Question

If the iron core of a collapsing star initially spins with a rotational frequency of \(f_{0}=3.2 \mathrm{~s}^{-1},\) 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) \(0.46 \mathrm{kHz}\) e) \(5.2 \mathrm{kHz}\)

Short answer

The initial rotational frequency is 3.2 s^{-1}. Answer: b) \(1.66 \mathrm{kHz}\)

Step-by-step solution

Identify the relevant concepts for the problem

In this problem, we need to find the new rotational frequency of an iron core after its radius decreases. To do this, we will use the conservation of angular momentum, which states that the initial and final angular momenta of the system must be equal.

Recall the formula for angular momentum

Angular momentum (L) is given by the product of the moment of inertia (I) and the angular velocity (ω). Thus, \(L = Iω\). For a solid sphere, the moment of inertia is given by: \(I = \dfrac{2}{5}MR^2\), where M is the mass and R is the radius of the sphere.

Apply conservation of angular momentum

Since the initial and final angular momenta are equal, we can write the conservation of angular momentum as: \(I_{0}ω_{0} = I_{f}ω_{f}\) Where \(I_{0}\) and \(I_{f}\) are the initial and final moments of inertia, and \(ω_{0}\) and \(ω_{f}\) are the initial and final angular velocities, respectively. Since the mass of the iron core is constant during the collapse, we can substitute the formula for the moment of inertia of a solid sphere: \((\dfrac{2}{5}MR_{0}^2)ω_{0} = (\dfrac{2}{5}MR_{f}^2)ω_{f}\) We can cancel out the mass (\(M\)) and the constant factor \(\dfrac{2}{5}\): \(R_{0}^2ω_{0} = R_{f}^2ω_{f}\)

Solve for the final angular velocity and convert to frequency

We are given that the initial rotational frequency is \(f_{0} = 3.2 s^{-1}\). Since \(ω = 2πf\), we can find the initial angular velocity: \(ω_{0} = 2πf_{0} = 2π (3.2 s^{-1})\) We are also given that the radius of the iron core decreases by a factor of 22.7 during the collapse. If we let \(R_{0}\) be the initial radius and \(R_{f}\) be the final radius, then we have: \(R_{f} = \dfrac{R_{0}}{22.7}\) Now, we can substitute the given information into the equation derived from the conservation of angular momentum: \(R_{0}^2(2π (3.2 s^{-1})) = (\dfrac{R_{0}}{22.7})^2ω_{f}\) We can cancel out the initial radius \(R_{0}^2\): \(2π(3.2 s^{-1}) = (\dfrac{1}{22.7^2})ω_{f}\) Solve for the final angular velocity, \(ω_{f}\): \(ω_{f} = 22.7^2(2π(3.2 s^{-1}))\) Now we need to convert the final angular velocity to rotational frequency by using \(f = \dfrac{ω}{2π}\): \(f_{f} = \dfrac{ω_{f}}{2π} = \dfrac{22.7^2(2π(3.2 s^{-1}))}{2π} \approx 1.66 \times 10^3 s^{-1}\)

Choose the correct answer

The final rotational frequency is approximately \(1.66 \times 10^3 s^{-1}\) or \(1.66 kHz\). Therefore, the correct answer is: b) \(1.66 \mathrm{kHz}\)

Key concepts

Rotational Frequency

Rotational frequency, often represented as 'f', is the number of full rotations that an object makes in one second. It is a measure of how fast an object is spinning. In the context of our exercise, we are examining the change in rotational frequency of an iron core star as it collapses. The collapsing action typically increases rotational frequency, akin to how a figure skater speeds up when pulling their arms in. This phenomenon occurs because of the conservation of angular momentum. To comprehend the alteration in rotational frequency, one must consider the initial frequency, which gives a substantial foundation for predicting the final rotational frequency following the reduction in radius of the star's core.

Understanding how rotational frequency is affected by changes in an object’s moment of inertia (a concept we’ll delve into shortly) is essential for physics students studying dynamic systems. When an object conserves its angular momentum, as is the case here, any decrease in radius brings about a proportional increase in rotational frequency. The formula connecting frequency (f) with angular velocity (ω) is given by the relationship:
\( f = \frac{\omega}{2\pi} \).
Using this, with the knowledge of an unchanged angular momentum, allows for the calculation of the new frequency after such a transformation.

Moment of Inertia

The moment of inertia, symbolized by 'I', is the rotational equivalent of mass in linear motion. It quantifies an object's resistance to changes in its rotational motion. The greater the distance of the object's mass from the axis of rotation, the larger its moment of inertia will be. For our collapsing star, the moment of inertia changes as the radius of the iron core decreases. The formula for the moment of inertia 'I' of a solid sphere is:
\( I = \frac{2}{5}MR^2 \),
where 'M' is the object's mass and 'R' its radius. It’s crucial for students to note that while the star collapses, its mass remains constant, but the moment of inertia decreases because the radius decreases.

In the exercise, we utilize the moment of inertia to equate the initial and final states of angular momentum. As the moment of inertia decreases, our angular velocity must increase to preserve the angular momentum (assuming no external torques act on the system). This interplay between moment of inertia and angular velocity is at the crux of understanding rotational dynamics and pivotal for solving this type of problem.

Angular Velocity

Angular velocity, denoted by the Greek letter 'ω' (omega), is the rate at which an object rotates around an axis. It is the angular displacement per unit time and is usually expressed in radians per second (rad/s). Angular velocity can be related to the rotational frequency by the formula:
\( \omega = 2\pi f \).
Importantly, while the rotational frequency describes the number of rotations per second, angular velocity complements this by supplying the rate of rotation irrespective of the size of the circle or the object rotating.

In our star collapse scenario, angular velocity increases dramatically as the radius of the iron core diminishes. By utilizing conservation of angular momentum (which can be intuitively understood as the object’s 'rotational oomph'), we deduce that if the moment of inertia drops, the angular velocity must rise correspondingly to keep the system’s angular momentum unchanged. This surge in angular velocity is subsequently translated into a significant increase in rotational frequency, explaining the observed speed up in our collapsed iron core.