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Calculate the rotational inertia of a meter stick, with mass \(0.56 \mathrm{~kg}\), about an axis perpendicular to the stick and located at the \(20 \mathrm{~cm}\) mark. (Treat the stick as a thin rod.)

Short Answer

Expert verified
The rotational inertia is approximately 0.0974 kg·m².

Step by step solution

01

Identify the Formula for Rotational Inertia

For a thin rod rotating about an axis not at its center, we use the parallel axis theorem. The formula for the rotational inertia, \(I\), is \(I = I_{CM} + md^2\), where \(I_{CM}\) is the inertia about the center of mass, \(m\) is the mass of the rod, and \(d\) is the distance from the center of mass to the new axis.
02

Calculate the Inertia About the Center of Mass

For a uniform rod of length \(L\), the moment of inertia about its center is \(I_{CM} = \frac{1}{12}mL^2\). Here, \(m = 0.56\, \text{kg}\) and \(L = 1\, \text{m}\). Plugging in these values, \(I_{CM} = \frac{1}{12} \times 0.56 \times 1^2 = 0.0467\, \text{kgm}^2\).
03

Find the Distance from the Center of Mass to the New Axis

The center of mass of the meter stick is at the 50 \(\text{cm}\) mark. The new axis is at the 20 \(\text{cm}\) mark. Therefore, the distance \(d\) from the center of mass to the new axis is \(50\, \text{cm} - 20\, \text{cm} = 30\, \text{cm} = 0.3\, \text{m}\).
04

Apply the Parallel Axis Theorem

Using the formula \(I = I_{CM} + md^2\), substitute \(I_{CM} = 0.0467\, \text{kgm}^2\), \(m = 0.56\, \text{kg}\), and \(d = 0.3\, \text{m}\). Calculate \(I = 0.0467 + 0.56 \times (0.3)^2 = 0.0974\, \text{kgm}^2\).

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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 handy tool used to find the moment of inertia of an object around an axis that is not through its center of mass. It's especially useful when dealing with rotational motions where the axis is shifted but remains parallel to the original axis through the center of mass. The theorem states: - If you know the moment of inertia of an object about an axis through its center of mass (denoted as \( I_{CM} \)), you can calculate the moment of inertia about any parallel axis by adding \( md^2 \) to \( I_{CM} \).- Here, \( m \) represents the mass of the object, and \( d \) is the perpendicular distance between the two axes. So, essentially, the formula becomes: \[ I = I_{CM} + md^2 \]Understanding this theorem is crucial when analyzing objects like rods, discs, or any rigid bodies not rotating around their centers. By applying this theorem, we can more easily solve complex physics problems involving off-center rotations.
Moment of Inertia
The Moment of Inertia, often symbolized as \( I \), is a fundamental concept in rotational dynamics that measures how much torque is needed for a desired angular acceleration about a rotational axis. Think of it as a rotational equivalent of mass in linear motion. For different shapes and axis positions, the calculation of the moment of inertia varies: - Rods: When revolving around their end, use \( \frac{1}{3}mL^2 \) (for ends) or \( \frac{1}{12}mL^2 \) (for centers).- Discs and Wheels: These use \( \frac{1}{2}mr^2 \), where \( r \) is the radius. In the case of a uniform rod, if rotating about the center, the formula is: \[ I_{CM} = \frac{1}{12}mL^2 \]Understanding how to calculate the moment of inertia helps in predicting how different shapes will react under rotational forces, guiding design and analysis in engineering and physics.
Center of Mass
The Center of Mass is the unique point in an object or system of particles where all the mass can be considered to be concentrated when analyzing mechanical systems. For symmetric objects, the center of mass is often located at the geometric center. When dealing with rods or any linear objects, the center of mass is halfway along its length. For example, in a uniform meter stick, the center of mass would be at the 50 cm mark. Key points about center of mass: - It helps in simplifying calculations for dynamics and statics by treating the object as if all mass is concentrated there. - It's crucial in applying the Parallel Axis Theorem, as calculations often revolve around the center of mass. - In systems or structures, the stability and balance are deeply connected to the position of this point. By understanding the center of mass, you simplify many complex physics problems, making them manageable by breaking them into simpler parts.

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Most popular questions from this chapter

Between 1911 and 1990 , the top of the leaning bell tower at Pisa, Italy, moved toward the south at an average rate of \(1.2 \mathrm{~mm} / \mathrm{y}\). The tower is \(55 \mathrm{~m}\) tall. In radians per second, what is the average angular speed of the tower's top about its base?

Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of \(200 \pi \mathrm{rad} / \mathrm{s}\). One such flywheel is a solid, uniform cylinder with a mass of \(500 \mathrm{~kg}\) and a radius of \(1.0 \mathrm{~m} .\) (a) What is the kinetic energy of the flywheel after charging? (b) If the truck uses an average power of \(8.0 \mathrm{~kW}\), for how many minutes can it operate between chargings?

A disk, with a radius of \(0.25 \mathrm{~m}\), is to be rotated like a merry-go- round through 800 rad, starting from rest, gaining angular speed at the constant rate \(\alpha_{1}\) through the first \(400 \mathrm{rad}\) and then losing angular speed at the constant rate \(-\alpha_{1}\) until it is again at rest. The magnitude of the centripetal acceleration of any portion of the disk is not to exceed \(400 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the least time required for the rotation? (b) What is the corresponding value of \(\alpha_{1} ?\)

Two uniform solid spheres have the same mass of \(1.65 \mathrm{~kg}\), but one has a radius of \(0.226 \mathrm{~m}\) and the other has a radius of \(0.854 \mathrm{~m}\). Each can rotate about an axis through its center. (a) What is the magnitude \(\tau\) of the torque required to bring the smaller sphere from rest to an angular speed of \(317 \mathrm{rad} / \mathrm{s}\) in \(15.5 \mathrm{~s} ?(\mathrm{~b})\) What is the magnitude \(F\) of the force that must be applied tangentially at the sphere's equator to give that torque? What are the corresponding values of (c) \(\tau\) and (d) \(F\) for the larger sphere?

During the launch from a board, a diver's angular speed about her center of mass changes from zero to \(6.20 \mathrm{rad} / \mathrm{s}\) in \(220 \mathrm{~ms}\). Her rotational inertia about her center of mass is \(12.0\) \(\mathrm{kg} \cdot \mathrm{m}^{2}\). During the launch, what are the magnitudes of (a) her average angular acceleration and (b) the average external torque on her from the board?

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