/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 A uniform bar has two small ball... [FREE SOLUTION] | 91Ó°ÊÓ

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A uniform bar has two small balls glued to its ends. The bar is 2.00 \(\mathrm{m}\) long and has mass 4.00 \(\mathrm{kg}\) , while the balls each have mass 0.500 \(\mathrm{kg}\) and can be treated as point masses. Find the moment of inertia of this combination about each of the following axes: (a) an axis perpendicular to the bar through its center; (b) an axis perpendicular to the bar through one of the balls; (c) an axis parallel to the bar through both balls; (d) an axallel to the bar and 0.500 m from it.

Short Answer

Expert verified
(a) 2.33 kg m², (b) 6.33 kg m², (c) 0 kg m², (d) Consider reformulating.

Step by step solution

01

Determine the Moment of Inertia for Reference Axes

The moment of inertia of the bar about its center (point A) is given by \( I_A = \frac{1}{12} mL^2 \), where \( m = 4.00 \, \mathrm{kg} \) and \( L = 2.00 \, \mathrm{m} \). Calculate \( I_A = \frac{1}{12} \times 4.00 \times (2.00)^2 = \frac{1}{12} \times 4 \times 4 = 1.33 \, \mathrm{kg \, m^2} \).
02

Moment of Inertia for the Ball Centers through Bar's Center

For point masses (balls at the ends of the bar), the moment of inertia is the sum of \( m_br^2 \) for each ball: \( I = 2 \times (0.500 \, \mathrm{kg}) \times (1.00 \, \mathrm{m})^2 = 1.00 \, \mathrm{kg \, m^2} \). The distance from the center of the bar to each ball is \( 1.00 \, \mathrm{m} \).
03

Total Moment of Inertia about Bar's Center

Add the moments of inertia calculated from the bar and balls. \( I_{\text{center}} = 1.33 \, \mathrm{kg \, m^2} + 1.00 \, \mathrm{kg \, m^2} = 2.33 \, \mathrm{kg \, m^2} \).
04

Axis Perpendicular to Bar through One Ball

Use the parallel axis theorem: \( I = I_{\text{center}} + Md^2 \), where \( d = 1.00 \, \mathrm{m} \) and \( M = 0.500 \, \mathrm{kg} \). Therefore, \( I = 2.33 + 4.00 \times (1.00)^2 = 6.33 \, \mathrm{kg \, m^2} \).
05

Axis Parallel to Bar through Balls

The moment of inertia about an axis along the bar through both balls is zero. Hence, \( I = 0 \) since no rotation occurs about this axis parallel to the length of the bar.
06

Axis Parallel to Bar and Offset 0.5 m

This configuration is tautological because an axis parallel at 0.5 m effectively acts as a translation of the entire system. Therefore, modify the part of Step 4 to account for this offset by considering the entirety of the parallelism.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Bar
A uniform bar in physics means a bar with mass distributed evenly along its length. Each part of the bar has the same amount of mass per unit length.
For instance, in our exercise, the uniform bar is 2 meters long and has a total mass of 4 kg, implying that its weight is equally distributed across its entire 2-meter span.
- When calculating properties like the moment of inertia for such a bar, we usually consider its shape and mass distribution.
- The bar's mass and dimensions are fundamental in determining its rotational characteristics, making calculations more straightforward since its density and mass distribution remain consistent.
Parallel Axis Theorem
The parallel axis theorem is crucial when finding the moment of inertia about an axis that is parallel to one passing through the object's center of mass.
This theorem is a valuable tool because it allows you to calculate the moment of inertia for any parallel axis once you know the inertia relative to a central axis.
- Formula: The theorem states that: \[ I = I_{ ext{center of mass}} + Md^2 \] where: - \( I \) is the moment of inertia about the new axis. - \( I_{ ext{center of mass}} \) is the moment of inertia through the center of mass. - \( M \) is the total mass of the object. - \( d \) is the distance between the new axis and the central axis.- In our exercise, we used it to calculate the moment of inertia through one of the balls by considering the bar's center and the distance to the ball.
Point Masses
Point masses are simplified models representing masses concentrated at a single point in space. This idealization is especially helpful in simplifying calculations in physics, especially when dealing with systems of particles or objects.
- The moment of inertia for a point mass is calculated by using: \[ I = mr^2 \] where: - \( m \) is the mass of the object (here, small balls glued to the bars). - \( r \) is the distance from the point mass to the axis of rotation.- In the exercise, the small balls are treated as point masses, which allows the inertia calculation to become more manageable by simply using their masses and their distances from the axis.
Rotational Motion
Rotational motion refers to the movement of an object around a center or an axis. This type of motion is all about how these objects rotate and spin.
- It's characterized by parameters like angular displacement, angular velocity, and angular acceleration.
- The moment of inertia plays a crucial role because it tells us how much torque is needed for a desired angular acceleration. - In our example, calculating the moment of inertia aids in understanding how the uniform bar with point masses behaves under rotation when different axes are considered. By knowing the inertia, one can predict how easy or hard it is to spin the bar around specific points, giving insights into the dynamics of rotational systems.

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Most popular questions from this chapter

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