Problem 80 - University Physics with Modern Physics

Question

A uniform, solid disk with mass \(m\) and radius \(R\) is pivoted about a horizontal axis through its center. A small object of the same mass \(m\) is glued to the rim of the disk. If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.

Step-by-Step Solution

Answer

The angular speed \(\omega\) is \(\sqrt{\frac{4g}{3R}}\).

Step 1: Identify the Initial Conditions

The system consists of a disk and a small object, both with mass \(m\). The disk is released from rest, meaning the initial angular speed \(\omega_0 = 0\). The small object is positioned at a horizontal distance \(R\) from the center of the disk.

Step 2: Calculate the Moment of Inertia

The moment of inertia \(I\) of the system is the sum of the moment of inertia of the disk and that of the small mass. \(I_{disk} = \frac{1}{2}mR^2\) for the disk and \(I_{object} = mR^2\) for the small object. Therefore, \(I = \frac{1}{2}mR^2 + mR^2 = \frac{3}{2}mR^2\).

Step 3: Apply Conservation of Energy

The initial gravitational potential energy \(U_i\) is converted into rotational kinetic energy \(K\). Initially, \(U_i = mgR\) and \(U_f = 0\) when the mass is directly below the axis. The rotational kinetic energy is given by \(K = \frac{1}{2}I\omega^2\).

Step 4: Set Up the Energy Equation

Using conservation of energy: \(U_i = K\), or \(mgR = \frac{1}{2}I\omega^2\). Plug in the expression for \(I\): \(mgR = \frac{1}{2}\left(\frac{3}{2}mR^2\right)\omega^2\).

Step 5: Solve for Angular Speed \(\omega\)

Rearrange the equation to solve for \(\omega\): \(\omega^2 = \frac{2mgR}{\frac{3}{2}mR^2}\). Simplify to \(\omega^2 = \frac{4g}{3R}\). Finally, take the square root to find \(\omega = \sqrt{\frac{4g}{3R}}\).

Key Concepts

Conservation of Energy

Conservation of Energy is a vital principle in physics that states energy cannot be created or destroyed, only transformed from one form to another. In this exercise, the initial potential energy of the system is transformed into kinetic energy as it moves. The disk and the mass attached to its rim begin at a position where the small mass is elevated. At this point, all the energy is in the form of gravitational potential energy, given by:- \( U_i = mgR \), where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( R \) is the radius of the disk.As the disk is released, this energy turns into rotational kinetic energy as it begins to spin. When the mass reaches its lowest point directly below the axis, the entire gravitational potential energy is converted into rotational kinetic energy:- \( K = \frac{1}{2}I\omega^2 \).This relationship demonstrates how energy conservation helps us calculate the angular speed \( \omega \) by equating the initial potential energy to the rotational kinetic energy of the system.

Moment of Inertia

Moment of Inertia is the rotational equivalent of mass in linear motion and plays a significant role in rotational dynamics. It determines how difficult it is to change an object's rotational speed.For rotational systems, the moment of inertia \( I \) depends on the distribution of mass concerning the axis of rotation. In this exercise, the system comprises a disk of moment of inertia \( I_{disk} = \frac{1}{2}mR^2 \), and a small object attached to its edge, contributing \( I_{object} = mR^2 \). The total moment of inertia is the sum: - \( I = \frac{3}{2}mR^2 \).Knowing \( I \) is crucial because it influences how the entire system rotates once released. The larger the moment of inertia, the harder it is to spin the object. This makes finding \( I \) a necessary step in computing the angular speed \( \omega \) when employing the conservation of energy principle.

Rotational Kinetics

Rotational Kinetics deals with the motion of objects as they spin and the forces acting on them. It involves using angular velocity and moment of inertia. Angular velocity \( \omega \) is a measure of how fast an object rotates. In this problem, we need to find \( \omega \) when the small object on the disk reaches a specific position. Using the energy considerations from the conservation of energy, we set: - \( mgR = \frac{1}{2}I\omega^2 \).To solve for \( \omega \), we first substitute the expression for the moment of inertia found earlier:- \( mgR = \frac{1}{2} \left(\frac{3}{2}mR^2\right)\omega^2 \).Rearrange and simplify:- \( \omega^2 = \frac{4g}{3R} \).Taking the square root gives:- \( \omega = \sqrt{\frac{4g}{3R}} \).This calculation showcases how rotational kinetics involves not only understanding how fast things spin but also linking speed with other rotational quantities like moment of inertia.