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Chapter 11: Problem 48

A uniform stick \(1.0 \mathrm{~m}\) long with a total mass of \(270 \mathrm{~g}\) is pivoted at its center. A 3.0 -g bullet is shot through the stick midway between the pivot and one end (Fig. \(11-36\) ). The bullet approaches at \(250 \mathrm{~m} / \mathrm{s}\) and leaves at \(140 \mathrm{~m} / \mathrm{s}\). With what angular speed is the stick spinning after the collision?

Short Answer

Expert verified
The angular speed of the stick is \(7.33 \, \text{rad/s}\).

Step by step solution

01

Determine the Moment of Inertia of the Stick

The moment of inertia (I) for a uniform rod pivoted at its center is given by the formula:\[ I = \frac{1}{12} m L^2 \]where \(m\) is the mass of the stick in kilograms and \(L\) is the length of the stick in meters. Here, \(m = 0.27\, \text{kg}\) and \(L = 1.0\, \text{m}\):\[ I = \frac{1}{12} \times 0.27 \times (1.0)^2 = 0.0225 \, \text{kg} \cdot \text{m}^2 \]
02

Calculate Angular Momentum Before the Collision

Angular momentum is conserved in the system. Calculate the angular momentum of the bullet before the collision. The angular momentum (L) is given by:\[ L = m v r \]where \(m = 0.003 \, \text{kg}\), \(v = 250 \, \text{m/s}\), and \(r = 0.5 \, \text{m}\) (half the length of the stick):\[ L_\text{initial} = 0.003 \times 250 \times 0.5 = 0.375 \, \text{kg} \cdot \text{m}^2/s \]
03

Calculate Angular Momentum After the Collision

Angular momentum of the bullet after the collision is:\[ L_\text{final} = m v_\text{final} r \]where \(v_\text{final} = 140 \, \text{m/s}\):\[ L_\text{final} = 0.003 \times 140 \times 0.5 = 0.21 \, \text{kg} \cdot \text{m}^2/s \]
04

Calculate Change in Angular Momentum

The change in angular momentum due to the bullet colliding with the stick is the difference between initial and final angular momenta:\[ \Delta L = L_\text{initial} - L_\text{final} = 0.375 - 0.21 = 0.165 \, \text{kg} \cdot \text{m}^2/s \]
05

Calculate Angular Speed of the Stick

The change in angular momentum is equal to the angular momentum of the stick after the collision. Using the formula \( \Delta L = I \omega \), solve for \( \omega \), which is the angular speed:\[ \omega = \frac{\Delta L}{I} = \frac{0.165}{0.0225} = 7.33 \, \text{rad/s} \]

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Key Concepts

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

Moment of Inertia
The moment of inertia is a fundamental concept in rotational dynamics, similar to mass in linear motion. It describes how difficult it is to change the rotational motion of an object. For objects of simple shape like a uniform rod, the moment of inertia depends on both the mass of the object and how that mass is distributed relative to the axis of rotation.
In our exercise, the uniform stick rotates around its center. To find its moment of inertia (I), we use the formula: \[ I = \frac{1}{12} m L^2 \]where \(m\) is the mass of the stick and \(L\) is its length. Plugging in the given values, we have a moment of inertia of 0.0225 \ ext{kg} \cdot \ ext{m}^2. The smaller this value, the easier it is to spin the stick.
Angular Speed
Angular speed is the rate at which an object rotates or spins. It tells us how quickly the angle is changing over time. In our context, after the bullet collides and passes through the stick, the stick begins to spin. This spin is characterized by its angular speed (\omega).

