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 34b

Chapter 9, Problem 34b

You are a project manager for a manufacturing company. One of the machine parts on the assembly line is a thin, uniform rod that is 60.0 cm long and has mass 0.400 kg. One of your engineers has proposed to reduce the moment of inertia by bending the rod at its center into a V-shape, with a 60.0^o angle at its vertex. What would be the moment of inertia of this bent rod about an axis perpendicular to the plane of the V at its vertex?

Verified step by step guidance

Understand the problem: The moment of inertia depends on the mass distribution relative to the axis of rotation. For a uniform rod bent into a V-shape, the mass is distributed along two equal-length segments, each forming an angle of 30° with the axis perpendicular to the plane of the V at its vertex.

Determine the length of each segment: Since the rod is bent at its center, each segment of the V will have half the total length of the rod. Thus, each segment is 30.0 cm (or 0.300 m) long.

Set up the formula for the moment of inertia: For a thin rod of length \( L \) and mass \( m \), the moment of inertia about an axis perpendicular to the rod and passing through one end is \( I = \frac{1}{3}mL^2 \). However, in this case, the rod is bent, so we need to account for the angle and calculate the contribution of each segment separately.

Calculate the perpendicular distance of each segment's center of mass from the axis: The center of mass of each segment is located at its midpoint. The perpendicular distance from the axis to the center of mass of each segment is \( d = \frac{L}{2} \sin(30°) \), where \( L = 0.300 \) m.

Combine the contributions of both segments: The total moment of inertia is the sum of the moments of inertia of the two segments. For each segment, the moment of inertia is \( I_{segment} = m_{segment} d^2 \), where \( m_{segment} = \frac{m}{2} \) is the mass of each segment. Add the contributions from both segments to find the total moment of inertia.

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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; the further the mass is from the axis, the greater the moment of inertia. For a uniform rod, the moment of inertia can be calculated using the formula I = (1/3)ml² for rotation about an end, or I = (1/12)ml² for rotation about its center.

Bending and Geometry

When a uniform rod is bent into a V-shape, its geometry changes, affecting how mass is distributed relative to the axis of rotation. The angle of the bend and the length of the segments of the rod must be considered to accurately calculate the new moment of inertia. The V-shape introduces two segments of the rod, each contributing differently to the overall moment of inertia based on their distances from the axis.

Parallel Axis Theorem

The parallel axis theorem is a principle used to determine 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 is the moment of inertia about the new axis, I_cm is the moment of inertia about the center of mass, m is the mass of the object, and d is the distance between the two axes. This theorem is essential when calculating the moment of inertia for complex shapes or configurations.

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Parallel Axis Theorem

Related Practice

Textbook Question

You are a project manager for a manufacturing company. One of the machine parts on the assembly line is a thin, uniform rod that is 60.0 cm long and has mass 0.400 kg. What is the moment of inertia of this rod for an axis at its center, perpendicular to the rod?

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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 uniform bar has two small balls glued to its ends. The bar is 2.00 m long and has mass 4.00 kg, while the balls each have mass 0.300 kg and can be treated as point masses. Find the moment of inertia of this combination about an axis parallel to the bar through both balls;

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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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Textbook Question

A wagon wheel is constructed as shown in Fig. E9.33. The radius of the wheel is 0.300 m, and the rim has mass 1.40 kg. Each of the eight spokes that lie along a diameter and are 0.300 m long has mass 0.280 kg. What is the moment of inertia of the wheel about an axis through its center and perpendicular to the plane of the wheel? (Use Table 9.2.)

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