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About what axis will a uniform, balsa-wood sphere have the same moment of inertia as does a thin-walled, hollow, lead sphere of the same mass and radius, with the axis along a diameter?

Short Answer

Expert verified
The axis for the balsa-wood sphere should be located at a distance equal to the radius of the sphere divided by square root of 3 from its center to have the same moment of inertia as the hollow lead sphere.

Step by step solution

01

Calculate Moment of Inertia for Different Axes

Knowing the formulas for the moment of inertia, it is calculated for a solid sphere rotating about an axis through its center as \( \frac{2}{5} m r^{2} \) and for a hollow sphere, it is \( \frac{2}{3} m r^{2} \). Now considering an arbitrary axis at a distance x from the center of the solid sphere, we use parallel axis theorem to get the moment of inertia as \( I = \frac{2}{5} m r^{2} + m x^{2} \)
02

Equate the Two Moments

The two moments of inertia must be the same for the two spheres, so equate the two expressions derived in the previous step to get \( \frac{2}{5} m r^{2} + m x^{2} = \frac{2}{3} m r^{2} \). Simplifying, it results in \( x^{2} = \frac{1}{3} r^{2} \). Thus, the sphere would have the same moment of inertia when revolved around an axis that is sqrt(1/3) times the radius away from the center.
03

Express the result in terms of the radius

Take the square root of both sides of the equation, to get \( x = \frac{r}{\sqrt{3}} \). This means the axis for the balsa-wood sphere should be located at a distance equal to the radius of the sphere divided by square root of 3 from its center to obtain the same moment of inertia as the hollow lead sphere.

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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 critical concept in physics that helps in calculating the moment of inertia of a body about any axis, given its moment of inertia about a parallel axis that passes through the body's center of mass.

It's important to remember that the rotational movement of an object depends on its mass distribution relative to the rotation axis. The theorem states that if you know an object's moment of inertia about an axis through its center of mass, you can find the moment of inertia about any parallel axis by adding the product of the mass of the object and the square of the distance between the two axes.

Mathematically, the theorem is expressed as:
\[ I = I_{cm} + md^{2} \]
where \( I \) is the moment of inertia about the parallel axis, \( I_{cm} \) is the moment of inertia about the center of mass axis, \( m \) is the mass of the object, and \( d \) is the distance between the two axes. Properly understanding and applying this theorem is crucial in solving complex problems involving rotational dynamics.
Rotational Dynamics
Rotational dynamics is the study of the motion of objects that rotate about an axis. This field of physics is analogous to linear dynamics but focuses on rotational motion. The moment of inertia plays a role similar to mass in linear dynamics. It is a measure of an object's resistance to changes in its rotational motion, or angular acceleration.

Several factors affect an object's rotational motion, including torque, angular velocity, and the aforementioned moment of inertia. Just as force is involved in changing an object's linear motion, torque is the equivalent for rotational motion. The second law of rotational dynamics, which is akin to Newton's second law of motion, can be expressed as:
\[ \tau = I\frac{d\theta}{dt} \]
Where \( \tau \) stands for the net torque acting on the object, \( I \) is the moment of inertia, and \( \frac{d\theta}{dt} \) is the angular acceleration. This relationship is central to understanding how different forces and moments of inertia affect the rotation of an object.
Moment of Inertia Calculation
Calculating the moment of inertia (\( I \)) is fundamental in understanding how an object will behave when it is rotating. The moment of inertia is dependent on the object's mass distribution with respect to the axis of rotation. For simple geometric shapes, there are standard formulas to calculate the moment of inertia.

For example, for a solid sphere and a hollow sphere, the basic moment of inertia formulas about an axis through their centers are \( \frac{2}{5} mr^2 \) and \( \frac{2}{3} mr^2 \), respectively. These formulas assume uniform density and provide a starting point for more complex calculations involving different axes of rotation.

For complex or composite bodies, the total moment of inertia can be calculated by summing the moments for each part, considering the mass distribution and geometry. The calculations can also be integrated for bodies with non-uniform density or irregular shapes to achieve a more precise moment of inertia. By mastering these calculations, students can tackle a wide array of problems in rotational dynamics.

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Most popular questions from this chapter

Spin cycles of washing machines remove water from clothes by producing a large radial acceleration at the rim of the cylindrical tub that holds the water and clothes. Estimate the diameter of the tub in a typical home washing machine. (a) What is the rotation rate, in rev/min, of the tub during the spin cycle if the radial acceleration of points on the tub wall is \(3 g ?\) (b) At this rotation rate, what is the tangential speed in \(\mathrm{m} / \mathrm{s}\) of a point on the tub wall?

The Crab Nebula is a cloud of glowing gas about 10 light-years across, located about 6500 light-years from the earth (Fig. P9.86). It is the remnant of a star that underwent a supernova \(e x\) plosion, seen on earth in 1054 A.D. Energy is released by the Crab Nebula at a rate of about \(5 \times 10^{31} \mathrm{~W},\) about \(10^{5}\) times the rate at which the sun radiates energy. The Crab Nebula obtains its energy from the rotational kinetic energy of a rapidly spinning neutron star at its center. This object rotates once every \(0.0331 \mathrm{~s},\) and this period is increasing by \(4.22 \times 10^{-13} \mathrm{~s}\) for each second of time that elapses. (a) If the rate at which energy is lost by the neutron star is equal to the rate at which energy is released by the nebula, find the moment of inertia of the neutron star. (b) Theories of supernovae predict that the neutron star in the Crab Nebula has a mass about 1.4 times that of the sun. Modeling the neutron star as a solid uniform sphere, calculate its radius in kilometers. (c) What is the linear speed of a point on the equator of the neutron star? Compare to the speed of light. (d) Assume that the neutron star is uniform and calculate its density. Compare to the density of ordinary rock \(\left(3000 \mathrm{~kg} / \mathrm{m}^{3}\right)\) and to the density of an atomic nucleus (about \(\left.10^{17} \mathrm{~kg} / \mathrm{m}^{3}\right) .\) Justify the statement that a neutron star is essentially a large atomic nucleus.

A hollow spherical shell has mass \(8.20 \mathrm{~kg}\) and radius \(0.220 \mathrm{~m} .\) It is initially at rest and then rotates about a stationary axis that lies along a diameter with a constant acceleration of \(0.890 \mathrm{rad} / \mathrm{s}^{2}\). What is the kinetic energy of the shell after it has turned through 6.00 rev?

A high-speed flywheel in a motor is spinning at 500 rpm when a power failure suddenly occurs. The flywheel has mass \(40.0 \mathrm{~kg}\) and diameter \(75.0 \mathrm{~cm}\). The power is off for \(30.0 \mathrm{~s}\), and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 200 complete revolutions. (a) At what rate is the flywheel spinning when the power comes back on? (b) How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

A wheel of diameter \(40.0 \mathrm{~cm}\) starts from rest and rotates with a constant angular acceleration of \(3.00 \mathrm{rad} / \mathrm{s}^{2}\). Compute the radial acceleration of a point on the rim for the instant the wheel completes its second revolution from the relationship (a) \(a_{\mathrm{rad}}=\omega^{2} r\) and (b) \(a_{\mathrm{rad}}=v^{2} / r\)

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