptions about variations in thickness or density across the sheet do not complicate the math. Thus, when the sheet experiences rotation, the mass opposed to the change in rotation behaves predictably, based on the equation derived for its inertia using the given mass and dimensions.

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Most popular questions from this chapter

A light, flexible rope is wrapped several times around a hollow cylinder, with a weight of 40.0 \(\mathrm{N}\) and a radius of 0.25 \(\mathrm{m}\), that rotates without friction about a fixed horizontal axis. The cylinder is attached to the axle by spokes of a negligible moment of inertia. The cylinder is initially at rest. The free end of the rope is pulled with a constant force \(P\) for a distance of \(5.00 \mathrm{m},\) at which point the end of the rope is moving at 6.00 \(\mathrm{m} / \mathrm{s} .\) If the rope does not slip on the cylinder, what is the value of \(P ?\)

Electric Drill. According to the shop manual, when drilling a 12.7 -mm- diameter hole in wood, plastic, or aluminum, a drill should have a speed of 1250 rev/min. For a 12.7 -mm diameter drill bit turning at a constant 1250 \(\mathrm{rev} / \mathrm{min}\) , find (a) the maximum linear speed of any part of the bit and (b) the maximum radial acceleration of any part of the bit.

A metal sign for a car dealership is a thin, uniform right triangle with base length \(b\) and height \(h .\) The sign has mass \(M .\) (a) What is the moment of inertia of the sign for rotation about the side of length \(h ?\) (b) If \(M=5.40 \mathrm{kg}, b=1.60 \mathrm{m},\) and \(h=1.20 \mathrm{m},\) what is the kinetic energy of the sign when it is rotating about an axis along the \(1.20-\mathrm{m}\) side at 2.00 \(\mathrm{rev} / \mathrm{s} ?\)

A thin uniform rod of mass \(M\) and length \(L\) is bent at its center so that the two segments are now perpendicular to each other. Find its moment of inertia about an axis perpendicular to its plane and passing through (a) the point where the two segments meet and (b) the midpoint of the line connecting its two ends.

A sphere with radius \(R=0.200 \mathrm{m}\) has density \(\rho\) that decreases with distance \(r\) from the center of the sphere according to \(\rho=3.00 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}-\left(9.00 \times 10^{3} \mathrm{kg} / \mathrm{m}^{4}\right) r .\) (a) Calculate the total mass of the sphere. (b) Calculate the moment of inertia of the sphere for an axis along a diameter.
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