Problem 46 - Fundamentals of Physics

Question

The rotational inertia of a collapsing spinning star drops to \(\frac{1}{3}\) its initial value. What is the ratio of the new rotational kinetic energy to the initial rotational kinetic energy?

Step-by-Step Solution

Answer

The new kinetic energy is three times the initial energy.

Step 1: Understanding Rotational Inertia and Kinetic Energy

Rotational inertia (moment of inertia) decreases to \(\frac{1}{3}\) of its initial value when the star collapses. The rotational kinetic energy is given by \(K = \frac{1}{2} I \omega^2\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity. Since the system is isolated, angular momentum \(L = I\omega\) remains constant.

Step 2: Conserving Angular Momentum

Given the angular momentum \(L\) is constant, initially \(L_0 = I_0 \omega_0\) and after collapse \(L = I_f \omega_f\). So, \(I_0 \omega_0 = I_f \omega_f\), meaning \(\omega_f = \frac{I_0}{I_f} \omega_0\). Since \(I_f = \frac{1}{3} I_0\), \(\omega_f = 3 \omega_0\).

Step 3: Finding the New Rotational Kinetic Energy

The initial rotational kinetic energy is \(K_i = \frac{1}{2} I_0 \omega_0^2\). The new kinetic energy is \(K_f = \frac{1}{2} I_f \omega_f^2 = \frac{1}{2} \left(\frac{1}{3} I_0\right) (3 \omega_0)^2\). Simplifying this gives \(K_f = \frac{1}{2} \times \frac{1}{3} I_0 \times 9 \omega_0^2 = \frac{3}{2} I_0 \omega_0^2\).

Step 4: Calculating the Ratio of Kinetic Energies

The ratio of the new to the initial kinetic energy is \(\frac{K_f}{K_i} = \frac{\frac{3}{2} I_0 \omega_0^2}{\frac{1}{2} I_0 \omega_0^2} = 3\).

Key Concepts

Rotational Inertia

Rotational inertia, also known as the moment of inertia, measures how difficult it is to change the rotational motion of an object. Imagine trying to spin a heavy solid disc compared to a lightweight hoop. The solid disc resists changes to its rotation more because of its higher rotational inertia. In the context of a collapsing star, as the star shrinks, its mass distribution changes. The rotational inertia decreases, which means it's easier for the star to spin faster. In our example, the rotational inertia drops to \(\frac{1}{3}\) of its initial value. This significant change influences the star's angular velocity, highlighting the importance of rotational inertia in rotational dynamics.

Angular Momentum Conservation

Angular momentum is a key player in rotational dynamics. It combines rotational inertia and angular velocity, described by the formula \(L = I \omega\). One amazing property of angular momentum is its conservation in isolated systems. This means, unless acted on by an external force, the total angular momentum remains constant over time. For the collapsing star, its initial angular momentum \(I_0 \omega_0\) is the same as the final angular momentum \(I_f \omega_f\). When the star's rotational inertia decreases to \(\frac{1}{3} I_0\), its angular velocity must increase to conserve angular momentum. This results in \(\omega_f = 3 \omega_0\). Thus, a smaller inertial system spins much faster to balance out angular momentum.

Rotational Kinetic Energy

Rotational kinetic energy is the energy due to the rotation of an object. It is calculated using the equation \(K = \frac{1}{2} I \omega^2\). This formula shows that kinetic energy depends not only on how fast something spins (angle velocity) but also on how the mass is spread out from the axis of rotation (rotational inertia). For the star, we had to find the ratio of the new rotational kinetic energy to its initial energy. Initially, the energy \(K_i\) was \(\frac{1}{2} I_0 \omega_0^2\). After collapsing, the energy \(K_f\) becomes \(\frac{1}{2} \times \frac{1}{3} I_0 \times (3 \omega_0)^2 = \frac{3}{2} I_0 \omega_0^2\). The ratio \(\frac{K_f}{K_i} = 3\) indicates that the rotational kinetic energy has tripled. As the star spins faster, its energy of rotation greatly increases, demonstrating the effects of its enhanced angular velocity and reduced inertia.