/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 97 A bullet weighing 0.08 lb is fir... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 97

A bullet weighing 0.08 lb is fired with a horizontal velocity of \(1800 \mathrm{ft}\) s into the lower end of a slender \(15-\mathrm{lb}\) bar of length \(L=30 \mathrm{in}\). Knowing that \(h=12 \mathrm{in}\), and that the bar is initially at rest, determine \((a)\) the angular velocity of the bar immediately after the bullet becomes embedded, (b) the impulsive reaction at \(C,\) assuming that the bullet becomes embedded in \(0.001 \mathrm{s}\).

Short Answer

Expert verified
(a) Use conservation of angular momentum to find angular velocity; (b) calculate impulsive force using momentum change.

Step by step solution

01

Understand the System

The problem involves a bullet embedding in a bar, resulting in rotation. We need to find the angular velocity of the bar and the impulsive reaction at the point where the bar is pivoted (C).
02

Convert Units

Convert the units of the bar's length and height to feet so we have consistent units.- Bar Length, \( L = 30 \text{ in} = 2.5 \text{ ft} \).- Height, \( h = 12 \text{ in} = 1 \text{ ft} \).
03

Calculate Initial Linear Momentum

The initial linear momentum of the bullet is given by \( p = mv \), where:- Mass of bullet, \( m = 0.08 \text{ lb} / g = 0.08 \times \frac{1}{32.2} \text{ slug} \),- Velocity, \( v = 1800 \text{ ft/s} \).Calculate the initial momentum:\[ p = 0.08 \times \frac{1}{32.2} \times 1800 \text{ slug ft/s} \].
04

Apply Conservation of Angular Momentum

Since there is no external torque acting on the system about C, apply conservation of angular momentum about point C:The initial angular momentum of the bullet about C:\[ L_{ ext{initial}} = p \cdot h \].Final angular momentum involves the bar and embedded bullet:\[ \left( I_{bar} + m h^2 \right) \omega = L_{ ext{initial}} \],where \( I_{bar} = \frac{1}{3}m_bL^2\), \( m_b = 15/32.2 \text{ slug} \), and \( \omega \) is the angular velocity.
05

Solve for Angular Velocity \(\omega\)

Calculate moment of inertia of the bar:\[ I_{bar} = \frac{1}{3} \times \frac{15}{32.2} \times (2.5)^2 \].Plug values into the angular momentum equation and solve for \( \omega \):\[ \omega = \frac{ p h }{ I_{bar} + mh^2 } \].
06

