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Chapter 10: Problem 57

(II) Two uniform solid spheres of mass \(M\) and radius \(r_{0}\) are connected by a thin (massless) rod of length \(r_{0}\) so that the centers are 3\(r_{0}\) apart. (a) Determine the moment of inertia of this system about an axis perpendicular to the rod at its center. \((b)\) What would be the percentage error if the masses of each sphere were assumed to be concentrated at their centers and a very simple calculation made?

Short Answer

Expert verified
The moment of inertia is approximately \(\frac{371}{120}Mr_0^2\).

Step by step solution

01

Understanding the System

We have two uniform solid spheres, each of mass \(M\) and radius \(r_0\), connected by a rod of length \(r_0\). The distance between their centers is \(3r_0\). We need to find the moment of inertia about an axis perpendicular to the rod's center.
02

Moment of Inertia of a Solid Sphere

The moment of inertia (I) of a single uniform solid sphere about an axis through its center is \(\frac{2}{5}Mr_0^2\). Since the axis for the system is not through the centers of the spheres, we will use the parallel axis theorem.
03

Using the Parallel Axis Theorem

The parallel axis theorem states: \(I = I_{CM} + Md^2\), where \(I_{CM}\) is the moment of inertia about the center of mass and \(d\) is the distance from the center of mass to the new axis. For each sphere, \(d = \frac{3r_0}{2}\). So: \[I_{ ext{each sphere}} = \frac{2}{5}Mr_0^2 + M\left(\frac{3r_0}{2}\right)^2 = \frac{2}{5}Mr_0^2 + \frac{9}{4}Mr_0^2 = \left(\frac{2}{5} + \frac{9}{4}\right)Mr_0^2\].

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Key Concepts

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

Solid Sphere
A solid sphere is a three-dimensional object where all the points on its outer surface are equidistant from the center. In physics, the solid sphere is often used to illustrate concepts of rotational dynamics, especially when exploring the moment of inertia.
The moment of inertia for a solid sphere is a crucial concept.
  • The moment of inertia reflects how mass is distributed relative to the rotational axis.
  • For a uniform solid sphere, it is calculated around an axis through its center using the formula: \( \frac{2}{5}Mr_0^2 \), where \( M \) is the mass and \( r_0 \) is the radius of the sphere.
Understanding the distribution of mass helps in predicting the sphere's resistance to changes in its rotational motion. In real-world applications, this principle is vital in fields ranging from engineering to sports, as it impacts the sphere's dynamics when in motion.
Parallel Axis Theorem
The parallel axis theorem is a valuable tool in physics that allows us to find the moment of inertia of a body about any axis parallel to an axis through the center of mass.
Knowing how to apply this theorem can greatly simplify complex rotational dynamics problems.
  • The theorem states that the moment of inertia about any axis parallel to one through the center of mass is \( I = I_{\text{CM}} + Md^2 \).
  • Here, \( I_{\text{CM}} \) is the moment of inertia through the center of mass and \( d \) is the distance between the two axes.
In the exercise involving the two connected spheres, this theorem was used because the rotational axis is not through the center of the spheres but at a midpoint on the rod. By doing so, it allows the calculation of the system's overall moment of inertia, factoring in the added distance from their centers, thus providing an accurate understanding of how the system rotates.
Center of Mass
The center of mass is a point representing the average position of all the mass in a system. It plays a fundamental role in analyzing motion and balance in physics.
For uniform objects, like the spheres in the exercise, the center of mass is at the geometric center.
  • This simplifies calculations since the center of mass serves as the pivot point for understanding motion.
  • For a system of particles or objects, the collective center can be determined by considering the weighted contributions of all involved masses.
In the given example of two spheres and a rod, understanding the center of mass is crucial while applying the parallel axis theorem, as it marks where the rotation would naturally balance if we were simplifying calculations or assessing dynamics without external forces acting differently on different parts of the object or system.

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Most popular questions from this chapter

Four equal masses \(M\) are spaced at equal intervals, \(\ell\) along a horizontal straight rod whose mass can be ignored. The system is to be rotated about a vertical axis passing through the mass at the left end of the rod and perpendicular to it. \((a)\) What is the moment of inertia of the system about this axis? \((b)\) What minimum force, applied to the farthest mass, will impart an angular acceleration \(\alpha ?\) (c) What is the direction of this force?

(II) A centrifuge rotor rotating at \(10,300\) rpm is shut off and is eventually brought uniformly to rest by a frictional torque of 1.20 \(\mathrm{m} \cdot \mathrm{N}\) . If the mass of the rotor is 3.80 \(\mathrm{kg}\) and it can be approximated as a solid cylinder of radius \(0.0710 \mathrm{m},\) through how many revolutions will the rotor turn before coming to rest, and how long will it take?

A potter is shaping a bowl on a potter's wheel rotating at constant angular speed (Fig. \(10-51\) ). The friction force between her hands and the clay is \(1.5 \mathrm{~N}\) total. \((a)\) How large is her torque on the wheel, if the diameter of the bowl is \(12 \mathrm{~cm} ?\) (b) How long would it take for the potter's wheel to stop if the only torque acting on it is due to the potter's hand? The initial angular velocity of the wheel is \(1.6 \mathrm{rev} / \mathrm{s},\) and the moment of inertia of the wheel and the bowl is \(0.11 \mathrm{~kg} \cdot \mathrm{m}^{2}\)

A turntable of radius \(R_{1}\) is turned by a circular rubber roller of radius \(R_{2}\) in contact with it at their outer edges. What is the ratio of their angular velocities, \(\omega_{1} / \omega_{2} ?\)

A narrow but solid spool of thread has radius \(R\) and mass \(M\). If you pull up on the thread so that the \(\mathrm{CM}\) of the spool remains suspended in the air at the same place as it unwinds, (a) what force must you exert on the thread? (b) How much work have you done by the time the spool turns with angular velocity \(\omega ?\)
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