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Chapter 7: Problem 12

A solid cylinder of mass \(20 \mathrm{~kg}\) rotates about its axis with angular speed \(100 \mathrm{rad} \mathrm{s}^{-1}\). The radius of the cylinder is \(0.25 \mathrm{~m}\). What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

Short Answer

Expert verified
The rotational kinetic energy of the cylinder is 3125 J, and the angular momentum is 62.5 kg m虏/s.

Step by step solution

01

Understand the Problem

We need to calculate two things about a rotating solid cylinder: 1) Its rotational kinetic energy and 2) Its angular momentum. The given values are the mass, radius, and angular speed of the cylinder.
02

Write Down Known Values

The mass of the cylinder, \( m = 20 \, \text{kg} \); Radius of the cylinder, \( r = 0.25 \, \text{m} \); Angular speed, \( \omega = 100 \, \text{rad/s} \).
03

Formulate Rotational Kinetic Energy

The formula for the rotational kinetic energy \( KE \) of a solid cylinder is given by: \[ KE = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia, and \( \omega \) is the angular speed.
04

Calculate Moment of Inertia

The moment of inertia \( I \) for a solid cylinder about its axis is: \[ I = \frac{1}{2} m r^2 \] Substitute the known values: \( I = \frac{1}{2} \times 20 \, \text{kg} \times (0.25 \, \text{m})^2 \). Calculate \( I \) as: \( I = 0.625 \, \text{kg} \, \text{m}^2 \).
05

Calculate Rotational Kinetic Energy

Substitute \( I \) and \( \omega \) into the kinetic energy formula: \[ KE = \frac{1}{2} \times 0.625 \, \text{kg} \, \text{m}^2 \times (100 \, \text{rad/s})^2 \]. Calculate \( KE \) to find that \( KE = 3125 \, \text{J} \).
06

Formulate Angular Momentum

The formula for angular momentum \( L \) is: \[ L = I \omega \] Where \( \omega \) is the angular speed.
07

Calculate Angular Momentum

Substitute \( I \) and \( \omega \) into the formula: \[ L = 0.625 \, \text{kg} \, \text{m}^2 \times 100 \, \text{rad/s} \]. Compute \( L \) and find that \( L = 62.5 \, \text{kg} \, \text{m}^2/\text{s} \).

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Key Concepts

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

Angular Momentum
Angular momentum is an essential concept in physics that describes how much rotation an object has. It is similar to linear momentum but in a rotational context. For a rotating object like a solid cylinder, angular momentum depends on both its moment of inertia and its angular speed.
  • Formula: The angular momentum \( L \) of an object is calculated using the formula: \[ L = I \omega \], where \( I \) is the moment of inertia and \( \omega \) is the angular speed.
  • Variables: In the context of a solid cylinder, the angular momentum indicates how the mass distribution and the speed of rotation influence its rotational motion.
Angular momentum is conserved in isolated systems, meaning that without external torques, the total angular momentum remains constant. This concept is fundamental in fields such as astrophysics and mechanical engineering where rotating systems are studied.
Moment of Inertia
The moment of inertia is a crucial property that determines how difficult it is to change an object's rotational motion. For a solid cylinder, the moment of inertia signifies how its mass is distributed concerning its axis of rotation.
  • Definition: It is defined mathematically as \( I = \frac{1}{2} m r^2 \), where \( m \) is the mass and \( r \) is the radius of the cylinder.
  • Role: Higher moment of inertia indicates more resistance to changes in the rotational speed, similar to how mass affects linear motion.
For instance, a solid cylinder with a larger radius or mass has a higher moment of inertia, meaning it takes more effort to start or stop its rotation. Understanding this concept is vital for designing rotating machinery and analyzing dynamics in engineering systems.
Solid Cylinder
A solid cylinder is a three-dimensional object with a uniform and fixed shape that is commonly encountered in physics problems. Its characteristics make it particularly straightforward to analyze using the principles of rigid body rotation.
  • Shape and Symmetry: A solid cylinder has a consistent radius along its length, and its mass distribution is uniform, which simplifies the calculation of properties like moment of inertia.
  • Applications: Examples of solid cylinders include wheels, rollers, and rods used in engines, being critical components in numerous mechanical systems.
When analyzing the rotation of a solid cylinder, one takes advantage of its symmetry which leads to more accessible formulas that depend on its mass and dimensions. This makes learning about solid cylinders a fundamental step in understanding rotational dynamics.

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Most popular questions from this chapter

Read each statement below carefully, and state, with reasons, if it is true or false: (a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body. (b) The instantaneous speed of the point of contact during rolling is zero. (c) The instantaneous acceleration of the point of contact during rolling is zero. (d) For perfect rolling motion, work done against friction is zero. (e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.

Show that the area of the triangle contained between the vectors a and \(\mathbf{b}\) is one half of the magnitude of \(\mathbf{a} \times \mathbf{b}\).

In the HCl molecule, the separation between the nuclei of the two atoms is about \(1.27 \mathrm{~A}\left(1 \mathrm{~A}=10^{-10} \mathrm{~m}\right) .\) Find the approximate location of the \(\mathrm{CM}\) of the molecule, given that a chlorine atom is about \(35.5\) times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

From a uniform disk of radius \(R\), a circular hole of radius \(R / 2\) is cut out. The centre of the hole is at \(R / 2\) from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

Prove the result that the velocity \(v\) of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height \(h\) is given by \(v^{2}=\frac{2 g h}{\left(1+k^{2} / R^{2}\right)}\) using dynamical consideration (i.e. by consideration of forces and torques). Note \(k\) is the radius of gyration of the body about its symmetry axis, and \(\mathrm{R}\) is the radius of the body. The body starts from rest at the top of the plane.
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