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Chapter 11: Problem 82

A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is \(6.00 \mathrm{~m}\) long, weighs \(10.0 \mathrm{~N}\), and rotates at 240 rev/min. Calculate (a) its rotational inertia about the axis of rotation and (b) the magnitude of its angular momentum about that axis.

Short Answer

Expert verified
Rotational inertia: 12.24 kg路m虏, Angular momentum: 307.45 kg路m虏/s.

Step by step solution

01

Understand the Problem

We need to find the rotational inertia and angular momentum of a rod rotating about one end. Given: rod length \( L = 6.00 \ \mathrm{m} \), weight \( W = 10.0 \ \mathrm{N} \), and rotation speed \( 240 \ \mathrm{rev/min} \).
02

Convert Weight to Mass

Since the weight of the rod is given, we first find its mass. Use the formula: \( W = mg \), where \( g = 9.8 \ \mathrm{m/s^2} \) is the acceleration due to gravity.\[ m = \frac{W}{g} = \frac{10.0 \ \mathrm{N}}{9.8 \ \mathrm{m/s^2}} = 1.02 \ \mathrm{kg} \].
03

Find Rotational Inertia

The rotational inertia of a rod about an axis through one end is \( I = \frac{1}{3}mL^2 \).Substitute \( m = 1.02 \ \mathrm{kg} \) and \( L = 6.0 \ \mathrm{m} \):\[ I = \frac{1}{3} \times 1.02 \ \mathrm{kg} \times (6.0 \ \mathrm{m})^2 = 12.24 \ \mathrm{kg \cdot m^2} \].
04

Convert Rotational Speed to Radians per Second

The given rotational speed is in rev/min. Convert it to radians per second using the conversion: \( 1 \ \mathrm{rev} = 2\pi \ \mathrm{rad} \) and \( 1 \ \mathrm{min} = 60 \ \mathrm{s} \).\[ \omega = 240 \times \frac{2\pi}{60} = 8\pi \ \mathrm{rad/s} \].
05

Calculate Angular Momentum

The angular momentum \( L \) is given by \( L = I \omega \).Using \( I = 12.24 \ \mathrm{kg \cdot m^2} \) and \( \omega = 8\pi \ \mathrm{rad/s} \):\[ L = 12.24 \ \mathrm{kg \cdot m^2} \times 8\pi \ \mathrm{rad/s} = 307.45 \ \mathrm{kg \cdot m^2/s} \].

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Key Concepts

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

Rotational Inertia
Rotational inertia, sometimes called moment of inertia, is a measure of how difficult it is to change the spin of an object. Imagine it like this: if an object has high rotational inertia, it doesn't want to start spinning and it doesn't want to stop. The bigger and heavier an object is, especially if its mass is far from the axis, the more rotational inertia it will have.

In our case, we are dealing with a rod. For a rod rotating about one end, the rotational inertia is calculated with the formula \( I = \frac{1}{3}mL^2 \).
  • \( m \) is the mass of the rod. For our exercise, the mass is \( 1.02 \ \mathrm{kg} \).
  • \( L \) is the length of the rod, which is \( 6.0 \ \mathrm{m} \).
  • With these values, the rotational inertia turns out to be \( 12.24 \ \mathrm{kg \cdot m^2} \).
It's important to remember that rotational inertia depends not just on mass but also on how this mass is distributed relative to the axis of rotation.
Angular Momentum
Angular momentum is the rotational equivalent of linear momentum. Just like a car speed up means it has high linear momentum, fast spinning objects have high angular momentum. This property depends on both how fast an object is rotating and its rotational inertia.

For our uniform rod, the angular momentum \( L \) is given by the formula \( L = I \omega \), where:
  • \( I \) is the rotational inertia, which we previously calculated to be \( 12.24 \ \mathrm{kg \cdot m^2} \).
  • \( \omega \) is the angular velocity in radians per second, determined as \( 8\pi \ \mathrm{rad/s} \).
  • The resulting angular momentum of the rod is \( 307.45 \ \mathrm{kg \cdot m^2/s} \).
Angular momentum is a conserved quantity in a closed system, meaning it will not change unless an external force acts on it. This is a fundamental principle in rotational dynamics and is useful in many real-world applications like understanding how satellites stay in orbit.
Conversion of Units
Conversions are crucial in physics, as they ensure consistency and ease of calculation. In our exercise, we needed to convert the rod's rotational speed from revolutions per minute to radians per second because physics generally uses the radian unit for angles.

  • First, understand that \( 1 \ \mathrm{revolution} = 2\pi \ \mathrm{radians} \) since a full circle is \( 2\pi \ \), a natural measure in geometry.
  • Second, note that time in physics needs to be in seconds, so change \( 1 \ \mathrm{minute} = 60 \ \mathrm{seconds} \).
  • Thus, the conversion from revolutions per minute (rev/min) to radians per second (rad/s) is based on \( \omega = 240 \times \frac{2\pi}{60} \), giving us \( 8\pi \ \mathrm{rad/s} \).
Getting these conversions right is key to solving physics problems accurately. It helps in understanding the motion in a universally accepted unit.

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Most popular questions from this chapter

At time \(t=0\), a \(3.0 \mathrm{~kg}\) particle with velocity \(\vec{v}=(5.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(6.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) is at \(x=3.0 \mathrm{~m}, y=8.0 \mathrm{~m}\). It is pulled by a \(7.0 \mathrm{~N}\) force in the negative \(x\) direction. About the origin, what are (a) the particle's angular momentum, (b) the torque acting on the particle, and (c) the rate at which the angular momentum is changing?

A Texas cockroach of mass \(0.17 \mathrm{~kg}\) runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has radius \(15 \mathrm{~cm}\), rotational inertia \(5.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\), and frictionless bearings. The cockroach's speed (relative to the ground) is \(2.0 \mathrm{~m} / \mathrm{s}\), and the lazy Susan turns clockwise with angular speed \(\omega_{0}=2.8 \mathrm{rad} / \mathrm{s}\). The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?

A \(30 \mathrm{~kg}\) child stands on the edge of a stationary merry-go-round of radius \(2.0 \mathrm{~m}\). The rotational inertia of the merry-go-round about its rotation axis is \(150 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The child catches a ball of mass \(1.0 \mathrm{~kg}\) thrown by a friend. Just before the ball is caught, it has a horizontal velocity \(\vec{v}\) of magnitude \(12 \mathrm{~m} / \mathrm{s}\), at angle \(\phi=37^{\circ}\) with a line tangent to the outer edge of the merry-go-round, as shown. What is the angular speed of the merry-go-round just after the ball is caught?

A hollow sphere of radius \(0.15 \mathrm{~m}\), with rotational inertia \(I=0.040 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about a line through its center of mass, rolls without slipping up a surface inclined at \(30^{\circ}\) to the horizontal. At a certain initial position, the sphere's total kinetic energy is \(20 \mathrm{~J}\). (a) How much of this initial kinetic energy is rotational? (b) What is the speed of the center of mass of the sphere at the initial position? When the sphere has moved \(1.0 \mathrm{~m}\) up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass?

A small solid sphere with radius \(0.25 \mathrm{~cm}\) and mass \(0.56 \mathrm{~g}\) rolls without slipping on the inside of a large fixed hemisphere with radius \(15 \mathrm{~cm}\) and a vertical axis of symmetry. The sphere starts at the top from rest. (a) What is its kinetic energy at the bottom? (b) What fraction of its kinetic energy at the bottom is associated with rotation about an axis through its com? (c) What is the magnitude of the normal force on the hemisphere from the sphere when the sphere reaches the bottom?
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