Problem 28 - College Physics

Question

A slender metal rod has a mass \(M\) and length \(L\). The rod is first rotated about a perpendicular axis through its center with an angular velocity \(\omega\) (Figure \(9.28 \mathrm{a}\) ). It is then rotated about a perpendicular axis through its end at the same angular velocity (Figure \(9.28 \mathrm{~b}\) ). Find the ratio of the kinetic energy of the first case to that of the second case.

Step-by-Step Solution

Answer

The ratio of kinetic energies is \( \frac{1}{4} \).

Step 1: Understanding Rotational Kinetic Energy

Rotational kinetic energy for a rotating object is given by the formula \( KE = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. Our goal is to find the ratio of kinetic energies for the rod rotating about two different axes.

Step 2: Calculate Moment of Inertia for Axis Through Center

For a slender rod rotating about an axis through its center, the moment of inertia \( I_1 \) is given by \( I_1 = \frac{1}{12}ML^2 \). This comes from the standard formula for a rod rotating about an axis through its center.

Step 3: Calculate Moment of Inertia for Axis Through End

For a slender rod rotating about an axis through one end, the moment of inertia \( I_2 \) is given by \( I_2 = \frac{1}{3}ML^2 \). This is derived using the parallel axis theorem, which states \( I = I_{cm} + Md^2 \), where \( I_{cm} \) is the inertia about the center of mass and \( d \) is the distance from the center of mass to the new axis (\( L/2 \)).

Step 4: Calculate Kinetic Energy for Both Cases

Using the moment of inertia formulas in the kinetic energy equation, the kinetic energy when rotating about the center is \( KE_1 = \frac{1}{2} I_1 \omega^2 = \frac{1}{2} \left( \frac{1}{12}ML^2 \right) \omega^2 = \frac{1}{24} ML^2 \omega^2 \). Similarly, for the end rotation, \( KE_2 = \frac{1}{2} I_2 \omega^2 = \frac{1}{2} \left( \frac{1}{3}ML^2 \right) \omega^2 = \frac{1}{6} ML^2 \omega^2 \).

Step 5: Calculate the Ratio of Kinetic Energies

The ratio of the kinetic energy of the first case to that of the second case is given by \( \frac{KE_1}{KE_2} = \frac{\frac{1}{24} ML^2 \omega^2}{\frac{1}{6} ML^2 \omega^2} \). Notice that \( ML^2 \omega^2 \) cancels out, simplifying to \( \frac{1}{24} \div \frac{1}{6} = \frac{1}{24} \times \frac{6}{1} = \frac{1}{4} \).

Key Concepts

Moment of Inertia

The moment of inertia is a crucial concept in understanding rotational dynamics. It represents the distribution of an object's mass with respect to a rotational axis and plays a similar role to mass in linear motion. Essentially, it is a measure of an object's resistance to changing its rotation. For everyday objects, the moment of inertia depends not only on the mass but also on the shape of the object and the position of the axis of rotation.

For the slender metal rod considered in our problem:

• When the axis is through the center of the rod, the moment of inertia is given by the formula \( I_1 = \frac{1}{12}ML^2 \). This is due to the symmetrical distribution of mass.
• When the axis is at one end of the rod, the formula becomes \( I_2 = \frac{1}{3}ML^2 \). Here, the mass is distributed farther from the axis, increasing the moment of inertia.

Understanding these calculations helps us see how different axis placements impact the rod's rotation.

Angular Velocity

Angular velocity, denoted by \( \omega \), is a vector quantity that describes how fast an object rotates or revolves relative to another point. It is measured in radians per second (rad/s) and signifies the rate of change of the angular displacement. In the context of our problem, both scenarios consider the same angular velocity, which means the rate at which the rod spins is unchanged despite alterations in axis positioning.

However, it’s important to remember:

• Angular velocity remains constant in both configurations.
• This constancy allows us to directly compare the rotational kinetic energies without stressing over fluctuation in spinning speed.

Thus, although angular velocity isn’t varied in the exercise, it is an essential component of the rotational kinetic energy equation.

Parallel Axis Theorem

The Parallel Axis Theorem provides a handy tool in calculating the moment of inertia of an object about any axis, given the moment of inertia about a parallel axis through its center of mass. This theorem is particularly useful in engineering and physics when faced with objects of standard shapes rotated around different axes.

Mathematically, it is represented as:\[ I = I_{cm} + Md^2 \]where:

• \( I_{cm} \) is the moment of inertia through the object's center of mass.
• \( M \) is the mass of the object.
• \( d \) is the distance between the center of mass axis and the new axis.

In the case of the rod, applying the theorem confirmed our moment of inertia calculations for the axis through one end, as seen with \( \frac{1}{3}ML^2 \), where \( d = \frac{L}{2} \). This powerful theorem simplifies the mathematics involved in solving rotational dynamics problems.

Ratio of Kinetic Energy

The concept of the ratio of kinetic energy hinges on comparing energy states under varying conditions. Here, we determine how much more or less kinetic energy the rod possesses when rotated about different axes, considering the same angular velocity.

For rotational kinetic energy calculations:

• When the rod spins about its center, energy is calculated by \( KE_1 = \frac{1}{24} ML^2 \omega^2 \).
• For spinning about its end, \( KE_2 = \frac{1}{6} ML^2 \omega^2 \).

By evaluating the ratio \( \frac{KE_1}{KE_2} \), we find it equals \( \frac{1}{4} \). This reveals that:

• The kinetic energy in the first scenario is one-fourth of that in the second scenario.
• The difference arises because the rod's mass distribution further from the axis in the second case increases energy despite identical \( \omega \).

This approach clarifies energy transitions and the practical impact of shifting rotational axes.