Ch 09: Rotation of Rigid Bodies

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 09: Rotation of Rigid Bodies

Problem 51

Chapter 9, Problem 51

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.

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Start by recalling the formula for the moment of inertia of a rectangular sheet about an axis perpendicular to its plane and passing through its center of mass. This is given by \( I_{\text{cm}} = \frac{1}{12} M (a^2 + b^2) \), where \( M \) is the mass of the sheet, and \( a \) and \( b \) are the lengths of its sides.

Next, recall the parallel-axis theorem, which states that the moment of inertia about any axis parallel to the center of mass axis is given by \( I = I_{\text{cm}} + M d^2 \), where \( d \) is the perpendicular distance between the two axes.

In this problem, the axis of rotation passes through one corner of the sheet. The distance \( d \) between the center of mass and this corner is the diagonal of the rectangle divided by 2. Using the Pythagorean theorem, the diagonal is \( \sqrt{a^2 + b^2} \), so \( d = \frac{1}{2} \sqrt{a^2 + b^2} \).

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

Simplify the expression by squaring \( \frac{1}{2} \sqrt{a^2 + b^2} \) to get \( \frac{1}{4} (a^2 + b^2) \), and combine terms to find the total moment of inertia \( I \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Moment of Inertia

Moment of inertia is a measure of an object's resistance to rotational motion about an axis. It depends on the mass distribution relative to the axis of rotation. For a thin rectangular sheet, the moment of inertia can be calculated using specific formulas that account for its shape and mass.

Parallel-Axis Theorem

The parallel-axis theorem states that the moment of inertia of a body about any axis parallel to an axis through its center of mass can be found by adding the product of the mass and the square of the distance between the two axes to the moment of inertia about the center of mass axis. This theorem is essential for calculating the moment of inertia when the axis of rotation does not pass through the center of mass.

Rectangular Sheet Properties

A thin rectangular sheet has specific geometric properties that influence its moment of inertia. The dimensions (length a and width b) and mass M are critical in determining how the mass is distributed across the sheet. Understanding these properties allows for the application of the moment of inertia formulas relevant to rectangular shapes.

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If we multiply all the design dimensions of an object by a scaling factor f, its volume and mass will be multiplied by f³. By what factor will its moment of inertia be multiplied?

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Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass M and radius R about an axis perpendicular to the hoop's plane at an edge.

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Textbook Question

If we multiply all the design dimensions of an object by a scaling factor f, its volume and mass will be multiplied by f³. If a 1/48 scale model has a rotational kinetic energy of 2.5 J, what will be the kinetic energy for the full-scale object of the same material rotating at the same angular velocity?

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