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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 rectangular sheet about an axis perpendicular to the plane of the sheet and passing through one corner is \(\frac{M}{3}(a^2 + b^2)\).

Step by step solution

01

Identify Knowns and Unknowns

Here we know the mass (M), and the dimensions of the rectangle (a and b). We are asked to find the moment of inertia about an axis perpendicular to the plane of the sheet and passing through a corner.
02

Calculate Moment Of Inertia For Axis Through Center of Mass

First, calculate the moment of inertia for a thin rectangular sheet about an axis through the center of mass and perpendicular to the plane of the sheet using the formula \(I_{cm} = \frac{1}{12} M (a^2 + b^2)\). The result, \(I_{cm}\), will be used in the next step to find the moment of inertia about an axis through a corner.
03

Apply The Parallel-Axis Theorem

Now, apply the parallel-axis theorem, which allows one to calculate the moment of inertia about any axis parallel to and a distance, \(d\), away from an axis through the center of mass. This theorem is given by \(I = I_{cm} + M d^2\). In this case, the axis through the corner of the sheet is a distance \(d = \sqrt{(\frac{a}{2})^2 + (\frac{b}{2})^2}\) away from the center of mass. Replace \(d\) with this value in the parallel axis theorem formula.
04

Calculate Final Moment Of Inertia

After the substitution you can simplify the equation to calculate the moment of inertia for the sheet about the corner, which should yield: \(I = \frac{M}{12}(a^2 + b^2) + M(\frac{a^2}{4} + \frac{b^2}{4}) = \frac{M}{3}(a^2 + b^2)\).

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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 an important tool used in physics to calculate the moment of inertia of an object about an axis that is parallel to an axis through its center of mass. This theorem states that if you know the moment of inertia of the object through its center of mass (denoted as \( I_{cm} \)), you can find the moment of inertia about any parallel axis some distance \( d \) away, using the formula: \[ I = I_{cm} + Md^2 \]- \( I \) is the moment of inertia about the new axis.- \( M \) is the mass of the object.- \( d \) is the perpendicular distance between the center of mass axis and the new axis.This theorem is particularly useful for objects with complex shapes and when the axis of rotation is not through the center of mass. It simplifies the calculation by relating the moment of inertia to a known value at the center of mass and an adjustment based on mass and distance.
Rectangular Sheet
A rectangular sheet is a flat, two-dimensional object characterized by its mass \( M \) and its planar dimensions, typically noted as side lengths \( a \) and \( b \). In the context of moment of inertia, such a sheet assumes uniform mass distribution across its surface, which makes calculating its moment of inertia somewhat straightforward when dealing with axes through one of its principal symmetry lines or points. The formula for the moment of inertia of a rectangular sheet about an axis perpendicular to its plane and through its center of mass is given by: \[ I_{cm} = \frac{1}{12}M(a^2 + b^2) \]This formula arises because the mass is equally distributed, allowing each point's contribution around the axis to be neatly summed up through the integration of its dimensions.
Understanding this fundamental calculation is crucial before applying modifications such as those described by the parallel-axis theorem.
Axis Through a Corner
When calculating the moment of inertia for a rectangular sheet about an axis that goes through one of its corners, it differs from calculation through its center because this axis is no longer conveniently aligned with any symmetry of the object. The position of the axis through a corner introduces the need for using the parallel-axis theorem. Since the moment of inertia through the center of mass is known, you adjust for the shift to the corner by calculating the distance \(d\) from the center to the corner:\[ d = \sqrt{ \left( \frac{a}{2} \right)^2 + \left( \frac{b}{2} \right)^2 } \]Using this distance in the parallel-axis theorem allows one to effectively shift the axis and calculate the required moment of inertia correctly.
Center of Mass
The center of mass is a crucial concept when dealing with moments of inertia as it is the pivot point around which the mass of the object can be regarded as being uniformly distributed for motion analysis. For a symmetrical object like a rectangular sheet, the center of mass coincides with its geometric center. Mathematically, the center of mass \( (x_{cm}, y_{cm}) \) for a rectangular sheet with uniform density and dimensions \( a \) and \( b \) is located at:\[ x_{cm} = \frac{a}{2} \]\[ y_{cm} = \frac{b}{2} \]This positioning importantly affects calculations using the parallel-axis theorem by providing a reference point for determining distances \( d \). Understanding and locating the center of mass ensures accurate and effective application of physical theories related to rotation and balance.

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Most popular questions from this chapter

Measuring \(I\). As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis perpendicular to the wheel at its center. You measure the diameter of the wheel to be \(0.640 \mathrm{~m}\). Then you mount the wheel on frictionless bearings on a horizontal frictionless axle at the center of the wheel. You wrap a light rope around the wheel and hang an \(8.20 \mathrm{~kg}\) block of wood from the free end of the rope, as in Fig. E9.49. You release the system from rest and find that the block descends \(12.0 \mathrm{~m}\) in \(4.00 \mathrm{~s}\). What is the moment of inertia of the wheel for this axis?

The earth is approximately spherical, with a diameter of \(1.27 \times 10^{7} \mathrm{~m} .\) It takes 24.0 hours for the earth to complete one revolution. What are the tangential speed and radial acceleration of a point on the surface of the earth, at the equator?

A uniform disk with radius \(R=0.400 \mathrm{~m}\) and mass \(30.0 \mathrm{~kg}\) rotates in a horizontal plane on a frictionless vertical axle that passes through the center of the disk. The angle through which the disk has turned varies with time according to \(\theta(t)=(1.10 \mathrm{rad} / \mathrm{s}) t+\left(6.30 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2}\) What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.100 rev?

You need to design an industrial turntable that is \(60.0 \mathrm{~cm}\) in diameter and has a kinetic energy of \(0.250 \mathrm{~J}\) when turning at \(45.0 \mathrm{rpm}\) (rev/min). (a) What must be the moment of inertia of the turntable about the rotation axis? (b) If your workshop makes this turntable in the shape of a uniform solid disk, what must be its mass?

A new species of eel is found to have the same mass but onequarter the length and twice the diameter of the American eel. How does its moment of inertia for spinning around its long axis compare to that of the American eel? The new species has (a) half the moment of inertia as the American eel; (b) the same moment of inertia as the American eel; (c) twice the moment of inertia as the American eel; (d) four times the moment of inertia as the American eel.

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