After determining that the change in angular momentum due to the bullet is 0.165 \ ext{kg} \cdot \ ext{m}^2/s, and knowing the moment of inertia (I) of the stick, we can calculate the stick's angular speed post-collision using:\[ \omega = \frac{\Delta L}{I} \]Plugging in the values, \(\omega\) comes out to be 7.33 rad/s. This indicates the rapidity with which the stick's angle changes as a result of the collision.
Collision Dynamics
Collision dynamics involves the study of how objects interact when they collide and how these interactions affect their motion. In our exercise, the collision is between a bullet and a pivoted stick, and we focus on the rotational effects instead of linear ones.
  • Conservation of Angular Momentum: This principle states that in the absence of external torques, the angular momentum of a system remains constant. Thus, the initial angular momentum of the bullet-sticking system equals the final angular momentum.
  • Change of State: Initially, the stick is at rest and the bullet in motion. After the collision, both the bullet exits at a slower speed, and the stick starts to rotate.
  • Energy Distribution: Although the bullet loses speed as it passes through the stick, part of its kinetic energy gets transferred to the stick, causing it to spin.
Understanding these dynamics helps us predict final outcomes like the stick's angular speed, and efficiently utilize principles like conservation of momentum.

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Most popular questions from this chapter

(II) (a) Show that the cross product of two vectors, $$ \vec{\mathbf{A}}=A_{x} \hat{\mathbf{i}}+A_{y} \hat{\mathbf{j}}+A_{z} \hat{\mathbf{k}}, \text { and } \vec{\mathbf{B}}=B_{x} \hat{\mathbf{i}}+B_{y} \hat{\mathbf{j}}+B_{z} \hat{\mathbf{k}} $$ $$ \begin{aligned} \vec{\mathbf{A}} \times \vec{\mathbf{B}}=&\left(A_{y} B_{z}-A_{z} B_{y}\right) \hat{\mathbf{i}}+\left(A_{z} B_{x}-A_{x} B_{z}\right) \hat{\mathbf{j}} \\ &+\left(A_{x} B_{y}-A_{y} B_{x}\right) \hat{\mathbf{k}} \end{aligned} $$ (b) Then show that the cross product can be written $$ \vec{\mathbf{A}} \times \vec{\mathbf{B}}=\left| \begin{array}{ccc}{\hat{\mathbf{i}}} & {\hat{\mathbf{j}}} & {\hat{\mathbf{k}}} \\\ {A_{x}} & {A_{y}} & {A_{z}} \\ {B_{x}} & {B_{y}} & {B_{z}}\end{array}\right| $$ where we use the rules for evaluating a determinant. (Note, however, that this is not really a determinant, but a memory aid.)

(1) What are the \(x, y,\) and \(z\) components of the angular momentum of a particle located at \(\vec{\mathbf{r}}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}\) which has momentum \(\mathbf{p}=p_{x} \mathbf{i}+p_{y} \mathbf{j}+p_{z} \mathbf{k} ?\)

The Moon orbits the Earth such that the same side always faces the Earth. Determine the ratio of the Moon's spin angular momentum (about its own axis) to its orbital angular momentum. (In the latter case, treat the Moon as a particle orbiting the Earth.)

The position of a particle with mass \(m\) traveling on a helical path (see Fig. \(11-45\) ) is given by $$ \overrightarrow{\mathbf{r}}=R \cos \left(\frac{2 \pi z}{d}\right) \hat{\mathbf{i}}+R \sin \left(\frac{2 \pi z}{d}\right) \hat{\mathbf{j}}+z \hat{\mathbf{k}} $$ where \(R\) and \(d\) are the radius and pitch of the helix, respectively, and \(z\) has time dependence \(z=v_{z} t\) where \(v_{z}\) is the (constant) component of velocity in the \(z\) direction. Determine the time-dependent angular momentum \(\overrightarrow{\mathbf{L}}\) of the particle about the origin.

Figure \(11-39\) shows a thin rod of mass \(M\) and length \(\ell\) resting on a frictionless table. The rod is struck at a distance \(x\) from its CM by a clay ball of mass \(m\) moving at speed \(v\). The ball sticks to the rod. (a) Determine a formula for the rotational motion of the system after the collision. (b) Graph the rotational motion of the system as a function of \(x,\) from \(x=0\) to \(x=\ell / 2,\) with values of \(M=450 \mathrm{~g}, \quad m=15 \mathrm{~g}\) \(\ell=1.20 \mathrm{~m},\) and \(v=12 \mathrm{~m} / \mathrm{s} .\) (c) Does the translational motion depend on \(x ?\) Explain.
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