Ch 09: Rotation of Rigid Bodies

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 09: Rotation of Rigid Bodies

Problem 50

Chapter 9, Problem 50

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.

Verified step by step guidance

Step 1: Recall the formula for the moment of inertia of a hoop about an axis perpendicular to its plane and passing through its center. For a hoop of mass $ M $ and radius $ R $, the moment of inertia is $ I = M R^2 $. However, in this problem, the axis is at the edge of the hoop, not the center.

Step 2: Use the parallel axis theorem to account for the shift in the axis. The parallel axis theorem states $ I = I_{\text{center}} + M d^2 $, where $ I_{\text{center}} $ is the moment of inertia about the center, $ M $ is the mass, and $ d $ is the distance between the center and the new axis.

Step 3: Determine the distance $ d $ between the center of the hoop and the edge. Since the hoop has a radius $ R $, the distance $ d $ is equal to $ R $.

Step 4: Substitute $ I_{\text{center}} = M R^2 $ and $ d = R $ into the parallel axis theorem formula. This gives $ I = M R^2 + M R^2 $.

Step 5: Combine the terms to express the total moment of inertia. The result is $ I = 2 M R^2 $. This is the moment of inertia of the hoop about the specified axis.

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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 changes in its rotational motion about a specific axis. It depends on the mass distribution relative to that axis. For a hoop, the moment of inertia can be calculated using the formula I = Σ(m * r²), where m is the mass of each particle and r is the distance from the axis of rotation.

Parallel Axis Theorem

The parallel axis theorem allows us to calculate the moment of inertia of an object about any axis parallel to an axis through its center of mass. It states that I = I_cm + Md², where I_cm is the moment of inertia about the center of mass, M is the total mass, and d is the distance between the two axes. This theorem is essential for finding the moment of inertia of the hoop about an edge.

Geometry of the Hoop

A hoop is defined as a thin-walled, hollow ring with all its mass concentrated at a constant distance (the radius R) from the axis of rotation. Understanding the geometry of the hoop is crucial for applying the moment of inertia formulas correctly, as it directly influences the calculations and the distribution of mass relative to the chosen axis.

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Textbook Question

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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Textbook Question

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 the point where the two segments meet.

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Textbook Question

A uniform sphere with mass $28.0$ kg and radius $0.380$ m is rotating at constant angular velocity about a stationary axis that lies along a diameter of the sphere. If the kinetic energy of the sphere is $236$ J, what is the tangential velocity of a point on the rim of the sphere?

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Textbook Question

A wheel is turning about an axis through its center with constant angular acceleration. Starting from rest, at t = 0, the wheel turns through 8.20 revolutions in 12.0 s. At t = 12.0 s the kinetic energy of the wheel is 36.0 J. For an axis through its center, what is the moment of inertia of the wheel?

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Textbook Question

A slender rod with length L has a mass per unit length that varies with distance from the left end, where x = 0, according to dm/dx = γx, where γ has units of kg/m². Use Eq. (9.20) to calculate the moment of inertia of the rod for an axis at the left end, perpendicular to the rod. Use the expression you derived in part (a) to express I in terms of M and L. How does your result compare to that for a uniform rod? Explain.

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Textbook Question

An airplane propeller is 2.08 m in length (from tip to tip) with mass 117 kg and is rotating at 2400 rpm (rev/min) about an axis through its center. You can model the propeller as a slender rod.

(a) What is its rotational kinetic energy?

(b) Suppose that, due to weight constraints, you had to reduce the propeller's mass to 75.0% of its original mass, but you still needed to keep the same size and kinetic energy. What would its angular speed have to be, in rpm?

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