/*! 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 24 A bullet of mass \(10 \mathrm{~g... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A bullet of mass \(10 \mathrm{~g}\) and speed \(500 \mathrm{~m} / \mathrm{s}\) is fired into a door and gets embedded exactly at the centre of the door. The door is \(1.0 \mathrm{~m}\) wide and weighs \(12 \mathrm{~kg}\). It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it. (Hint: The moment of inertia of the door about the vertical axis at one end is \(M L^{2} / 3 .\) )

Short Answer

Expert verified
The angular speed of the door after the bullet embeds is \(0.625\, \text{rad/s}\).

Step by step solution

01

Identify the System

We have a bullet of mass \( m = 10\, \text{g} = 0.01\, \text{kg} \) and speed \( v = 500\, \text{m/s} \) embedding itself into the door. The door has mass \( M = 12\, \text{kg} \), width \( L = 1.0\, \text{m} \), and rotates about a hinge at one end.
02

Calculate Bullet's Linear Momentum

The linear momentum \( p \) of the bullet is given by \( p = mv \). Substituting the given values, we get \( p = 0.01 \, \text{kg} \times 500\, \text{m/s} = 5\, \text{kg}\,\text{m/s} \).
03

Convert Linear Momentum to Angular Momentum

As the bullet hits the door at its center (distance \( L/2 = 0.5\, \text{m} \) from the hinge), the angular momentum \( L_b \) about the hinge is \( L_b = p \times (L/2) = 5 \, \text{kg}\,\text{m/s} \times 0.5\, \text{m} = 2.5\, \text{kg}\,\text{m}^2/\text{s} \).
04

Calculate Moment of Inertia of the Door

The moment of inertia \( I \) of the door about the hinge is given by \( I = \frac{ML^2}{3} \). Substituting the values, \( I = \frac{12\, \text{kg} \times (1.0\, \text{m})^2}{3} = 4\, \text{kg}\,\text{m}^2 \).
05

Apply Conservation of Angular Momentum

With no external torques, angular momentum is conserved. Thus the angular momentum of the bullet is equal to the angular momentum of the door plus the bullet embedded: \( I \cdot \omega = L_b \). Therefore, \( 4\, \text{kg}\,\text{m}^2 \cdot \omega = 2.5\, \text{kg}\,\text{m}^2/\text{s} \).
06

Solve for Angular Speed

Solve the equation \( 4\omega = 2.5 \) for \( \omega \), the angular speed of the door: \( \omega = \frac{2.5}{4} = 0.625\, \text{rad/s} \).

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 91Ó°ÊÓ!

Key Concepts

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

Moment of Inertia
Imagine trying to spin a door open. If it's light, it's easy to move, but if it's heavy, it takes more effort. This is where the **moment of inertia** comes into play. It is a measure of how much resistance an object has to a change in its rotational motion.
In the case of a door, the moment of inertia depends on:
  • The mass of the door (the heavier it is, the greater the moment of inertia).
  • The distribution of mass relative to the axis of rotation (the further the mass is distributed from the axis, the harder it is to rotate).
For a door rotating about a hinge, the moment of inertia, denoted as \( I \), can be calculated using the formula: \[ I = \frac{ML^2}{3} \]Here, \( M \) is the mass and \( L \) is the width of the door. This formula tells us that not only the mass but also the **square** of its width influences how the door swings. A wider door is harder to push than a smaller one, showcasing how the moment of inertia influences rotational motion.
Conservation of Angular Momentum
Angular momentum is like a spinning dancer drawing her arms in to spin faster. It is a combination of rotational inertia and rotational velocity. Just as linear momentum explains the way objects move straight ahead, angular momentum explains how they spin.
A core principle of rotational motion is the **conservation of angular momentum**. This principle states that if no external forces act on a system, its angular momentum will stay constant. In our scenario, when the bullet embeds into the door, it causes the door to rotate. No external torque is acting on the system, so the angular momentum before and after the bullet hits is the same.
Mathematically, this is expressed as:\[ L_{ ext{initial}} = L_{ ext{final}} \]Where \( L \) is the angular momentum. Initially, the bullet's angular momentum is \( L_b \). When it embeds into the door, the door spins, creating a new combined angular momentum. This principle allows us to calculate the door's angular speed after the bullet's impact using conservation concepts.
Angular Speed
When we talk about how fast something is spinning, we often refer to its **angular speed**. This is different from the usual speed we think of because it's about rotation rather than linear movement.
Angular speed is how quickly an object spins around a central point and is usually measured in radians per second (rad/s). An easy way to understand it is to imagine how fast the hands of a clock move. A smaller angular speed means it moves slower, while a larger one means a quicker movement.
After the bullet hits the door, the door starts to rotate. By applying the concept of conservation of angular momentum, we can find its angular speed, denoted by \( \omega \). In our scenario, we calculated:\[ \omega = \frac{2.5}{4} = 0.625 \, \text{rad/s} \]This tells us exactly how fast the door spins right after the bullet embeds in it, showing how angular speed helps us understand dynamics in rotational scenarios.

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

Find the components along the \(x, y, z\) axes of the angular momentum 1 of a particle, whose position vector is \(\mathbf{r}\) with components \(x, y, z\) and momentum is \(\mathbf{p}\) with components \(p_{x} \cdot p_{y}\) and \(p_{z} .\) Show that if the particle moves only in the \(x-y\) plane the angular momentum has only a z-component.

The oxygen molecule has a mass of \(5.30 \times 10^{-26} \mathrm{~kg}\) and a moment of inertia of \(1.94 \times 10^{46} \mathrm{~kg} \mathrm{~m}^{2}\) about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is \(500 \mathrm{~m} / \mathrm{s}\) and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be \(2 M R^{2} / 5\), where \(M\) is the mass of the sphere and \(R\) is the radius of the sphere. (b) Given the moment of inertia of a disc of mass \(M\) and radius \(R\) about any of its diameters to be \(M R^{2} / 4\), find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

A metre stick is balanced on a knife edge at its centre. When two coins, each of mass \(5 \mathrm{~g}\) are put one on top of the other at the \(12.0 \mathrm{~cm}\) mark, the stick is found to be balanced at \(45.0 \mathrm{~cm}\). What is the mass of the metre stick?

A rope of negligible mass is wound round a hollow cylinder of mass \(3 \mathrm{~kg}\) and radius \(40 \mathrm{~cm}\). What is the angular acceleration of the cylinder if the rope is pulled with a force of \(30 \mathrm{~N}\) ? What is the linear acceleration of the rope ? Assume that there is no slipping.

See all solutions

Recommended explanations on Physics Textbooks

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.