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Chapter 4: Problem 10

A bullet of mass \(m\) is fired with a velocity \(v\) into a wooden disk of mass \(M\) and radius \(R\) that is free to rotate, as shown. The bullet lodges itself in the disk at a height \(b\) above the wheel's center and the system begins to rotate with an angular velocity \(\omega\). Assume \(M \gg m\). In order to predict \(\omega\), one needs (A) conservation of energy. (B) conservation of momentum. (C) conservation of angular momentum. (D) conservation of energy and conservation of momentum. (E) conservation of energy and conservation of angular momentum.

Short Answer

Expert verified
Answer: (C) conservation of angular momentum.

Step by step solution

01

Understand what physical quantity remains constant during the problem

In this case, we are asked to determine the final angular velocity of the wooden disk with the bullet lodged in it. Knowing what physical quantity remains constant throughout the system's movement will help us find the appropriate conservation law(s).
02

Analyze each conservation law

Now, let's analyze each conservation law and see if it applies to our scenario. (A) Conservation of energy states that the sum of kinetic, potential, and thermal energies remains constant. In this problem, we have no loss of energy accounted for, and thus energy is conserved. However, this alone may not be enough to determine the angular velocity ω. (B) Conservation of momentum states that the total momentum of a closed system remains constant. The initial momentum of the system is due to the incoming bullet only, while the final momentum of the system is shared between the bullet and the wooden disk. Since there are no external forces in the horizontal direction (parallel to the plane of rotation), the linear momentum must be conserved. (C) Conservation of angular momentum states that the total angular momentum of a closed system remains constant. The system is closed, and there is no net external torque acting on it, hence the angular momentum is conserved. Since we require a final angular velocity, ω, this conservation law is crucial. (D) Conservation of energy and conservation of momentum: While both energy and momentum are conserved, they alone may not provide enough information to determine the angular velocity ω. (E) Conservation of energy and conservation of angular momentum: In this case, the conservation of angular momentum is critical to determine the final angular velocity ω. The conservation of energy is not directly helpful, but it indicates that no energy is lost in the process.
03

Choose the appropriate conservation law

Based on our analysis, the conservation law needed for determining the angular velocity ω is: - (C) conservation of angular momentum. The answer is (C). Using conservation of angular momentum, we can predict the final angular velocity, ω, of the wooden disk and bullet system after the bullet lodges into the disk.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Velocity
Angular velocity, represented by the symbol \(\omega\), is a measure of the rate of rotation of an object. In the context of our given exercise, angular velocity refers to how quickly the wooden disk and the lodged bullet rotate around their axis after the bullet embeds itself in the disk. This physical quantity is directly related to the conservation of angular momentum.

For an object rotating about a fixed axis, angular velocity is expressed as the angle \(\theta\) rotated per unit of time. The formula to calculate angular velocity is \(\omega = \frac{\Delta \theta}{\Delta t}\), where \(\Delta \theta\) represents the change in the angular position and \(\Delta t\) is the time interval. It's important to note that the angular velocity must align with the axis of rotation and adhere to the right-hand rule, which determines the direction of positive angular velocity.
Physical Quantity Conservation
The principle of physical quantity conservation refers to the concept that certain measurable properties of isolated systems remain constant over time. Conservation laws are foundational in physics because they describe a property that does not change, even as the system evolves. For our problem, we are dealing with the conservation of angular momentum, which is the rotational equivalent of linear momentum.

Angular momentum, denoted \(L\), is calculated as the product of the moment of inertia \(I\) and angular velocity \(\omega\). It can also be viewed as the rotational momentum of an object, and it remains constant in an isolated system unless acted upon by an external torque. Mathematically, \(L = I\omega\) highlights the relationship between angular momentum, the distribution of mass in an object (\((I)\), and its angular velocity. With no external torques applied, \(L\) before an event equals \(L\) after, enforcing conservation.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. All moving objects have kinetic energy, which is dependent on their mass and the square of their velocity. In the linear form, kinetic energy \(KE\) can be expressed by the equation \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass of the object and \(v\) is its velocity.

In rotational dynamics, kinetic energy also exists and is defined similarly through angular variables. The rotational kinetic energy of a rigid body rotating about its axis is given by \(KE_{rot} = \frac{1}{2}I\omega^2\), where \(I\) is the object's moment of inertia and \(\omega\) is the angular velocity. This type of kinetic energy plays a significant role when analyzing situations where objects transition between rotational and translational motion, though in the problem from the textbook, the focus is more on the conservation of angular momentum than on kinetic energy alone.

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Most popular questions from this chapter

A gyroscope is essentially a spinning wheel attached to an axle that is free to rotate in any direction about a pivot. In the figure below, a gyro wheel of mass \(M\) and radius \(R\) spins with angular velocity \(\omega\) in a counterclockwise direction when viewed from the right. The distance between the wheel and the pivot \(O\) is \(D\). Assume that the axle is of negligible mass and that the wheel's entire mass is concentrated around the rim. a) Is there an axis of symmetry in the problem? If so what is it? b) What is the magnitude of the angular momentum of the gyroscope? Draw the direction the angular momentum vector is pointing on the diagram. c) Are there any external forces acting on the gyro wheel? If so, draw them on the diagram. d) What is the magnitude of any torque acting on the wheel? Draw the direction of the torque on the diagram. e) Does the gyroscope move in any direction? If so which?

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A figure skater can be idealized as composed of two solid cylinders of uniform density crossed at right angles to one another. Assume that the skater is spinning with angular speed \(\omega\) around a vertical axis passing through the center of her body. The mass of the large cylinder is \(M\); it has radius \(R\) and length \(L\). The mass of the small cylinder is \(m\); its radius is \(r\) and its length is \(\ell\). a) What is the moment of inertia of the large vertical cylinder in terms of the given quantities? b) What is the moment of inertia of the horizontal cylinder in terms of the given quantities? c) What is the figure skater's initial angular momentum? d) If \(m=0.1 M, \ell=8 R\) and the skater's arms can be totally retracted to her body, what is her new angular speed in terms of \(\omega\) ?

Which of the following could be a unit of angular momentum? i. \(\mathrm{Nm} \mathrm{S}\) ii. \(\sqrt{\mathrm{Jkg}} \mathrm{m}\) iii. \(\mathrm{W} \mathrm{s}^2\) iv. \(J / m\) v. \(W s^2 / m\) (A) \(\mathrm{i}\) (B) ii (C) iii (D) i, ii and iii (E) i, ii, iii, iv, V
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