Calculate Force at C using Impulse

Using the change in momentum over the short interval, find the impulsive reaction at C. The force \( F_C \):\[ F_C \cdot 0.001 = m_b \cdot v_f - p \],where \( v_f = \omega \cdot L \),and solve for \( F_C \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Conservation of Angular Momentum
Conservation of angular momentum is a fundamental principle in physics. It states that if no external torque acts on a system, the total angular momentum of the system remains constant. In the context of this exercise, the system involves a bullet embedding into a bar. Let's break down the essentials:
  • Angular momentum is the rotational counterpart to linear momentum.
  • It depends on the object's rotational speed and how mass is distributed around the axis of rotation.
  • The conservation law allows us to equate the initial angular momentum with the final angular momentum, simplifying complex rotational problems.
Initially, the bullet has linear momentum, which is transformed into angular momentum when it embeds into the bar. The absence of external torques at the pivot point C validates the use of this conservation principle. Thus, we equate the bullet's initial linear momentum to the system's final angular momentum to find the angular velocity post-collision.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotational motion. In this exercise, it plays a crucial role in determining how the bar reacts when the bullet becomes embedded.
  • The formula for calculating the moment of inertia of a slender rod rotating about the end is: \[ I_{bar} = \frac{1}{3} m_b L^2 \] where \( m_b \) is the mass of the bar and \( L \) is its length.
  • A higher moment of inertia means more resistance to changes in rotational speed.
For the bar in this problem, we first calculate its moment of inertia and then consider the added effect of the embedded bullet. By including the bullet's influence, we adjust the system's moment of inertia to reflect both the mass of the bar and the location and mass of the bullet. This helps us determine the new angular velocity after the impact.
Impulsive Reaction
An impulsive reaction refers to a force exerted over a short period of time, causing a change in momentum. In this problem, the impulsive reaction occurs at point C when the bullet embeds in the bar. Here's how we approach it:
  • Impulse is defined by the equation: \[ F_C \cdot \Delta t = \Delta p \] where \( F_C \) is the impulsive force at point C, \( \Delta t \) is the time duration, and \( \Delta p \) is the change in momentum.
  • The problem states the bullet gets embedded in 0.001 seconds, which gives us the \( \Delta t \) value.
  • The change in momentum (\( \Delta p \)) is calculated from the initial and final conditions of the bullet-bar system.
Since the bullet-bar interaction is brief, the impulse allows us to calculate the impulsive reaction force at C, giving crucial insights into the immediate forces experienced by the system as the bullet embeds.
Linear Momentum
Linear momentum is the product of an object's mass and velocity. It is a vector quantity, having both magnitude and direction. In this exercise, linear momentum plays a pivotal role right at the start, when the bullet is fired.
  • Defined by \( p = mv \), where \( m \) is mass and \( v \) is velocity.
  • In our problem, the bullet's initial linear momentum is calculated using its mass (converted from pounds to slugs) and its velocity.
  • Understanding the bullet's momentum is crucial; it sets the initial conditions for applying conservation of angular momentum.
This initial linear momentum transforms into angular momentum as the bullet comes to embed in the bar. By equating it to the bar's rotational dynamics post-impact, we ensure all momentum transfers are accounted for accurately, giving a clear picture of the system's behavior post-collision.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A satellite has a total weight (on Earth) of 250 lbs, and each of the solar panels weighs 15 lbs. The body of the satellite has mass moment of inertia about the z-axis of 6 slug-ft \(^{2}\), and the panels can be modeled as flat plates. The satellite spins with a rate of 10 rpm about the z-axis when the solar panels are positioned in the xy plane. Determine the spin rate about z after a motor on the satellite has rotated both panels to be positioned in the yz plane.

An adapted golf device attaches to a wheelchair to help people with mobility impairments play putt-putt. The stationary frame OD is attached to the wheelchair, and a club holder OB is attached to the pin at O. Holder OB is 6 in. long and weighs 8 oz, and the distance between \(O\) and \(D\) is \(x=1\) ft. The putter shaft has a length of \(f\) \(L=36\) in. and weighs 10 oz, while the putter head at \(A\) weighs 12 oz. Knowing that the 1 -lb/in. spring between \(D\) and \(B\) is unstretched when \(\theta=90^{\circ}\) and that the putter is released from rest at \(\theta=0\), determine the putter head speed when it hits the golf ball.

The rotor of an electric motor has an angular velocity of 3600 rpm when the load and power are cut off. The 110-lb rotor, which has a centroidal radius of gyration of 9 in., then coasts to rest. Knowing that the kinetic friction of the rotor produces a couple with a magnitude of 2.5 lb?ft, determine the number of revolutions that the rotor executes before coming to rest.

The rotor of an electric motor has a mass of 25 kg, and it is observed that 4.2 min is required for the rotor to coast to rest from an angular velocity of 3600 rpm. Knowing that kinetic friction produces a couple of magnitude 1.2 N?m, determine the centroidal radius of gyration for the rotor.

A small 4 -lb collar \(C\) can slide freely on a thin ring of weight 6 lb and radius 10 in. The ring is welded to a short vertical shaft, which can rotate freely in a fixed bearing. Initially, the ring has an angular velocity of \(35 \mathrm{rad} / \mathrm{s}\) and the collar is at the top of the ring \((\theta=0)\) when it is given a slight nudge. Neglecting the effect of friction, determine ( \(a\) ) the angular velocity of the ring as the collar passes through the position \(\theta=90^{\circ},(b)\) the corresponding velocity of the collar relative to the ring.
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Geometrical and Physical Optics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Solid State Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Turning Points in Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.