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What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal surface? (a) a uniform solid cylinder; (b) a uniform sphere; ( c) a thin walled, hollow sphere; (d) a hollow cylinder with outer radius \(R\) and inner radius \(R / 2\) .

Short Answer

Expert verified
(a) \(\frac{1}{3}\), (b) \(\frac{2}{7}\), (c) \(\frac{2}{5}\), (d) \(\frac{3}{7}\).

Step by step solution

01

Identify kinetic energy components

The total kinetic energy for an object rolling without slipping is the sum of translational kinetic energy and rotational kinetic energy. It can be expressed as \( KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \), where \( m \) is the mass, \( v \) is the linear velocity, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity.
02

Relate linear and angular velocities

Since the object is rolling without slipping, the linear velocity \( v \) is related to the angular velocity \( \omega \) by the equation \( v = \omega R \), where \( R \) is the radius of the object. Substitute \( \omega = \frac{v}{R} \) into the rotational kinetic energy term.
03

Define fraction of rotational kinetic energy

The fraction of rotational kinetic energy to the total kinetic energy is given by \( \text{Fraction}_{rotational} = \frac{\frac{1}{2}I\omega^2}{\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2} \). Substitute \( \omega = \frac{v}{R} \) for each object.
04

Calculate for solid cylinder

For a solid cylinder, the moment of inertia \( I = \frac{1}{2} mR^2 \). Substitute this value in:\[ \text{Fraction}_{solid, cylinder} = \frac{\frac{1}{2} \cdot \frac{1}{2} mR^2 \cdot \left( \frac{v}{R} \right)^2}{\frac{1}{2}mv^2 + \frac{1}{2} \cdot \frac{1}{2} mR^2 \cdot \left( \frac{v}{R} \right)^2} = \frac{1/4}{3/4} = \frac{1}{3} \]
05

Calculate for uniform sphere

For a uniform sphere, the moment of inertia \( I = \frac{2}{5}mR^2 \). Substitute this in:\[ \text{Fraction}_{uniform, sphere} = \frac{\frac{1}{2} \cdot \frac{2}{5} mR^2 \cdot \left( \frac{v}{R} \right)^2}{\frac{1}{2}mv^2 + \frac{1}{2} \cdot \frac{2}{5} mR^2 \cdot \left( \frac{v}{R} \right)^2} = \frac{1/5}{7/10} = \frac{2}{7} \]
06

Calculate for hollow sphere

For a thin, hollow sphere, the moment of inertia \( I = \frac{2}{3}mR^2 \). Substitute in:\[ \text{Fraction}_{hollow, sphere} = \frac{\frac{1}{2} \cdot \frac{2}{3} mR^2 \cdot \left( \frac{v}{R} \right)^2}{\frac{1}{2}mv^2 + \frac{1}{2} \cdot \frac{2}{3} mR^2 \cdot \left( \frac{v}{R} \right)^2} = \frac{1/3}{5/6} = \frac{2}{5} \]
07

Calculate for hollow cylinder

The hollow cylinder, with an outer radius \( R \) and inner radius \( \frac{R}{2} \), has an inertia of \( I = \frac{3}{4}mR^2 \). Substitute this in:\[ \text{Fraction}_{hollow, cylinder} = \frac{\frac{1}{2} \cdot \frac{3}{4} mR^2 \cdot \left( \frac{v}{R} \right)^2}{\frac{1}{2}mv^2 + \frac{1}{2} \cdot \frac{3}{4} mR^2 \cdot \left( \frac{v}{R} \right)^2} = \frac{3/8}{7/8} = \frac{3}{7} \]

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Key Concepts

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

Moment of Inertia
The moment of inertia is a crucial concept in rotational dynamics. Think of it as the rotational equivalent of mass in linear motion. It measures how much torque is required for a desired angular acceleration about a rotational axis. The moment of inertia depends not only on the mass but also on how that mass is distributed relative to the axis of rotation.

For different objects, the moment of inertia varies based on their shapes and mass distributions:
  • **Solid Cylinder:** The moment of inertia is calculated using the formula \( I = \frac{1}{2}mR^2 \). This reflects the mass being concentrated more towards the axis.
  • **Uniform Sphere:** Here, \( I = \frac{2}{5}mR^2 \), indicating a more even distribution of mass.
  • **Thin-Walled, Hollow Sphere:** With \( I = \frac{2}{3}mR^2 \), the mass is concentrated more towards the sphere's surface.
  • **Hollow Cylinder:** Having an outer and inner radius, its moment of inertia is \( I = \frac{3}{4}mR^2 \), reflecting the distributed mass between the shell and the hollow part.
Understanding the moment of inertia is vital in determining how different objects accelerate when subjected to rotational forces.
Rolling Without Slipping
Rolling without slipping is a condition where an object rolls on a surface without sliding. It's crucial for understanding the energy distribution in rolling objects, as it establishes a specific relationship between linear and angular velocities.

