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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) \(\frac{5mL^2}{8}\).

Step by step solution

01

Moment of Inertia Concept

The moment of inertia for a system of particles is given by \(I = \sum m_i r_i^2\), where \(m_i\) is the mass of each particle and \(r_i\) is the distance from the axis of rotation to the particle.
02

Calculate Moment of Inertia for Center Axis

For the axis passing through the center at \(L/2\), the distribution is symmetrical. The central block at the center has \(r = 0\) and thus contributes \(0\) to the moment of inertia. Each block at the ends of the rod is a distance \(L/2\) from this axis. So the moment of inertia is, \(I = 2m (L/2)^2 = 2m \frac{L^2}{4} = \frac{mL^2}{2}\).
03

Determine Distance for Off-Center Axis

For an axis one-fourth down the length of the rod \((L/4)\), compute each block's distance: \(L/4\) and \(3L/4\) for one side, and \(3L/4\) and \(L/4\) for the opposite side.
04

Calculate Moment of Inertia for Off-Center Axis

Using the calculated distances: \(I = m (L/4)^2 + m (3L/4)^2 + m (3L/4)^2 + m (L/4)^2 = 2m (L/4)^2 + 2m (3L/4)^2 = 2m \frac{L^2}{16} + 2m \frac{9L^2}{16} = 2m \left( \frac{1L^2}{16} + \frac{9L^2}{16} \right) = 2m *\frac{10L^2}{16} = \frac{5mL^2}{8}\).

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Key Concepts

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

Rotational Dynamics
Rotational dynamics is a fascinating area of physics that deals with the movement of bodies in rotation. Just as forces affect the motion of a particle moving in a straight line, torques affect objects that are rotating. One key aspect here is the concept of "moment of inertia," which is akin to mass in linear motion but for rotational motion.
  • The moment of inertia determines how much torque is needed for a desired angular acceleration.
  • It is a rotational analogue to mass—larger moments of inertia require more effort to change the rotational state.
  • In the problem given, small masses fixed to a rod demonstrate how different positions relative to the axis of rotation impact the overall moment of inertia.
Understanding rotational dynamics allows us to analyze how body structures, like the distribution of mass, influence dynamic behavior when rotating.
Physics Problem Solving
Physics problem solving is all about breaking down a problem into manageable parts and applying key principles. For rotational motion, this usually involves the moments of inertia and understanding the effects of changing axes.
To tackle such problems effectively:
  • Identify the physical quantities involved and their relationships.
  • Choose an appropriate axis of rotation, as different orientations lead to different solutions.
  • Be systematic in calculations—carefully compute distances and arrangements relative to chosen axes.
  • Ensure units are consistent to avoid miscalculation.
By carefully applying these steps, one can solve complex physics problems with confidence.
Particle Systems
In physics, a particle system refers to a collection of masses treated as discreet entities, while ignoring internal forces between them. This simplification allows us to focus on external forces and torques, which is critical in rotational dynamics.
  • Each mass in the system can contribute individually to the overall dynamics.
  • Understanding these contributions helps compute important properties like the total moment of inertia.
  • In the exercise, the system involves masses that are strategically placed at the rod's ends and center, mimicking a practical setup.
A particle system's analysis is useful in simplifying complex dynamics into more digestible problems by focusing on major elements affecting the motion.
Axis of Rotation
The axis of rotation is a crucial concept in rotational dynamics. It is an imaginary line through which a body rotates. The position of this axis can greatly influence the calculated moment of inertia.
  • The choice of axis is fundamental in determining how rotation dynamics are calculated.
  • An axis passing through the center of mass typically minimizes the moment of inertia.
  • Different axis positions, such as off-center ones, result in different inertia values, as seen in the given solution where two scenarios are explored.
Understanding the impact of an axis of rotation enables more accurate modeling of rotational behavior, which is essential in both engineering applications and theoretical physics.

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Most popular questions from this chapter

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 (a) perpendicular to the bar through its center; (b) perpendicular to the bar through one of the balls; (c) parallel to the bar through both balls; and (d) parallel to the bar and 0.500 m from it.

An advertisement claims that a centrifuge takes up only \(0.127 \mathrm{~m}\) of bench space but can produce a radial acceleration of \(3000 \mathrm{~g}\) at 5000 rev \(/ \mathrm{min}\). Calculate the required radius of the centrifuge. Is the claim realistic?

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. (a) What is the moment of inertia of this rod for an axis at its center, perpendicular to the rod? (b) One of your engineers has proposed to reduce the moment of inertia by bending the rod at its center into a V-shape, with a 60.0\(^\circ\) angle at its vertex. What would be the moment of inertia of this bent rod about an axis perpendicular to the plane of the V at its vertex?

An electric turntable \(0.750 \mathrm{~m}\) in diameter is rotating about a fixed axis with an initial angular velocity of \(0.250 \mathrm{rev} / \mathrm{s}\) and a constant angular acceleration of \(0.900 \mathrm{rev} / \mathrm{s}^{2}\). (a) Compute the angular velocity of the turntable after \(0.200 \mathrm{~s}\). (b) Through how many revolutions has the turntable spun in this time interval? (c) What is the tangential speed of a point on the rim of the turntable at \(t=0.200 \mathrm{~s} ?\) (d) What is the magnitude of the resultant acceleration of a point on the rim at \(t=0.200 \mathrm{~s} ?\)

The motor of a table saw is rotating at 3450 rev/min. A pulley attached to the motor shaft drives a second pulley of half the diameter by means of a V-belt. A circular saw blade of diameter 0.208 m is mounted on the same rotating shaft as the second pulley. (a) The operator is careless and the blade catches and throws back a small piece of wood. This piece of wood moves with linear speed equal to the tangential speed of the rim of the blade. What is this speed? (b) Calculate the radial acceleration of points on the outer edge of the blade to see why sawdust doesn’t stick to its teeth.

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