Problem 70 - 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 is \(\omega = \sqrt{\frac{4g}{3R}}\)."

Step 1: Identify the gravitational potential energy (GPE) loss

Initially, the small object is at the highest point, equidistant from the pivot. As it swings down to the lowest point directly beneath the pivot, it loses gravitational potential energy, which is converted into kinetic energy. The initial GPE of the small object is \(mgh\), where \(h = R\), so the initial GPE is \(mgR\). The disk itself remains unchanged in GPE as its center of mass does not move vertically.

Step 2: Set up the equation for conservation of energy

According to the conservation of energy, the loss in GPE of the small object is equal to the gain in kinetic energy of the system. Thus, we can set \(mgR = \frac{1}{2} I \omega^2\), where \(I\) is the moment of inertia of the system about the pivot axis, and \(\omega\) is the angular speed we want to find.

Step 3: Determine the moment of inertia of the system

The total moment of inertia \(I\) is the sum of the moment of inertia of the disk about its center, \(I_{disk} = \frac{1}{2} m R^2\), and the moment of inertia of the small object, \(I_{object} = m R^2\), since it's at a distance \(R\) from the axis. Therefore, \(I = \frac{1}{2} m R^2 + m R^2 = \frac{3}{2} m R^2\).

Step 4: Solve for the angular speed \(\omega\)

Substitute the total moment of inertia into the conservation of energy equation: \(mgR = \frac{1}{2} \left(\frac{3}{2} m R^2\right) \omega^2\). Simplifying gives \(mgR = \frac{3}{4} m R^2 \omega^2\). Cancel mass \(m\) and solve for \(\omega^2\): \(gR = \frac{3}{4} R^2 \omega^2\). Divide both sides by \(R\) and multiply by \(\frac{4}{3}\) to isolate \(\omega^2\): \(\omega^2 = \frac{4g}{3R}\). Finally, take the square root to find \(\omega\): \(\omega = \sqrt{\frac{4g}{3R}}\).

Key Concepts

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field. In our exercise, the small object at the rim of the disk initially has GPE because it is positioned at some height above its lowest point. As it moves downward during the disk's rotation, this potential energy is converted into kinetic energy. To express GPE mathematically, the formula used is

• \( GPE = mgh \)

where \(m\) is mass, \(g\) is acceleration due to gravity, and \(h\) is the height above the reference point. In this exercise, \(h\) is equal to the radius \(R\) of the disk because the object moves from its highest point at the top of the circle to its lowest point directly below the pivot axis. This conversion of energy is fundamental to solving the problem, as the loss of GPE is equal to the gain in the kinetic energy of the system.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotation and is an important part of rotational mechanics. It plays a similar role in rotational motion as mass does in linear motion. The total moment of inertia of a system is the sum of the moments of inertia of all the components in the system. For the given problem, we have two components: the disk and the small object attached to it.

• The disk's moment of inertia calculation is given by \( I_{disk} = \frac{1}{2} m R^2 \)
• The small object, since it’s at the rim, has a moment of inertia given by \( I_{object} = m R^2 \)

By summing these, we obtain the total moment of inertia:

• \( I = \frac{1}{2} m R^2 + m R^2 = \frac{3}{2} m R^2 \)

This value is inserted in the conservation of energy equation to solve for the angular speed.

Angular Speed

Angular speed, represented by the symbol \( \omega \), is a measure of how fast an object rotates or spins around a central point or axis. It is crucial in understanding how a rotational system behaves over time. For our problem, it is derived using the conservation of energy principle, where the potential energy loss equals the kinetic energy gain.
Starting from the equation:

• \( mgR = \frac{1}{2} I \omega^2 \)

We substitute the total moment of inertia to get:

• \( mgR = \frac{3}{4} m R^2 \omega^2 \)

By simplifying this equation, we find:

• \( \omega^2 = \frac{4g}{3R} \)
• \( \omega = \sqrt{\frac{4g}{3R}} \)

Angular speed gives insight into how fast the object is moving as it reaches the lowest point below the pivot axis.

Mechanics

Mechanics is the branch of physics concerned with the motion of objects and the forces acting on them. In this problem, understanding both the linear and rotational aspects of mechanics is essential.
Rotational mechanics involves principles like moment of inertia and angular speed, which are counterparts to mass and linear velocity in linear motion.

• The conservation of energy principle is a vital mechanics concept where the total energy within an isolated system remains constant, despite energy transformations from potential to kinetic energy.
• The equations governing rotational motion such as \( I \omega^2 \), demonstrate how energy and force interplay in turning objects.
• Knowing how these components work together helps in understanding and predicting the system’s behavior as forces and energies act on the rotating disk and object.

Through mechanics, we can predict that despite transformations and motions, the total mechanical energy of the object remains conserved throughout its motion.