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Ch 09: Rotation of Rigid Bodies
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoYoung & Freedman Calc 15th EditionCh 09: Rotation of Rigid BodiesProblema 33d
Capitolo 9, Problema 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.

Guida verificata passo dopo passo
1
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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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.
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Percorso guidato
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Intro to Moment of Inertia

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.
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Percorso guidato
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Parallel Axis Theorem

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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Intro to Center of Mass
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