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A solid sphere is rolling on a surface. What fraction of its total kinetic energy is in the form of rotational kinetic energy about the center of mass?

Short Answer

Expert verified
The fraction of kinetic energy as rotational energy is \( \frac{2}{7} \).

Step by step solution

01

Identify Types of Kinetic Energy

The total kinetic energy (KE) of a rolling object is the sum of its translational and rotational kinetic energy. For a sphere of mass M radius R, the translational kinetic energy is \( KE_{trans} = \frac{1}{2} Mv^2 \), where \( v \) is the linear velocity of the center of mass. The rotational kinetic energy is \( KE_{rot} = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
02

Calculate Moment of Inertia

For a solid sphere, the moment of inertia about its center is \( I = \frac{2}{5} MR^2 \).
03

Link Angular and Linear Velocity

The rolling condition ensures that \( v = \omega R \), so we can express the angular velocity \( \omega \) as \( \omega = \frac{v}{R} \).
04

Substitute and Simplify Rotational Kinetic Energy

Substitute the expression for \( I \) and \( \omega \) into the relation for rotational kinetic energy: \( KE_{rot} = \frac{1}{2} \left(\frac{2}{5} MR^2\right)\left(\frac{v}{R}\right)^2 = \frac{1}{5} Mv^2 \).
05

Calculate Total Kinetic Energy

Combine the translational and rotational kinetic energies to find total kinetic energy: \( KE = KE_{trans} + KE_{rot} = \frac{1}{2} Mv^2 + \frac{1}{5} Mv^2 = \frac{7}{10} Mv^2 \).
06

Find Fraction of Rotational Kinetic Energy

The fraction of total kinetic energy that is rotational is \( \frac{KE_{rot}}{KE} = \frac{\frac{1}{5} Mv^2}{\frac{7}{10} Mv^2} = \frac{2}{7} \).

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Key Concepts

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

Rotational Kinetic Energy
When an object spins around an axis, it possesses rotational kinetic energy. This kind of energy is part of an object's total kinetic energy when it rolls as it involves spinning as well as moving along a path. For a rigid body like a solid sphere, the rotational kinetic energy is calculated by the formula:
  • \( KE_{rot} = \frac{1}{2} I \omega^2 \)
Here, \( I \) stands for the moment of inertia and \( \omega \) is the angular velocity.
Understanding rotational kinetic energy helps us analyze how energy is distributed within a rolling sphere. Knowing this fraction tells us how efficiently an object taps into its rotational motion.
Moreover, this concept is crucial in the study of dynamics, playing a pivotal role in system equilibrium and motion prediction.
Translational Kinetic Energy
Translational kinetic energy describes how energy is stored in an object due to its motion through space. It is calculated using the formula:
  • \( KE_{trans} = \frac{1}{2} Mv^2 \)
Where \( M \) is the mass and \( v \) is the linear velocity of the object.
This form of kinetic energy highlights the straight-line movement aspect of motion, as opposed to the spinning involved in rotational kinetic energy.
To understand a rolling object's total kinetic energy, both translational and rotational components must be considered.
Translational kinetic energy gives insight into how the speed of the entire object's center of mass contributes to its total energy expression.
Moment of Inertia
Moment of inertia, often referred to as the rotational equivalent of mass in linear motion, quantifies an object's resistance to changes in its rotation. For a solid sphere, the moment of inertia about its axis through its center is given by:
  • \( I = \frac{2}{5} MR^2 \)
This value dictates how an object will react under rotational forces, similarly to how mass dictates response under linear forces.
The moment of inertia depends on both the mass of the object and how that mass is distributed relative to the axis of rotation.
A deeper grasp of this concept allows for better understanding of energy partitioning during rotation, necessary for analyzing rolling bodies like the sphere in the exercise.
Rolling Motion
Rolling motion combines rotation and translation, making it complex yet fascinating. It occurs when an object moves on a surface, rotating about its axis and translating in a direction perpendicular to that axis. For something like a sphere, the motion can be described by the condition:
  • \( v = \omega R \)
This reflects a balance between linear and angular physical descriptions.
This connection implies that for every point on the rolling sphere that touches the ground, it momentarily matches the ground speed, creating a smooth, no-slip condition.
The study of rolling motion can apply to many real-world phenomena, like wheels and gears, demonstrating how interconnected linear and rotational dynamics are in everyday physics.

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Most popular questions from this chapter

Just after a motorcycle rides off the end of a ramp and launches into the air, its engine is turning counterclockwise at 7700 rev/min. The motorcycle rider forgets to throttle back, so the engine"s angular speed increases to 12.500 rev/min. As a result, the rest of the motorcycle (including the rider) begins to rotate clockwise about the engine at 3.8 rev/min. Calculate the ratio \(I_{\mathrm{L}} / I_{\mathrm{M}}\) of the moment of inertia of the engine to the moment of inertia of the rest of the motorcycle (and the rider). Ignore torques due to gravity and air resistance.

Starting from rest, a basketball rolls from the top of a hill to the bottom, reaching a translational speed of \(6.6 \mathrm{m} / \mathrm{s}\). Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?

Two thin rods of length \(L\) are rotating with the same angular speed \(\omega(\) in \(\mathrm{rad} / \mathrm{s})\) about axes that pass perpendicularly through one end. Rod \(\mathrm{A}\) is massless but has a particle of mass \(0.66 \mathrm{kg}\) attached to its free end. Rod B has a mass of \(0.66 \mathrm{kg}\), which is distributed uniformly along its length. The length of each rod is \(0.75 \mathrm{m},\) and the angular speed is \(4.2 \mathrm{rad} / \mathrm{s}\). Find the kinetic energies of rod \(A\) with its attached particle and of rod \(B\).

A solid disk rotates in the horizontal plane at an angular velocity of \(0.067 \mathrm{rad} / \mathrm{s}\) with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is \(0.10 \mathrm{kg} \cdot \mathrm{m}^{2} .\) From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of \(0.40 \mathrm{m}\) from the axis. The sand in the ring has a mass of \(0.50 \mathrm{kg}\). After all the sand is in place, what is the angular velocity of the disk?

As seen from above, a playground carousel is rotating counterclockwise about its center on frictionless bearings. A person standing still on the ground grabs onto one of the bars on the carousel very close to its outer edge and climbs aboard. Thus, this person begins with an angular speed of zero and ends up with a nonzero angular speed, which means that he underwent a counterclockwise angular acceleration. The carousel has a radius of \(1.50 \mathrm{m},\) an initial angular speed of \(3.14 \mathrm{rad} / \mathrm{s},\) and a moment of inertia of \(125 \mathrm{kg} \cdot \mathrm{m}^{2} .\) The mass of the person is \(40.0 \mathrm{kg} .\) Find the final angular speed of the carousel after the person climbs aboard.

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