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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

A uniform disk has radius \(R_{0}\) and mass \(M_{0}\). Its moment of inertia for an axis perpendicular to the plane of the disk at the disk's center is \(\frac{1}{2} M_{0} R_{0}^{2}\). You have been asked to halve the disk's moment of inertia by cutting out a circular piece at the center of the disk. In terms of \(R_{0}\), what should be the radius of the circular piece that you remove?

A wheel rotates from rest with constant angular acceleration. If it rotates through 8.00 revolutions in the first \(2.50 \mathrm{~s}\), how many more revolutions will it rotate through in the next 5.00 s?

Two metal disks, one with radius \(R_{1}=2.50 \mathrm{~cm}\) and mass \(M_{1}=0.80 \mathrm{~kg}\) and the other with radius \(R_{2}=5.00 \mathrm{~cm}\) and mass \(M_{2}=1.60 \mathrm{~kg},\) are welded together and mounted on a frictionless axis through their common center (Fig. \(\mathbf{P 9 . 7 9}\) ). (a) What is the total moment of inertia of the two disks? (b) A light string is wrapped around the edge of the smaller disk, and a \(1.50 \mathrm{~kg}\) block is suspended from the free end of the string. If the block is released from rest at a distance of \(2.00 \mathrm{~m}\) above the floor, what is its speed just before it strikes the floor? (c) Repeat part (b), this time with the string wrapped around the edge of the larger disk. In which case is the final speed of the block greater? Explain.

An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev \(/\) min to 200 rev \(/ \min\) in 4.00 s. (a) Find the angular acceleration in rev/s \(^{2}\) and the number of revolutions made by the motor in the 4.00 s interval. (b) How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in part (a)?

A child is pushing a merry-go-round. The angle through which the merry-go- round has turned varies with time according to \(\theta(t)=\gamma t+\beta t^{3}, \quad\) where \(\quad \gamma=0.400 \mathrm{rad} / \mathrm{s} \quad\) and \(\quad \beta=0.0120 \mathrm{rad} / \mathrm{s}^{3}\) (a) Calculate the angular velocity of the merry-go-round as a function of time. (b) What is the initial value of the angular velocity? (c) Calculate the instantaneous value of the angular velocity \(\omega_{z}\) at \(t=5.00 \mathrm{~s}\) and the average angular velocity \(\omega_{\mathrm{av}-z}\) for the time interval \(t=0\) to \(t=5.00 \mathrm{~s}\) Show that \(\omega_{\mathrm{av}-z}\) is \(n o t\) equal to the average of the instantaneous angular velocities at \(t=0\) and \(t=5.00 \mathrm{~s},\) and explain.

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