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 33d

Chapter 9, Problem 33d

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 and 0.500 m from it.

Verified step by step guidance

Step 1: Recall the formula for the moment of inertia of a system of point masses about a given axis. For a system of discrete masses, the moment of inertia is given by \( I = \sum m_i r_i^2 \), where \( m_i \) is the mass of the \( i \)-th object and \( r_i \) is its perpendicular distance from the axis of rotation.

Step 2: For the uniform bar, treat it as a continuous object. The moment of inertia of a uniform bar about an axis parallel to the bar and at a distance \( d \) from it is given by \( I_{\text{bar}} = M d^2 \), where \( M \) is the mass of the bar and \( d \) is the distance from the axis to the center of mass of the bar. Here, \( M = 4.00 \; \text{kg} \) and \( d = 0.500 \; \text{m} \).

Step 3: For the two small balls, treat them as point masses. The moment of inertia for each ball is \( I_{\text{ball}} = m r^2 \), where \( m \) is the mass of the ball and \( r \) is the perpendicular distance from the axis of rotation. The distance \( r \) for each ball is the sum of the distance from the axis to the bar (0.500 m) and half the length of the bar (1.00 m). Thus, \( r = 0.500 + 1.00 = 1.50 \; \text{m} \).

Step 4: Calculate the total moment of inertia by summing the contributions from the bar and the two balls. The total moment of inertia is \( I_{\text{total}} = I_{\text{bar}} + 2 \cdot I_{\text{ball}} \). Substitute the expressions for \( I_{\text{bar}} \) and \( I_{\text{ball}} \) into this equation.

Step 5: Substitute the given values into the expressions: \( M = 4.00 \; \text{kg} \), \( d = 0.500 \; \text{m} \), \( m = 0.300 \; \text{kg} \), and \( r = 1.50 \; \text{m} \). Simplify the terms 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. For point masses, it is calculated as the product of the mass and the square of the distance from the axis of rotation. In this problem, the moment of inertia will be calculated for both the bar and the balls attached to its ends.

Parallel Axis Theorem

The parallel axis theorem allows us to calculate the moment of inertia of a body about any axis parallel to an axis through its center of mass. It states that the moment of inertia about the new axis is equal to the moment of inertia about the center of mass axis plus the product of the mass and the square of the distance between the two axes. This theorem is essential for determining the moment of inertia of the bar and the balls about the specified axis.

Center of Mass

The center of mass is the point at which the mass of a system is concentrated and can be considered to act for translational motion. For composite systems, like the bar with attached balls, the center of mass can be found by considering the masses and their positions. Understanding the center of mass is crucial for applying the parallel axis theorem and calculating the moment of inertia accurately.

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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

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?

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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;

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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