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Chapter 11: Problem 6

(II) A uniform horizontal rod of mass \(M\) and length \(\ell\) rotates with angular velocity \(\omega\) about a vertical axis through its center. Attached to cach end of the rod is a small mass \(m\) . Determine the angular momentum of the system about the axis.

Short Answer

Expert verified
The angular momentum is \( \left(\frac{1}{12}M + \frac{1}{2}m\right)\ell^2 \omega \).

Step by step solution

01

Understand the System Components

The system consists of a uniform horizontal rod of mass \( M \) and length \( \ell \), and two small masses, \( m \), one attached to each end of the rod. The entire setup rotates with an angular velocity \( \omega \) about a vertical axis through the center of the rod.
02

Identify the Contribution to Angular Momentum

The angular momentum \( L \) of the system has contributions from both the rod and the two masses. The rod provides a continuous distribution, while each mass provides a point contribution.
03

Calculate the Rod's Moment of Inertia

The moment of inertia \( I_{rod} \) of the uniform rod about its center is given by: \[ I_{rod} = \frac{1}{12}M\ell^2. \]
04

Contribution of the Rod to Angular Momentum

The angular momentum \( L_{rod} \) due to the rod is calculated using the formula: \[ L_{rod} = I_{rod} \cdot \omega = \left(\frac{1}{12}M\ell^2\right) \omega. \]
05

Calculate the Point Masses' Moment of Inertia

Since each mass \( m \) is a distance \( \frac{\ell}{2} \) from the axis of rotation, the moment of inertia for each mass is \( m\left(\frac{\ell}{2}\right)^2 \). Hence, the total moment of inertia due to both masses is: \[ I_{masses} = 2 \times m\left(\frac{\ell}{2}\right)^2 = \frac{1}{2}m\ell^2. \]
06

Contribution of the Point Masses to Angular Momentum

The angular momentum \( L_{masses} \) due to the point masses is calculated using: \[ L_{masses} = I_{masses} \cdot \omega = \left(\frac{1}{2}m\ell^2\right) \omega. \]
07

Combine the Contributions

The total angular momentum \( L_{total} \) is the sum of the angular momentum contributions from the rod and the masses: \[ L_{total} = L_{rod} + L_{masses} = \left(\frac{1}{12}M\ell^2 + \frac{1}{2}m\ell^2\right) \omega. \]
08

Simplify the Formula

Combine terms to simplify the expression for the total angular momentum: \[ L_{total} = \left(\frac{1}{12}M + \frac{1}{2}m\right)\ell^2 \cdot \omega. \]

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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 understanding rotational motion. It can be thought of as a measure of how much resistance an object has to changes in its rotation. Just as mass influences linear motion with inertia, the distribution of mass relative to an axis of rotation determines the moment of inertia in rotational dynamics.

In the scenario with the rotating rod and masses, we calculate the moment of inertia differently for the rod and the small point masses attached to it:
  • For the rod, a continuous mass distribution, the moment of inertia about its center is given by the formula \( \frac{1}{12}M\ell^2 \). This accounts for the spread of mass over the length of the rod.
  • For each point mass at the ends of the rod, which can be considered as separate from the rod itself, the moment of inertia is calculated by \( m\left(\frac{\ell}{2}\right)^2 \). Hence, the total for the two masses becomes \( \frac{1}{2}m\ell^2 \) because there are two masses contributing to the inertia.
Combining these values provides the total moment of inertia for the entire system, which influences how the system behaves when we apply torque or analyze angular momentum.
Rotational Dynamics
Rotational dynamics deals with how different forces and torques affect the motion of rotating objects. It's a branch of classical mechanics that examines the actions leading to rotational motion, much like how Newton's laws describe linear motion.

In this problem, the key points to understand are:
  • Torque: It鈥檚 the rotational equivalent of force. Though not directly calculated in this specific problem, torque causes the rod and masses to rotate.
  • Angular Momentum: For our rotating system, the angular momentum must be calculated by considering both the rod and the additional masses. Each element's contribution to the overall rotational motion helps us know how the system will respond under different conditions.
  • Angular Velocity: All parts of the system rotate with the same angular velocity \( \omega \), indicating the rotational speed and direction around the axis.
Understanding these components allows us to predict how the system will spin, how changes might occur with different forces, and how much energy is stored in its motion.
Rigid Body Rotation
Rigid body rotation is a simplified model that assumes an object does not deform when it rotates. This means distances between points in the body remain constant, even as the body spins around an axis.

In the given exercise, the rod and attached masses are considered a rigid body rotating around a central axis:
  • The assumption of rigidity simplifies calculations, as we don't need to account for changes in shape or internal movements within the object while it spins.
  • This simplification is helpful in problems like this, where we calculate the total angular momentum by addressing individual parts 鈥� like the rod and point masses 鈥� separately.
  • Using the moment of inertia, we can effectively describe the system's resistance to changes in rotational speed, relying on the properties of a rigid body to maintain consistency.
Rigid body rotation is an essential concept, providing a foundation for analyzing many mechanical systems where internal deformation isn't a concern, allowing focusing solely on the dynamics of rotation itself.

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Most popular questions from this chapter

The directions of vectors \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\) are given below for several cases. For each case, state the direction of \(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}\). (a) \(\overrightarrow{\mathbf{A}}\) points east, \(\overrightarrow{\mathbf{B}}\) points south. \((b) \overrightarrow{\mathbf{A}}\) points east, \(\overrightarrow{\mathbf{B}}\) points straight down. ( \(c\) ) \(\overrightarrow{\mathbf{A}}\) points straight up, \(\overrightarrow{\mathbf{B}}\) points north. (d) \(\overrightarrow{\mathbf{A}}\) points straight up, \(\overrightarrow{\mathbf{B}}\) points straight down.

A 230 -kg beam \(2.7 \mathrm{~m}\) in length slides broadside down the ice with a speed of \(18 \mathrm{~m} / \mathrm{s}\) (Fig. 11-38). A 65-kg man at rest grabs one end as it goes past and hangs on as both he and the beam go spinning down the ice. Assume frictionless motion. ( \(a\) ) How fast does the center of mass of the system move after the collision? (b) With what angular velocity does the system rotate about its CM?

A person stands on a platform, initially at rest, that can rotate freely without friction. The moment of inertia of the person plus the platform is \(I_{P}\). The person holds a spinning bicycle wheel with its axis horizontal. The wheel has moment of inertia \(I_{\mathrm{W}}\) and angular velocity \(\omega_{\mathrm{W}}\). What will be the angular velocity \(\omega_{\mathrm{P}}\) of the platform if the person moves the axis of the wheel so that it points \((a)\) vertically upward, (b) at a \(60^{\circ}\) angle to the vertical, \((c)\) vertically downward? (d) What will \(\omega_{\mathrm{P}}\) be if the person reaches up and stops the wheel in part \((a) ?\)

If vector \(\overrightarrow{\mathbf{A}}\) points along the negative \(x\) axis and vector \(\overrightarrow{\mathbf{B}}\) along the positive \(z\) axis, what is the direction of \((a) \overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}\) and (b) \(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}}\) ? (c) What is the magnitude of \(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}\) and \(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}} ?\)

Two identical particles have equal but opposite momenta, \(\overrightarrow{\mathbf{p}}\) and \(-\overrightarrow{\mathbf{p}},\) but they are not traveling along the same line. Show that the total angular momentum of this system does not depend on the choice of origin.
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