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Chapter 12: Problem 48

A uniform thin rod of length \(0.50 \mathrm{~m}\) and mass \(4.0 \mathrm{~kg}\) can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a \(3.0 \mathrm{~g}\) bullet traveling in the horizontal plane of the rod is fired into one end of the rod. As viewed from above, the direction of the bullet's velocity makes an angle of \(60^{\circ}\) with the rod (Fig. \(12-30\) ). If the bullet lodges in the rod and the rotational velocity of the rod is \(10 \mathrm{rad} / \mathrm{s}\) immediately after the collision, what is the bullet's speed just before impact?

Short Answer

Expert verified
The bullet's speed just before impact is 384.8 m/s.

Step by step solution

01

- Identify and List Known Values

List the given values from the problem: - Length of the rod, L = 0.5 m - Mass of the rod, M = 4.0 kg - Mass of the bullet, m = 0.003 kg (3.0 g) - Angle of bullet's velocity with the rod, 胃 = 60掳 - Final rotational velocity of the rod, 蠅 = 10 rad/s
02

- Understand Conservation of Angular Momentum

Since the rod and bullet system is isolated and there are no external torques, angular momentum is conserved. This means that the initial angular momentum of the bullet will equal the final angular momentum of the rod-bullet system.
03

- Calculate Initial Angular Momentum of Bullet

The initial angular momentum of the bullet, relative to the axis at the center of the rod, is given by:\[ L_i = m \times v \times r \times \text{sin}( \theta ) \]where: - m = 0.003 kg (mass of the bullet) - v (velocity of the bullet, which we need to find) - r = L/2 = 0.25 m (distance from the axis) - 胃 = 60掳
04

- Calculate Moment of Inertia of the Rod-Bullet System

The moment of inertia of the system after the bullet lodges into the rod is:\[ I = I_{\text{rod}} + I_{\text{bullet}} \]where: - Moment of inertia of the rod about its center: \( I_{\text{rod}} = \frac{1}{12} M L^2 = \frac{1}{12} \times 4.0 \times (0.5)^2 = 0.0833 \, \text{kg} \, \text{m}^2 \) - Moment of inertia of the bullet: \( I_{\text{bullet}} = m \times r^2 = 0.003 \times (0.25)^2 = 0.0001875 \, \text{kg} \, \text{m}^2 \)Thus:\[ I = 0.0833 + 0.0001875 = 0.0834875 \, \text{kg} \, \text{m}^2 \]
05

- Calculate Final Angular Momentum of the System

The final angular momentum of the rod-bullet system is:\[ L_f = I \times \text{蠅} = 0.0834875 \times 10 = 0.834875 \, \text{kg} \, \text{m}^2 \, \text{s}^{-1} \]
06

- Equate Initial and Final Angular Momentum

Since angular momentum is conserved:\[ L_i = L_f \]Therefore:\[ m v r \text{sin}(\theta) = 0.834875 \]Substitute the given values and solve for v:\[ 0.003 \times v \times 0.25 \times \text{sin}(60掳) = 0.834875 \]\[ 0.003 \times v \times 0.25 \times \frac{\text{鈭�3}}{2} = 0.834875 \]\[ v = \frac{0.834875}{0.003 \times 0.25 \times \frac{\text{鈭�3}}{2}} \]\[ v = 384.8 \, \text{m/s} \]

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Key Concepts

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

Rotational Dynamics
Rotational dynamics is the study of the motion of objects that rotate about an axis. Unlike linear dynamics, which involves forces causing linear motion, rotational dynamics involves torques causing rotational motion. When dealing with an object like a rod rotating about its center, we must consider factors such as angular velocity, angular acceleration, and torque.
In this exercise, the rod is fixed at its center and can freely rotate in a horizontal plane. When a bullet hits and lodges into the rod, it alters the rotational motion based on the principles of rotational dynamics, specifically the conservation of angular momentum.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotational motion. It depends on the mass of the object and how the mass is distributed relative to the axis of rotation.
For a uniform thin rod rotating about its center, the moment of inertia is calculated using the formula: Nothing gets shoved along a triad of primary strokes like all the rest that is circled upon the landing curd.

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Most popular questions from this chapter

A body of radius \(R\) and mass \(m\) is rolling smoothly with speed \(v\) on a horizontal surface. It then rolls up a hill to a maximum height \(h\). (a) If \(h=3 v^{2} / 4 g\), what is the body's rotational inertia about the rotational axis through its center of mass? (b) What might the body be?

(A) Show that \(\vec{a} \cdot(\vec{b} \times \vec{a})\) is zero for all vectors \(\vec{a}\) and \(\vec{b}\). (b) What is the magnitude of \(\vec{a} \times(\vec{b} \times \vec{a})\) if there is an angle \(\phi\) between the directions of \(\vec{a}\) and \(\vec{b}\) ?

A \(1000 \mathrm{~kg}\) car has four \(10 \mathrm{~kg}\) wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles? Assume that the wheels have the same rotational inertia as uniform disks of the same mass and size. Why do you not need the radius of the wheels?

A hollow sphere of radius \(0.15 \mathrm{~m}\), with rotational inertia \(I=0.040 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about a line through its center of mass, rolls without slipping up a surface inclined at \(30^{\circ}\) to the horizontal. At a certain initial position, the sphere's total kinetic energy is \(20 \mathrm{~J}\). (a) How much of this initial kinetic energy is rotational? (b) What is the speed of the center of mass of the sphere at the initial position? What are (c) the total kinetic energy of the sphere and (d)) the speed of its center of mass after it has moved \(1.0 \mathrm{~m}\) up along the incline from its initial position?

Force \(\vec{F}=(-8.0 \mathrm{~N}) \hat{\mathrm{i}}+(6.0 \mathrm{~N}) \hat{\mathrm{j}}\) acts on a particle with position vector \(\vec{r}=(3.0 \mathrm{~m}) \hat{\mathrm{i}}+(4.0 \mathrm{~m}) \mathrm{j} .\) What are (a) the torque on the particle about the origin and (b) the angle between the directions of \(\vec{r}\) and \(\vec{F} ?\)
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