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Chapter 11: Problem 78

Rotational Vs. Translational Energy of Motion (a) Describe how a solid ball can move so that i. Its total kinetic energy is just the energy of motion of its center of mass ii. Its total kinetic energy is the energy of its motion relative to its center of mass (b) Two bowling balls are moving down a bowling alley so that their centers of mass have the same velocity, but one just slides down the alley, while the other rolls down the alley. Which ball has more energy? Explain your reasoning.

Short Answer

Expert verified
The ball rolling has more energy because it possesses both translational and rotational kinetic energy, while the sliding ball only has translational kinetic energy.

Step by step solution

01

Understanding Kinetic Energy of Center of Mass

When a solid ball moves such that its total kinetic energy is just the energy of its motion of center of mass, the ball should be sliding without any rotation. This means every point on the ball has the same velocity as the center of mass. Hence, its kinetic energy is purely translational: \[ K_{translational} = \frac{1}{2} m v_{cm}^2 \]
02

Understanding Kinetic Energy Relative to Center of Mass

When a solid ball moves so that its total kinetic energy is the energy of its motion relative to its center of mass, the ball should be rotating with respect to its center of mass. Thus, its kinetic energy is purely rotational: \[ K_{rotational} = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
03

Compare Bowling Balls' Energies

For the bowling balls: one slides (translational kinetic energy only) and the other rolls (both translational and rotational kinetic energies). The sliding ball's energy is \( K_{sliding} = \frac{1}{2} m v_{cm}^2 \). The rolling ball has both translational and rotational energy. Its total energy is: \[ K_{rolling} = \frac{1}{2} m v_{cm}^2 + \frac{1}{2} I \omega^2 \] Given that the center of mass velocity is the same for both, but the rolling ball also has rotational energy, it has more total kinetic energy.

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Key Concepts

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

Translational Kinetic Energy
Translational kinetic energy is the energy that an object possesses due to its motion along a path. It's the type of kinetic energy found in objects moving in a straight line. The formula for translational kinetic energy is:

\[ K_{translational} = \frac{1}{2} m v^2 \]
where:
  • \( m \): Mass of the object
  • \( v \): Velocity of the object's center of mass

Imagine a ball sliding down a smooth alley without spinning. Every part of the ball moves with the same speed as its center of mass. This is a typical example of translational motion where the only kinetic energy involved is due to the linear motion.

Understanding translational kinetic energy helps explain which scenarios involve purely linear motion, such as a sliding ball.
Rotational Kinetic Energy
Rotational kinetic energy refers to the energy an object has because it鈥檚 rotating around an axis. When an object spins, every point on it moves at a different velocity, depending on its distance from the axis of rotation. The formula for rotational kinetic energy is:

\[ K_{rotational} = \frac{1}{2} I \omega^2 \]
where:
  • \( I \): Moment of inertia
  • \( \omega \): Angular velocity

Consider a solid ball rolling down an alley; it鈥檚 both moving forward and spinning around its own axis. This ball鈥檚 energy isn't just from its forward motion but also from its rotation around the center of mass.

Rotational kinetic energy is crucial for understanding motions where rotation plays a role, such as rolling balls or spinning tops.
Moment of Inertia
The moment of inertia is a property of an object that measures how difficult it is to change its rotational motion. It's analogous to mass in linear motion. Essentially, it depends on how mass is distributed relative to the axis of rotation. The formula used to calculate the moment of inertia is typically:

\[ I = \sum m_i r_i^2 \]
where:
  • \( m_i \): Mass of individual particles
  • \( r_i \): Distance of each particle from the axis of rotation

Think of a solid ball and a hollow ball of the same mass and size. The hollow ball has more mass distributed farther from the center, making it harder to spin, thus having a larger moment of inertia.

Understanding the moment of inertia is key to comprehending how different shapes and mass distributions impact an object's rotational motion.

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Most popular questions from this chapter

Polar Axis of Earth (a) What is the rotational speed \(\omega\) about the polar axis of a point on Earth's surface at a latitude of \(40^{\circ} \mathrm{N} ?\) (Earth rotates about that axis.) (b) What is the translational speed \(v\) of the point? What are (c) \(\omega\) and \((\mathrm{d}) v\) for a point at the equator?

Oxygen Molecule The oxygen molecule \(\mathrm{O}_{2}\) has a mass of \(5.30 \times\) \(10^{-26} \mathrm{~kg}\) and a rotational inertia of \(1.94 \times 10^{-46} \mathrm{~kg} \cdot \mathrm{m}^{2}\) about an axis through the center of the line joining the atoms and perpendicular to that line. Suppose the center of mass of an \(\mathrm{O}_{2}\) molecule in a gas has a translational speed of \(500 \mathrm{~m} / \mathrm{s}\) and the molecule has a rotational kinetic energy that is \(\frac{2}{3}\) of the translational kinetic energy of its center of mass. What then is the molecule's rotational speed about the center of mass?

Thin Hoop A \(32.0 \mathrm{~kg}\) wheel, essentially a thin hoop with radius \(1.20 \mathrm{~m}\), is rotating at \(280 \mathrm{rev} / \mathrm{min}\). It must be brought to a stop in \(15.0 \mathrm{~s}\). (a) How much work must be done to stop it? (b) What is the required average power?

Uniform Spherical Shell A uniform spherical shell of mass \(M\) and radius \(R\) rotates about a vertical axis on frictionless bearings (Fig. 11-39). A massless cord passes around the equator of the shell. over a pulley of rotational inertia \(I\) and radius \(r\), and is attached to a small object of mass \(m\). There is no friction on the pulley's axle; the cord does not slip on the pulley. What is the speed of the object after it falls a distance \(h\) from rest? Use energy considerations.

Starting from Rest A wheel, starting from rest, rotates with a constant rotational acceleration of \(2.00 \mathrm{rad} / \mathrm{s}^{2} .\) During a certain \(3.00 \mathrm{~s}\) interval, it turns through \(90.0\) rad. (a) How long is the wheel turning before the start of the \(3.00 \mathrm{~s}\) interval? (b) What is the rotational velocity of the wheel at the start of the \(3.00 \mathrm{~s}\) interval?
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