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\(\bullet\) Calculate the angular momentum and kinetic energy of a solid uniform sphere with a radius of 0.120 \(\mathrm{m}\) and a mass of 14.0 \(\mathrm{kg}\) if it is rotating at 6.00 \(\mathrm{rad} / \mathrm{s}\) about an axis through its center.

Short Answer

Expert verified
The angular momentum is \(4.83264 \, \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-1}\) and the kinetic energy is \(14.49792 \, \text{J}\).

Step by step solution

01

Determine Moment of Inertia

For a solid uniform sphere rotating about an axis through its center, the moment of inertia \(I\) is given by the formula \(I = \frac{2}{5}mR^2\), where \(m\) is the mass and \(R\) is the radius. Substituting the given values, \(m = 14.0 \ kg\) and \(R = 0.120 \ m\), we can calculate \(I\) as follows: \[I = \frac{2}{5} \times 14.0 \times (0.120)^2 = 0.80544 \, \text{kg} \cdot \text{m}^2\].
02

Calculate Angular Momentum

The angular momentum \(L\) of a rotating object can be found using the formula \(L = I \cdot \omega\), where \(\omega\) is the angular velocity. Given \(\omega = 6.00 \ \text{rad/s}\) and using the moment of inertia from Step 1, we calculate: \[L = 0.80544 \cdot 6.00 = 4.83264 \, \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-1}\].
03

Calculate Kinetic Energy

The rotational kinetic energy \(KE\) of the sphere can be calculated with the formula \(KE = \frac{1}{2}I \omega^2\). Using the moment of inertia \(I\) from Step 1 and \(\omega = 6.00 \ \text{rad/s}\), we find:\[KE = \frac{1}{2} \times 0.80544 \times (6.00)^2 = 14.49792 \, \text{J}\].

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Key Concepts

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

Moment of Inertia
Moment of Inertia is a crucial concept in understanding rotational dynamics. It is essentially the rotational counterpart of mass in linear motion. Moment of Inertia, often denoted by the symbol \(I\), quantifies how difficult it is to change the rotational speed of an object. This property depends on two primary factors:

  • Mass of the object \(m\) - more mass means a higher moment of inertia.
  • Distribution of mass relative to the axis of rotation - the further the mass is from the axis, the higher the moment of inertia.
This can be understood by imagining spinning a dumbbell around different axes. When mass is further out, it takes more effort to spin. In mathematical terms, the Moment of Inertia for various solids can be calculated using specific formulas. For example, for a solid uniform sphere, the formula is \(I = \frac{2}{5} mR^2\), where \(m\) stands for mass and \(R\) for radius. This formula reflects how both the mass and the radius affect an object's rotational inertia. Adjusting either attribute will change how the object reacts to rotational force.
Rotational Kinetic Energy
Rotational Kinetic Energy is the energy an object possesses due to its rotation around an axis. It uses the principles of kinetic energy, which in linear motion equates to \(\frac{1}{2}mv^2\), and adapts them for rotational motion. For objects in rotational motion, the formula becomes:

\[KE = \frac{1}{2} I \omega^2\]

Here, \(KE\) is the kinetic energy, \(I\) is the moment of inertia, and \(\omega\) (omega) symbolizes the angular velocity. Angular velocity is measured in radians per second and signifies how fast the object is spinning.

Consider a solid sphere spinning in place; its rotational kinetic energy depends on how much mass it has, how this mass is distributed relative to its spinning axis (moment of inertia), and how fast it spins (angular velocity). This relationship illustrates that both the distribution of mass and the speed of rotation are significant in determining how much kinetic energy the object holds. Understanding this helps in analyzing systems like spinning wheels or planets moving in space, providing insights into the forces and energies involved in their movements.
Rotational Motion
Rotational Motion refers to the motion of an object around a central axis. Unlike linear motion, where paths are straight, rotational motion follows circular paths, characterized by angles and rotations. Key parameters that define rotational motion include:

  • Angular Displacement: The angle rotated by an object in a given direction.
  • Angular Velocity (\(\omega\)): The rate at which an object rotates, measured in radians per second.
  • Angular Acceleration: Describes how quickly the rotational speed changes.
This type of motion is essential for understanding phenomena where rotation is involved, such as wheels, gears, planets, and even atoms in certain contexts. Using angular instead of linear measures allows for proper analysis of objects moving in circles.

Conceptually, forces acting to cause rotation form what's called a "torque." Torque, akin to force in linear motion, determines how effective a rotation-producing force is based on its direction and point of application. Understanding these components provides a framework for analyzing rotational systems, ensuring a comprehensive grasp of dynamics in various engineering, physical, and natural systems.

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Most popular questions from this chapter

What would be the radial (centripetal) acceleration of an astronaut standing on the inner surface of the space station, which is at a distance of 240 \(\mathrm{m}\) from the axis of rotation, once the space station reaches its final angular velocity of 0.20 \(\mathrm{rad} / \mathrm{s}\) ? A. 0 \(\mathrm{m} / \mathrm{s}^{2}\) B. 4.8 \(\mathrm{m} / \mathrm{s}^{2}\) \(\mathrm{C}, 9.6 \mathrm{m} / \mathrm{s}^{2}\) D. 48 \(\mathrm{m} / \mathrm{s}^{2}\)

Two people are carrying a uniform wooden board that is 3.00 \(\mathrm{m}\) long and weighs 160 \(\mathrm{N}\) . If one person applies an upward force equal to 60 \(\mathrm{N}\) at one end, at what point and with what force does the other person lift? Start with a free-body diagram of the board.

You are trying to raise a bicycle wheel of mass \(m\) and radius \(R\) up over a curb of height \(h .\) To do this, you apply a horizontal force \(\vec{F}\) (Fig. \(10.81 ) .\) What is the smallest magnitude of the force \(\vec{F}\) that will succeed in raising the wheel onto the curb when the force is applied (a) at the center of the wheel, and (b) at the top of the wheel? (c) In which case is less force required?

\(\cdot\) What is the power output in horsepower of an electric motor turning at 4800 rev/min and developing a torque of 4.30 \(\mathrm{N} \cdot \mathrm{m}\) ?

A playground merry-go-round has a radius of 4.40 \(\mathrm{m}\) and a moment of inertia of 245 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) and turns with negligible friction about a vertical axle through its center. (a) A child applies a 25.0 \(\mathrm{N}\) force tangentially to the edge of the merry-go-round for 20.0 s. If the merry-go-round is initially at rest, what is its angular velocity after this 20.0 s interval? (b) How much work did the child do on the merry-go- round? (c) What is the average power supplied by the child?

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