91Ó°ÊÓ

A slender rod with length \(L\) has a mass per unit length that varies with distance from the left end, where \(x=0\) , according to \(d m / d x=\gamma x,\) where \(\gamma\) has units of \(\mathrm{kg} / \mathrm{m}^{2}\) . (a) Calculate the total mass of the rod in terms of \(\gamma\) and \(L .\) (b) Use Eq. \((9.20)\) to calculate the moment of inertia of the rod for an axis at the left end, perpendicular to the rod. Use the expression you derived in part (a) to express \(I\) in terms of \(M\) and \(L .\) How does your result compare to that for a uniform rod? Explain this comparison. (c) Repeat part (b) for an axis at the right end of the rod. How do the results for parts (b) and (c) compare? Explain this result.

Step by step solution

01

Understand Mass Distribution

Given the mass per unit length as \( \frac{dm}{dx} = \gamma x \), integrate to find the mass over the entire length of the rod. We need to integrate from 0 to \( L \), this represents the total mass \( M \) of the rod.

02

Integrate to Find Total Mass

The total mass \( M \) can be found by integrating the mass density function: \[ M = \int_0^L \gamma x \, dx = \gamma \int_0^L x \, dx. \] This integral simplifies to \( \gamma \left[ \frac{x^2}{2} \right]_0^L = \gamma \frac{L^2}{2} \). So, \( M = \frac{\gamma L^2}{2} \).

03

Calculate Moment of Inertia (Left End)

Using the definition of moment of inertia \( I = \int x^2 dm \), substitute \( dm = \gamma x dx \): \[ I = \int_0^L x^2 (\gamma x) \, dx = \gamma \int_0^L x^3 \, dx. \] This integral simplifies to \( \gamma \left[ \frac{x^4}{4} \right]_0^L = \gamma \frac{L^4}{4} \).

04

Express Moment of Inertia in Terms of Mass and Length

Substitute \( \gamma = \frac{2M}{L^2} \) into the expression for \( I \): \[ I = \frac{2M}{L^2} \cdot \frac{L^4}{4} = \frac{1}{2} ML^2.\] This result compares similarly to a solid rod's inertia of \( \frac{1}{3} ML^2 \), indicating the effect of mass distribution.

05

Calculate Moment of Inertia (Right End)

Use parallel axis theorem for the right end, which says \( I' = I + Md^2 \), where \( d \) is the distance from the left end. Since the original moment of inertia relative to the left end is \( \frac{1}{2} ML^2 \), from the right end it's \( L \), so \( I' = \frac{1}{2} ML^2 + ML^2 = \frac{3}{2} ML^2.\)

06

Comparison between Moments of Inertia

In the first case, moment of inertia from the left end is \( \frac{1}{2} ML^2 \), while from the right end it is \( \frac{3}{2} ML^2 \). The difference is due to the mass distribution being greater towards the right end, increasing inertia from there.

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Key Concepts

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

Mass Distribution

In the context of physics, mass distribution refers to how mass is spread across an object. It influences the object's mechanical properties, such as its moment of inertia. In the given problem, the rod's mass distribution is non-uniform. It varies along its length according to the given mass per unit length: \( \frac{dm}{dx} = \gamma x \).
This equation indicates that the mass density increases linearly with the distance \( x \) from the left end of the rod. For practical purposes:

• Mass is greater at positions further from the starting point \( x=0 \).
• This non-uniform distribution affects how the rod rotates around different axes.

This characteristic is important for finding the rod's mass and its rotational attributes because these depend on how mass is spread. Integrating this distribution over the rod's entire length provides the total mass.

Integration

Integration is a crucial mathematical tool used to sum infinitesimally small quantities over a defined range. In this exercise, integration helps calculate the total mass of the rod. From the mass density equation \( \frac{dm}{dx} = \gamma x \), integration from 0 to \( L \) gives the total mass \( M \).
This can be visualized as accumulating all the small masses along the rod:

• Symbolically: \[ M = \int_0^L \gamma x \, dx \]
• Simplifies to: \( M = \frac{\gamma L^2}{2} \)

The application of integration here reveals that the total rod mass is directly related to \( \gamma \) and the square of its length. Integration therefore plays a vital role in formulating mass and inertia expressions from non-uniform mass distributions.

Parallel Axis Theorem

The Parallel Axis Theorem is a principle that enables the calculation of an object's moment of inertia about any axis, given its moment of inertia about a parallel axis through the object's center of mass. This theorem is particularly useful in part (c) of the problem.
Here is why the theorem matters:

• Calculates inertia in scenarios where the rotational axis is not through the center of mass.
• Formula: \( I' = I + Md^2 \), where \( d \) is the distance between the axes.

In this exercise,

• Original inertia from the left end is \( \frac{1}{2} ML^2 \)
• Using the theorem for the right end (distance \( L \)), it becomes: \[ I' = \frac{1}{2} ML^2 + ML^2 = \frac{3}{2} ML^2 \]

This increase reflects more mass being concentrated away from the calculated axis, enhancing its inertia.

Rod Mechanics

Rod mechanics involves studying how rods, specifically their structure and material, respond to forces and motion. This aspect of mechanics considers the distribution of mass along the rod, affecting how it turns or resists rotation. In this problem:

• The rod's unique mass distribution influences its moment of inertia significantly.
• As we see from inertia calculations, a non-uniform rod like this one behaves differently in rotational motion than a uniform rod.

This examination juxtaposes the physics of uniform rods, whose inertia about an end is generally \( \frac{1}{3} ML^2 \). The given rod, with more mass towards the far end, has greater inertia values (\( \frac{1}{2} ML^2 \) and \( \frac{3}{2} ML^2 \) from the left and right ends respectively). This difference highlights how the mechanics of rods play a role in real-world applications, from simple mechanical systems to complex engineering designs.