In this case, the point of contact between the object and the surface is momentarily at rest. This condition is mathematically expressed as:
  • \( v = \omega R \)
where \( v \) is the linear velocity, \( \omega \) is the angular velocity, and \( R \) is the radius of the object.

This means the linear speed of the object is dictated by its rate of rotation and its size. When an object rolls without slipping:
  • The translational kinetic energy is \( \frac{1}{2}mv^2 \).
  • The rotational kinetic energy is \( \frac{1}{2}I\omega^2 \).
This relationship lets us calculate the energy associated with both the movement and the spinning of the object, ensuring calculations accurately reflect real-world motions.
Fraction of Kinetic Energy
The concept of the fraction of kinetic energy aims at understanding how much of the total kinetic energy of a rolling object is due to its rotation. It's significant for evaluating energy distribution and is useful for design and analysis in mechanical systems.

To find this fraction, you typically start with the expression:
  • Total Kinetic Energy, \( KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \)
  • Fraction of rotational kinetic energy, \( \text{Fraction}_{rotational} = \frac{\frac{1}{2}I\omega^2}{\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2} \)
By substituting \( \omega = \frac{v}{R} \), you can calculate this fraction for different shapes:
  • For a **solid cylinder**, it is \( \frac{1}{3} \).
  • A **uniform sphere** has a fraction of \( \frac{2}{7} \).
  • The **hollow sphere** results in \( \frac{2}{5} \).
  • Lastly, a **hollow cylinder** gives \( \frac{3}{7} \).
These ratios tell us how much of their motion energy is spinning compared to moving along a path. Recognizing this helps in fields like biomechanics, vehicle dynamics, and robotics, where efficient energy utilization is critical.

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Most popular questions from this chapter

A bicycle racer is going downhill at 11.0 \(\mathrm{m} / \mathrm{s}\) when, to his horror, one of his \(2.25-\mathrm{kg}\) wheels comes off as he is 75.0 \(\mathrm{m}\) above the foot of the hill. We can model the wheel as a thin- walled cylinder 85.0 \(\mathrm{cm}\) in diameter and neglect the small mass of the spokes. (a) How fast is the wheel moving when it reaches the foot of the hill if it rolled without slipping all the way down? (b) How much total kinetic energy does the wheel have when it reaches the bottomof the hill?

A small 10.0-g bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass 50.0 \(\mathrm{g}\) and is 100 \(\mathrm{cm}\) in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of 20.0 \(\mathrm{cm} / \mathrm{s}\) relative to the table. (a) What is the angular speed of the bar just after the frisky insect leaps? (b) What is the total kinetic energy of the system just after the bug leaps? (c) Where does this energy come from?

A woman with mass 50 \(\mathrm{kg}\) is standing on the rim of a large disk that is rotating at 0.50 \(\mathrm{rev} / \mathrm{s}\) about an axis through its center. The disk has mass 110 \(\mathrm{kg}\) and radius 4.0 \(\mathrm{m} .\) Calculate the magnitude of the total angular momentum of the woman-disk system. (Assume that you can treat the woman as a point.)

Tarzan and Jane in the 21 st Century. Tarzan has foolishly gotten himself into another scrape with the animals and must be rescued once again by Jane. The 60.0 -kg Jane starts from rest at a height of 5.00 \(\mathrm{m}\) in the trees and swings down to the ground using a thin, but very rigid, 30.0 -kg vine 8.00 \(\mathrm{m}\) long. She arrives just in time to snatch the 72.0 -kg Tarzan from the jaws of an angry hippopotamus. What is Jane's (and the vine's) angular speed (a) just before she grabs Tarzan and (b) just after she grabs him? (c) How high will Tarzan and Jane go on their first swing after this daring rescue?

A uniform marble rolls down a symmetrical bowl, starting from rest at the top of the left side. The top of each side is a distance \(h\) above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil. (a) How far up the smooth side will the marble go, measured vertically from the bottom? (b) How high would the marble go if both sides were as rough as the left side? (c) How do you account for the fact that the marble goes higher with friction on the right side than without friction?

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