/*! 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 25 A small 4.0 kg brick is released... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A small 4.0 kg brick is released from rest 2.5 \(\mathrm{m}\) above a horizontal seesaw on a fulcrum at its center, as shown in Figure 10.52 . Find the angular momentum of this brick about a horizontal axis through the fulcrum and perpendicular to the plane of the figure (a) the instant the brick is released and (b) the instant before it strikes the seesaw.

Short Answer

Expert verified
(a) 0; (b) 70 kg⋅m²/s

Step by step solution

01

Understanding Angular Momentum at Release

Angular momentum when the brick is released from rest is zero. Since there is no initial velocity (and thus no initial motion), the angular momentum is zero. Formula at rest: \( L = mvr \). Since velocity \( v = 0 \), \( L = 0 \).
02

Calculating Velocity Just Before Impact

To find the velocity just before impact, use the energy conservation principle. The gravitational potential energy at height translates to kinetic energy right before the impact: \( mgh = \frac{1}{2}mv^2 \). Solving for \( v \), we have: \( v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 2.5} \approx 7 \text{ m/s} \).
03

Calculating Angular Momentum Before Impact

Angular momentum before the brick strikes the seesaw is calculated using \( L = mvr \), where \( m = 4.0 \text{ kg} \), \( v \) is the velocity found, and \( r \) is the seesaw's distance from the axis of rotation. Assuming a perpendicular distance equal to the initial drop height, \( L = 4.0 \times 7 \times 2.5 = 70 \text{ kg}\cdot\text{m}^2/\text{s} \).

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.

Conservation of Energy
The Conservation of Energy principle is fundamental in solving physics problems, especially those involving motion and mechanics. This principle states that energy cannot be created or destroyed; it can only be transformed from one form to another. In the problem of the brick being released on a seesaw, we start by considering potential energy and kinetic energy.
  • Potential Energy: Initially, the brick possesses gravitational potential energy, which is given by the formula: \( E_p = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height above the ground.
  • Kinetic Energy: As the brick falls, this potential energy converts into kinetic energy, described by \( E_k = \frac{1}{2}mv^2 \).
To determine the velocity before the brick strikes the seesaw, we set the gravitational potential energy equal to the kinetic energy, which helps us calculate the speed at which the brick impacts the seesaw. This use of energy conservation is the key step for the velocity calculation and is crucial for solving the entire problem.
Physics Problem Solving
Physics problem solving involves breaking down the exercise into manageable parts. Here, we follow a systematic approach:
  • Start by understanding the problem statement and identifying what you need to find. In this case, it's the angular momentum at two different instances of the brick's motion.
  • Use relevant physics formulas and concepts—for angular momentum, the formula \( L = mvr \) is crucial.
  • Step-by-step calculations help ensure accuracy and allow you to track your progress. Don't skip any steps to avoid making simple mistakes.
  • Double-check unit consistency and assumptions, such as considering the seesaw as a rigid body in this problem.
Following these steps not only helps in obtaining the correct solution but also in comprehending the physical processes and reinforcing learning.
Rotational Dynamics
Rotational dynamics is a branch of physics that deals with objects rotating around an axis. In this problem, the axis is through the seesaw's fulcrum, perpendicular to the plane. Understanding these dynamics is pivotal for problems involving angular momentum.
  • Angular Momentum: It represents the amount of rotation an object has, calculated using \( L = mvr \) for point masses.
  • Axis and Distance: Here, \( r \) is particularly important as it represents the distance from the axis of rotation to the point mass.
  • No Initial Angular Momentum: Initially, since the brick is at rest, the angular momentum is zero. This changes as the brick falls and gains velocity.
Grasping rotational dynamics allows us to predict how objects behave under rotation, which is indispensable in understanding real-world mechanics.
Kinetic Energy
Kinetic Energy is the energy an object possesses due to its motion. Before the brick hits the seesaw, all the gravitational potential energy has been converted into kinetic energy. This transformation is quantified by the formula \( E_k = \frac{1}{2}mv^2 \).
  • This energy increases as the brick accelerates under gravity, illustrating the change from potential to kinetic energy as it moves.
  • Kinetic energy is directly linked to the velocity of the brick, and plays a pivotal role in the angular momentum formula \( L = mvr \) because velocity is a component of both energy and momentum.
  • Understanding kinetic energy helps predict the effect of forces on moving objects, such as the impact's effect on the seesaw.
By mastering these concepts, you can solve a wide range of physics problems, linking the movement of objects to their energy and momentum dynamics.

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

\(\bullet\) Two people carry a heavy electric motor by placing it on a light board 2.00 m long. One person lifts at one end with a force of \(400.0 \mathrm{N},\) and the other lifts at the opposite end with a force of 600.0 \(\mathrm{N}\) . (a) Start by making a free-body diagram of the motor. (b) What is the weight of the motor? (c) Where along the board is its center of gravity located?

While exploring a castle, Exena the Exterminator is spotted by a dragon who chases her down a hallway. Exena runs into a room and attempts to swing the heavy door shut before the dragon gets her. The door is initially perpendicular to the wall, so it must be turned through \(90^{\circ}\) to close. The door is 3.00 \(\mathrm{m}\) tall and 1.25 \(\mathrm{m}\) wide, and it weighs 750 \(\mathrm{N} .\) You can ignore the friction at the hinges. If Exena applies a force of 220 \(\mathrm{N}\) at the edge of the door and perpendicular to it, how much time does it take her to close the door?

A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotation axis of 18 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) She then tucks into a small ball, decreasing this moment of inertia to 3.6 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) . While tucked, she makes two complete revolutions in 1.0 s. If she hadn't tucked at all, how many revolutions would she have made in the 1.5 s from board to water?

\(\cdot\) On an old-fashioned rotating piano stool, a woman sits holding a pair of dumbbells at a distance of 0.60 m from the axis of rotation of the stool. She is given an angular velocity of 3.00 rad/s, after which she pulls the dumbbells in until they are only 0.20 m distant from the axis. The woman's moment of inertia about the axis of rotation is 5.00 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) and may be considered constant. Each dumbbell has a mass of 5.00 \(\mathrm{kg}\) and may be considered a point mass. Neglect friction. (a) What is the initial angular momentum of the system? (b) What is the angular velocity of the system after the dumbbells are pulled in toward the axis? (c) Compute the kinetic energy of the system before and after the dumbbells are pulled in. Account for the difference, if any.

(a) Compute the torque developed by an industrial motor whose output is 150 \(\mathrm{kW}\) at an angular speed of 4000.0 rev/min. (b) A drum with negligible mass and 0.400 \(\mathrm{m}\) in diameter is attached to the motor shaft, and the power output of the motor is used to raise a weight hanging from a rope wrapped around the drum. How heavy a weight can the motor lift at constant speed? (c) At what constant speed will the weight rise?

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.