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Chapter 9: Problem 36

Small blocks, each with mass \(m\) , are clamped at the ends and at the center of a rod of length \(L\) and negligible mass. Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through (a) the center of the rod and (b) a point one-fourth of the length from one end.

Short Answer

Expert verified
(a) \(\frac{mL^2}{2}\); (b) \(\frac{13mL^2}{16}\).

Step by step solution

01

Understand the Problem

We are tasked with finding the moment of inertia of a system consisting of a rod with three equal small masses located at specific positions along the rod. The rod itself has negligible mass. The goal is to compute the moment of inertia about two specific axes.
02

Identify Key Concepts

The moment of inertia, denoted as \(I\), of a system about an axis is calculated by summing \(mr^2\) for each mass, where \(m\) is the mass and \(r\) is the distance from the axis. We will use this formula for our calculations.
03

Set Positions of Masses

The rod has three masses at positions: one at the left end (\(x=0\)), one in the center (\(x=\frac{L}{2}\)), and one at the right end (\(x=L\)). These positions will help calculate the distance for each mass from the specified axes.
04

Calculate Moment of Inertia Through Center

For an axis through the center of the rod at \(x=\frac{L}{2}\), calculate the distances of the masses: the end masses are at \(r=\frac{L}{2}\) and the central mass is at \(r=0\).The moment of inertia is given by:\[I_{center} = m\left(\frac{L}{2}\right)^2 + m(0)^2 + m\left(\frac{L}{2}\right)^2 = 2m\left(\frac{L}{2}\right)^2 = \frac{mL^2}{2}.\]
05

Calculate Moment of Inertia Through One-Fourth from One End

For an axis one-fourth of the length from one end, at \(x=\frac{L}{4}\), calculate the distances: left mass is at \(r=\frac{L}{4}\), center mass is at \(r=\frac{L}{4}\), and right mass is at \(r=\frac{3L}{4}\).The moment of inertia is:\[I_{one\text{-}fourth} = m\left(\frac{L}{4}\right)^2 + m\left(\frac{L}{4}\right)^2 + m\left(\frac{3L}{4}\right)^2 = 2m\left(\frac{L}{4}\right)^2 + m\left(\frac{3L}{4}\right)^2,\]which simplifies to \[I_{one\text{-}fourth} = \frac{mL^2}{8} + \frac{mL^2}{8} + \frac{9mL^2}{16} = \frac{13mL^2}{16}.\]

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Key Concepts

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

Rigid Body Systems
A rigid body system consists of objects where the relative distances between different parts do not change. In this problem, the system comprises a rod with three small masses attached at fixed points. One at each end and one in the center. The rod, due to negligible mass, primarily serves as a connector. The masses are clamped, meaning their positions are fixed during rotation.
This arrangement simplifies the moment of inertia calculation. In essence, the system acts like individual masses rotating about a common axis. Because the distances between the masses and the axis remain consistent, it provides a convenient setup for problem-solving.
  • Each mass acts individually, contributing to the overall moment of inertia.
  • Understanding that these fixed positions do not alter helps predict behavior during rotational movement.
  • The negligible mass of the rod implies that only the masses affect calculations, making it easier to apply the necessary formulas.
Rotational Dynamics
Rotational dynamics involves the study of objects in rotational motion and how they are affected by torques and other forces. A key component of this subject is the moment of inertia, which determines how difficult it is to change an object's rotational state. In our exercise, the system's moment of inertia must be computed for different axes.
The moment of inertia ( I ) is calculated by the sum of each mass multiplied by the square of its distance from the axis of rotation ( r^2 ). For the axis through the center of the rod:
  • Both end masses are equally distant from the axis, contributing equally to the moment of inertia.
  • The central mass does not contribute, as it is on the axis ( r=0 ).
For the second axis, located one-fourth length of the rod from one end:
  • The distances of each mass from this axis vary, leading to a more complex calculation.
  • Understanding these distances is crucial for determining individual contributions to the total inertia.
Physics Problem Solving
In physics problem-solving, it is vital to thoroughly understand the problem statement before proceeding. Here, knowing the positions of the masses and the axis locations was essential information.
Approaching the calculation step-by-step:
  • Identify what is given: masses, lengths, and axis positions.
  • Set up the problem: Understand the position of each mass relative to the specified axis.
  • Apply the correct formula: The formula for moment of inertia, I = sum of ( m r^2 ), where r is the distance from the axis.
Breaking down complex physics problems into smaller tasks reinforces conceptual understanding. Every step, from identifying axis locations to calculating the distances of masses, contributes to solving the problem correctly.

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Most popular questions from this chapter

When a toy car is rapidly scooted across the floor, it stores energy in a flywheel. The car has mass 0.180 \(\mathrm{kg}\) , and its flywheel has moment of inertia \(4.00 \times 10^{-5} \mathrm{kg} \cdot \mathrm{m}^{2} .\) The car is 15.0 \(\mathrm{cm}\) long. An advertisement claims that the car can travel at a scale speed of up to 700 \(\mathrm{km} / \mathrm{h}(440 \mathrm{mi} \mathrm{h}) .\) The scale speed is the speed of the toy car multiplied by the ratio of the length of an actual car to the length of the toy. Assume a length of 3.0 \(\mathrm{m}\) for a real car. (a) For a scale speed of 700 \(\mathrm{km} / \mathrm{h}\) , what is the actual translational speed of the car? (b) If all the kinetic energy that is initially in the flywheel is converted to the translational kinetic energy of the toy, how much energy is originally stored in the flywheel? (c) What initial angular velocity of the flywheel was needed to store the amount of energy calculated in part (b)?

(a) Derive an equation for the radial acceleration that includes \(v\) and \(\omega,\) but not \(r .\) (b) You are designing a merry-go-round for which a point on the rim will have a radial acceleration of 0.500 \(\mathrm{m} / \mathrm{s}^{2}\) when the tangential velocity of that point has magnitude 2.00 \(\mathrm{m} / \mathrm{s}\) . What angular velocity is required to achieve these values?

A thin, rectangular sheet of metal has mass \(M\) and sides of length \(a\) and \(b\) . Use the parallel-axis theorem to calculate the moment of inertia of the sheet for an axis that is perpendicular to the plane of the sheet and that passes through one comer of the sheet.

Electric Drill. According to the shop manual, when drilling a \(12.7-\mathrm{mm}\) -diameter hole in wood, plastic, or aluminum, a drill should have a speed of 1250 rev/min. For a 12.7 -mm-diameter drill bit turning at a constant 1250 rev \(/ \mathrm{mm}\) , find (a) the maximum linear speed of any part of the hit and (b) the maximum radial acceleration of any part of the bit.

A wheel is rotating about an axis that is in the \(z\) -direction. The angular velocity \(\omega_{z}\) is \(-6.00 \mathrm{rad} / \mathrm{s}\) at \(t=0,\) increases linearly with time, and is \(+8.00 \mathrm{m} / \mathrm{s}\) at \(t=7.00 \mathrm{s}\) . We have taken counterclockwise rotation to be positive. (a) Is the angular acceleration during this time interval positive or negative? (b) During what time interval is the speed of the wheel increasing? Decreasing? (c) What is the angular displacement of the wheel at \(t=7.00 \mathrm{s} ?\)
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