Question

If the iron core of a collapsing star initially spins with a rotational frequency of \(f_{0}=3.20 \mathrm{~s}^{-1},\) and if the core's radius decreases during the collapse by a factor of \(22.7,\) what is the rotational frequency of the iron core at the end of the collapse? a) \(10.4 \mathrm{kHz}\) b) \(1.66 \mathrm{kHz}\) c) \(65.3 \mathrm{kHz}\) d) \(0.460 \mathrm{kHz}\) e) \(5.20 \mathrm{kHz}\)

Short answer

Answer: 1.66 kHz

Step-by-step solution

1. Identify angular momentum conservation in the system

The angular momentum of a system remains conserved if external torque doesn't act on it. Since this is a star's core collapsing, we can assume that there's no external torque acting on it. Therefore, the angular momentum of the core remains conserved during the process of collapsing.

2. Write the formula for angular momentum of the core

For a rotating object, angular momentum "L" can be written as: \[L = I \omega\] Where "I" is the moment of inertia and "ω" is the angular velocity. Since the moment of inertia for a solid sphere is given as: \[I = \frac{2}{5}mr^2\] And the relation between angular velocity and frequency is: \[\omega = 2\pi f\] We can rewrite the equation of angular momentum as: \[L = \frac{2}{5}m(2\pi f)r^2\]

3. Apply conservation of angular momentum to the system

According to the conservation of angular momentum, the initial and final angular momenta should be equal. So we can write: \[L_0 = L_f\] Substituting the equation for angular momentum, we get: \[\frac{2}{5}m(2\pi f_0)r_0^2 = \frac{2}{5}m(2\pi f_f)r_f^2\]

4. Simplify the equation and isolate the final frequency

Since we are only looking for the final frequency \(f_f\), we can cancel out the common terms and rearrange the equation to find \(f_f\). Dividing both sides by \(\frac{2}{5}m(2\pi)\) and rearranging the equation we get: \[f_f = f_0 \frac{r_0^2}{r_f^2}\]

5. Calculate the final frequency using the given values

Now, we can plug the given values into the equation. The initial frequency \(f_0 = 3.20 \thinspace s^{-1}\), and the radius decreases by a factor of \(22.7\), which means \(r_f = \frac{r_0}{22.7}\). Substituting these values to the equation and calculating the final frequency \(f_f\), we get: \[f_f = 3.20 \frac{r_0^2}{\left(\frac{r_0}{22.7}\right)^2}\] \[f_f = 3.20 \cdot 22.7^2\] \[f_f = 1.66 \times 10^3 \thinspace s^{-1} \] The final rotational frequency of the iron core at the end of the collapse is \(1.66 \mathrm{kHz}\). So, the correct answer is (b) \(1.66 \mathrm{kHz}\).

Key concepts

Angular Momentum Conservation

One fundamental principle in physics that keeps the cosmos in check is angular momentum conservation. When a star collapses, the awe-inspiring process is governed by this natural law. Simply put, angular momentum is a measure of how much rotation an object has. It's like when a figure skater pulls in her arms and spins faster; she's not pulling some magic trick, she's conserving her angular momentum. This happens because when she pulls her arms in, her moment of inertia decreases but her angular momentum doesn't, so her angular velocity must increase to compensate. This is the same principle that dictates the accelerated spin of a collapsing star's core.

In a star's core collapse, there's no hidden hand in space applying extra twists or turns (external torque), so the core's angular momentum before and after the collapse must remain the same. To calculate how fast the core will be spinning after collapsing, we put our trust in this conservation law.

Moment of Inertia

The moment of inertia might sound daunting, but think of it as simply the rotational version of mass. It's a property that defines how much resistance an object has to change in its spinning rate. Imagine trying to push a merry-go-round; the more it weighs and the further you are from the center, the harder it is to spin. That's because its moment of inertia is higher.

For a star's core, which is roughly spherical in shape, the moment of inertia depends on its mass and radius, growing with the square of the radius. That's why when the star collapses and its radius shrinks, the core's moment of inertia drops dramatically. This reduction allows the core's angular velocity to skyrocket to maintain the angular momentum conserved, as seen in the step-by-step solution of our collapsing star's core.

Angular Velocity

Angular velocity, denoted by the symbol \(\omega\), is a vector quantity that describes the rate at which an object rotates about an axis. If you've ever watched a top spin or a planet revolve, you've observed angular velocity in action. It's measured in radians per second and can be related to the more familiar concept of frequency, which tells you how many rotations an object completes in one second.

The faster the angular velocity, the faster the object is spinning. For the iron core of the star in our exercise, as it collapses and its radius decreases, the core needs to spin more rapidly to keep the angular momentum unchanged. The formula that connects angular velocity and frequency helps us calculate just how much faster the core will spin at the end of its collapse. This is pivotal for understanding not only spinning stars but also many rotating systems like wheels, planets, and even galaxies.