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Chapter 9: Problem 30

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鈥攍arger 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鈥攃arefully 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 thin, uniform rod is bent into a square of side length \(a\). If the total mass is \(M\), find the moment of inertia about an axis through the center and perpendicular to the plane of the square. (\(Hint\): Use the parallel-axis theorem.)

A wheel is rotating about an axis that is in the \(z\)-direction.The angular velocity \(\omega_z\) is \(-\)6.00 rad/s at \(t =\) 0, increases linearly with time, and is \(+\)4.00 rad/s at \(t =\) 7.00 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 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鈥檛 stick to its teeth.

You need to design an industrial turntable that is 60.0 cm in diameter and has a kinetic energy of 0.250 J when turning at 45.0 rpm 1rev/min2. (a) What must be the moment of inertia of the turntable about the rotation axis? (b) If your workshop makes this turntable in the shape of a uniform solid disk, what must be its mass?

A thin uniform rod of mass \(M\) and length \(L\) is bent at its center so that the two segments are now perpendicular to each other. Find its moment of inertia about an axis perpendicular to its plane and passing through (a) the point where the two segments meet and (b) the midpoint of the line connecting its two ends.
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