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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
All textbooksYoung & Freedman Calc 14th EditionCh 09: Rotation of Rigid BodiesProblem 31b
Chapter 9, Problem 31b

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 perpendicular to the bar through one of the balls;

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Step 1: Recall the formula for the moment of inertia. The moment of inertia (I) for a system of point masses is given by \( I = \sum m_i r_i^2 \), where \( m_i \) is the mass of each object and \( r_i \) is the distance of the object from the axis of rotation. For a uniform bar, the moment of inertia can also be calculated using its mass distribution.
Step 2: Break the system into components. The system consists of a uniform bar and two point masses (the balls). The axis of rotation is perpendicular to the bar and passes through one of the balls. This means we need to calculate the contributions to the moment of inertia from the bar and the other ball separately.
Step 3: Calculate the contribution of the bar. Since the axis of rotation is at one end of the bar, the moment of inertia of the bar about this axis is given by \( I_{\text{bar}} = \frac{1}{3} M L^2 \), where \( M \) is the mass of the bar (4.00 kg) and \( L \) is its length (2.00 m).
Step 4: Calculate the contribution of the second ball. The second ball is located at the opposite end of the bar, so its distance from the axis of rotation is equal to the length of the bar (2.00 m). The moment of inertia of the ball is given by \( I_{\text{ball}} = m r^2 \), where \( m \) is the mass of the ball (0.300 kg) and \( r \) is the distance from the axis (2.00 m).
Step 5: Add the contributions to find the total moment of inertia. The total moment of inertia is the sum of the contributions from the bar, the ball at the axis (which has zero contribution since \( r = 0 \)), and the second ball. Thus, \( I_{\text{total}} = I_{\text{bar}} + I_{\text{ball}} \). Substitute the values and simplify to find the result.

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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 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 case, the moment of inertia will include contributions from both the bar and the balls at the ends.
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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 useful when dealing with composite objects like the bar and balls in this problem.
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Rotational Dynamics

Rotational dynamics is the study of the motion of objects that are rotating and the forces that cause this motion. It encompasses concepts such as torque, angular momentum, and the relationship between linear and angular quantities. Understanding rotational dynamics is essential for analyzing how the moment of inertia affects the rotational behavior of the bar and balls when subjected to forces.
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Related Practice
Textbook Question

Calculate the moment of inertia of each of the following uniform objects about the axes indicated. Consult Table 9.2 as needed. A thin 2.50-kg rod of length 75.0 cm, about an axis perpendicular to it and passing through (i) one end and (ii) its center, and (iii) about an axis parallel to the rod and passing through it.

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Textbook Question

Four small spheres, each of which you can regard as a point of mass 0.200 kg, are arranged in a square 0.400 m on a side and connected by extremely light rods (Fig. E9.28). Find the moment of inertia of the system about an axis that passes through the centers of the upper left and lower right spheres and through point O.

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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. 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 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

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.

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 perpendicular to the bar through its center;

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