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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 \(\frac{9}{5}Mr_0^2\). The percentage error is 150\%.

Step by step solution

01

Understand the Problem Setup

We need to calculate the moment of inertia of a system consisting of two uniform solid spheres connected by a rod. The total distance between the centers of the spheres is \(3r_0\), with the axis of rotation perpendicular to the rod at its center.
02

Calculate Moment of Inertia for One Sphere

The moment of inertia of a solid sphere about its own center is given by \(\frac{2}{5}MR^2\), where \(R\) is its radius. For our spheres, this value is \(I_{sphere} = \frac{2}{5}Mr_0^2\).
03

Apply Parallel Axis Theorem

Since the rotation axis is not about the center of each sphere, use the Parallel Axis Theorem. The theorem states \(I = I_c + Md^2\), where \(d\) is the distance from the center of the sphere to the new axis. Here, each sphere is \(\frac{3r_0}{2} - r_0 = \frac{r_0}{2}\) away from the center axis. Hence, \(I_{sphere} = \frac{2}{5}Mr_0^2 + M\left(\frac{r_0}{2}\right)^2\).
04

Calculate Moment of Inertia for Both Spheres

Since there are two spheres and they are identical, the total moment of inertia \(I_{total}\) is twice the moment of inertia of one sphere: \[I_{total} = 2 \times \left(\frac{2}{5}Mr_0^2 + M\left(\frac{r_0}{2}\right)^2\right)\].
05

Simplify the Expression

Simplifying the expression: \[I_{total} = 2 \times \left(\frac{2}{5}Mr_0^2 + \frac{M}{4}r_0^2\right) = 2 \times \left(\frac{18}{20}Mr_0^2\right) = \frac{36}{20}Mr_0^2 = \frac{9}{5}Mr_0^2\].
06

Simplified Calculation Assumption

Assuming the mass of each sphere is concentrated at its center, the moment of inertia for each would be \(I = M(\frac{3r_0}{2})^2 = \frac{9}{4}Mr_0^2\). The total for both spheres is \(2 \times \frac{9}{4}Mr_0^2 = \frac{9}{2}Mr_0^2\).
07

Calculate the Percentage Error

Calculate the error as \[\text{Error} = \left|\frac{\text{Approximated Moment of Inertia} - \text{Exact Moment of Inertia}}{\text{Exact Moment of Inertia}} \right| \times 100\%\]. Substituting, we get \[\text{Error} = \left|\frac{\frac{9}{2}Mr_0^2 - \frac{9}{5}Mr_0^2}{\frac{9}{5}Mr_0^2} \right| \times 100\% = 150\%\].

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Key Concepts

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

Parallel Axis Theorem
The parallel axis theorem is a useful tool in calculating the moment of inertia when an object is not rotating around its center of mass. The theorem states that the moment of inertia of an object about any axis parallel to an axis through its center of mass can be found using the formula:\[ I = I_c + Md^2 \]where:
  • \( I \) is the moment of inertia about the new axis,
  • \( I_c \) is the moment of inertia about the center of mass axis,
  • \( M \) is the mass of the object, and
  • \( d \) is the perpendicular distance between the two axes.
Applying this theorem to a system involving solid spheres and rods is important because it allows us to correctly calculate the rotational motion properties about non-standard axes. In our exercise, each sphere's axis through its center does not coincide with the axis of rotation, which requires adjustments using this theorem. Thus, for each sphere, we add \( M\left(\frac{r_0}{2}\right)^2 \) to its inertia calculated at its center, reflecting shifting the axis halfway down the rod connecting the spheres.
Solid Sphere
A solid sphere is a three-dimensional object where every point on its surface is equidistant from the center. In physics, it’s essential to understand the behavior of solid spheres concerning rotation to solve problems involving their motion.The moment of inertia for a solid sphere about an axis through its center is calculated using the formula:\[ I_{sphere} = \frac{2}{5}MR^2 \]where:
  • \( M \) is the mass of the sphere, and
  • \( R \) is the radius of the sphere.
In the exercise, each sphere has a moment of inertia of \( \frac{2}{5}Mr_0^2 \) when rotating about its own center. However, this base calculation changes when a non-central axis is involved, requiring the use of the parallel axis theorem. Understanding the basic inertial properties of a solid sphere aids in mastering more complex rotational dynamics.
Percentage Error
Percentage error is an important concept in experimental physics and problem-solving that helps in assessing the deviation of an approximated or measured value from the actual value. It gives us an idea of how accurate or precise our calculations are in relation to the true values.The formula for calculating percentage error can be expressed as:\[\text{Percentage Error} = \left|\frac{\text{Approximated Value} - \text{Exact Value}}{\text{Exact Value}}\right| \times 100\%\]In applying this concept to our exercise, we aimed to determine the difference between the simplified estimation where the sphere's masses are concentrated at their centers, and the more accurate measurement that accounts for the actual distribution of mass. This resulted in a significant percentage error of 150%, highlighting the degree to which simplifications can lead to inaccuracies. Understanding this can help us decide when it is appropriate to use approximations and when it's crucial to perform a detailed calculation.

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Most popular questions from this chapter

The \(1100-\mathrm{kg}\) mass of a car includes four tires, each of mass (including wheels) \(35 \mathrm{~kg}\) and diameter \(0.80 \mathrm{~m}\). Assume each tire and wheel combination acts as a solid cylinder. Determine \((a)\) the total kinetic energy of the car when traveling \(95 \mathrm{~km} / \mathrm{h}\) and \((b)\) the fraction of the kinetic energy in the tires and wheels. ( \(c\) ) If the car is initially at rest and is then pulled by a tow truck with a force of \(1500 \mathrm{~N}\), what is the acceleration of the car? Ignore frictional losses. \((d)\) What percent error would you make in part \((c)\) if you ignored the rotational inertia of the tires and wheels?

A large spool of rope rolls on the ground with the end of the rope lying on the top edge of the spool. A person grabs the end of the rope and walks a distance \(\ell\) , holding onto it, Fig. \(62 .\) The spool rolls behind the person without slipping. What length of rope unwinds from the spool? How far does the spool's center of mass move?

A thin \(7.0-\mathrm{kg}\) wheel of radius \(32 \mathrm{~cm}\) is weighted to one side by a \(1.50-\mathrm{kg}\) weight, small in size, placed \(22 \mathrm{~cm}\) from the center of the wheel. Calculate \((a)\) the position of the center of mass of the weighted wheel and (b) the moment of inertia about an axis through its \(\mathrm{CM},\) perpendicular to its face.

The density (mass per unit length) of a thin rod of length \(\ell\) increases uniformly from \(\lambda_{0}\) at one end to \(3 \lambda_{0}\) at the other end. Determine the moment of inertia about an axis perpendicular to the rod through its geometric center.

(II) A sphere of radius \(r_{0}=24.5 \mathrm{cm}\) and mass \(m=1.20 \mathrm{kg}\) starts from rest and rolls without slipping down a \(30.0^{\circ}\) incline that is 10.0 \(\mathrm{m}\) long. (a) Calculate its translational and rotational speeds when it reaches the bottom. (b) What is the ratio of translational to rotational kinetic energy at the bottom? Avoid putting in numbers until the end so you can answer: \((c)\) do your answers in \((a)\) and \((b)\) depend on the radius of the sphere or its mass?

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