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Chapter 9: Problem 42

A solid cylinder of mass \(20 \mathrm{~kg}\) rotates about its axis with angular speed \(100 \mathrm{rad} / \mathrm{s}\). 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? (a) \(3125 \mathrm{~J}, 62.5 \mathrm{~J}-\mathrm{s}\) (b) \(72.5 \mathrm{~J} \mathrm{~s}\) and \(62.5 \mathrm{~J}-\mathrm{s}\) (c) \(3125 \mathrm{~J}, 82.5 \mathrm{~J}-\mathrm{s}\) (d) None of the above

Short Answer

Expert verified
The correct option is (a) \( 3125 \mathrm{~J}, 62.5 \mathrm{~J}-\mathrm{s} \).

Step by step solution

01

Calculate Moment of Inertia

The moment of inertia \( I \) for a solid cylinder rotating about its axis is given by the formula \( I = \frac{1}{2}mr^2 \). Substituting the given values: \( m = 20 \ kg \) and \( r = 0.25 \ m \), we get: \[ I = \frac{1}{2} \times 20 \times (0.25)^2 = 0.625 \ kg \cdot m^2 \]
02

Calculate Rotational Kinetic Energy

The kinetic energy \( KE \) associated with the rotation is given by the formula \( KE = \frac{1}{2} I \omega^2 \). Using \( I = 0.625 \ kg \cdot m^2 \) and the angular speed \( \omega = 100 \ rad/s \), we find: \[ KE = \frac{1}{2} \times 0.625 \times (100)^2 = 3125 \ J \]
03

Calculate Angular Momentum

The angular momentum \( L \) of the cylinder is given by the formula \( L = I \omega \). Using \( I = 0.625 \ kg \cdot m^2 \) and \( \omega = 100 \ rad/s \), we calculate: \[ L = 0.625 \times 100 = 62.5 \ J \cdot s \]
04

Determine the Correct Answer

Comparing the calculated values of kinetic energy and angular momentum with the options provided, option (a) matches: Rotational kinetic energy = \( 3125 \ J \), Angular momentum = \( 62.5 \ J \cdot s \).

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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, sometimes known as the rotational inertia, can be thought of as the rotational equivalent of mass in linear motion. It is crucial in understanding how difficult it is to change the rotational state of an object. For a solid cylinder, like the one in our problem, the moment of inertia is calculated using the formula:
  • \( I = \frac{1}{2}mr^2 \)
In this formula:
  • \( m \) is the mass of the cylinder
  • \( r \) is the radius
In our specific example, with a mass of 20 kg and a radius of 0.25 m, the moment of inertiabecomes 0.625 kg路m虏. This quantifies how the mass and shape of the cylinder affect its tendency to resist angular acceleration. The larger the moment of inertia, the harder it is to spin the object. This concept helps engineers and scientists design objects with desired motion properties, whether in fields like automotive engineering or biomechanical applications.
Rotational Kinetic Energy
Rotational kinetic energy refers to the energy an object possesses due to its rotation. Just like objects in linear motion have kinetic energy, rotating objects do as well. The kinetic energy for rotational motion is given by the formula:
  • \( KE = \frac{1}{2} I \omega^2 \)
Where \( \omega \) represents the angular velocity, and \( I \) is the moment of inertia calculated earlier.For our solid cylinder, substituting the values \( I = 0.625 \ kg \, \cdot \, m^2 \) and \( \omega = 100 \ rad/s \) gives us a rotational kinetic energy of 3125 J. This value is significant because it tells us how much work is needed to bring the cylinder to its current rotational speed from a state of rest. Understanding this is essential in many practical applications from calculating the energy output of turbines to assessing the performance of sports equipment.
Angular Momentum
Angular momentum is a measure of the amount of rotation an object has, taking into account its moment of inertia and angular velocity. It is an important concept in rotational dynamics as it is conserved in isolated systems, much like linear momentum.The formula for angular momentum is:
  • \( L = I \omega \)
For the solid cylinder scenario, using \( I = 0.625 \ kg \, m^2 \) and \( \omega = 100 \ rad/s \), the angular momentum comes out as 62.5 J路s. This means that the cylinder, with its given rotation, has 62.5 Joule seconds of angular momentum. In physics, angular momentum quantifies the rotational analogue of linear momentum. It is crucial in understanding phenomena in both classical mechanics and quantum mechanics, from how ice skaters spin to the behavior of electrons in atoms.

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Most popular questions from this chapter

Two thin uniform circular rings each of radius \(10 \mathrm{~cm}\) and mass \(0.1 \mathrm{~kg}\) are arranged such that they have a common centre and their planes are perpendicular to each other. The moment of inertia of this system about an axis passing through their common centre and perpendicular to the plane of one of the rings in \(\mathrm{kgm}^{2}\) is (a) \(15 \times 10^{-3}\) (b) \(5 \times 10^{-3}\) (c) \(1.5 \times 10^{-3}\) (d) \(18 \times 10^{-4}\)

The net external torque on a system of particles about an axis is zero. Which of the following are compatible with it? \(\quad\) INCERT Exemplar] (a) The forces may be acting radially from a point on the axis (b) The forces may be acting on the axis of rotation (c) The forces may be acting parallel to the axis of rotation (d) The torque caused by some forces may be equal and opposite to that caused by other forces

A wheel of mass \(8 \mathrm{~kg}\) and radius \(40 \mathrm{~cm}\) is rolling on a horizontal road with angular velocity of \(15 \mathrm{rad} \mathrm{s}^{-1}\). The moment of inertia of the wheel about its axis is \(0.64 \mathrm{~kg} \mathrm{~m}^{-2}\). Total kinetic energy of wheel is (a) \(288 \mathrm{~J}\) (b) \(216 \mathrm{~J}\) (c) \(72 \mathrm{~J}\) (d) \(144 \mathrm{~J}\)

A wheel has angular acceleration of \(3.0 \mathrm{rads}^{-2}\) and an initial angular speed of \(2.00 \mathrm{rads}^{-2}\). In a time of \(2 \mathrm{~s}\) it has rotated through an angle (in radian) of [UP SEE 2007] (a) 6 (b) 10 [c) 12 (d) 4

A thin circular ring of mass \(M\) and radius \(r\) is rotating about its axis with a constant angular velocity @. Two objects each of mass \(M\) are attached gently to the opposite ends of a diameter of the ring. The ring will now rotate with an angular velocity (a) \(\frac{\omega(M-2 m)}{M+2 m}\) (b) \(\frac{\omega M}{M+2 m}\) (c) \(\frac{\omega M}{M+m}\) (d) \(\frac{\omega(M+2 M)}{M}\)
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