/*! 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 38 A long ladder of length \(l,\) m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A long ladder of length \(l,\) mass \(m,\) and centroidal mass moment of inertia \(\bar{I}\) is placed against a house at an angle \(\theta=\theta_{0} .\) Knowing that the ladder is released from rest, determine the angular velocity of the ladder when \(\theta=\theta_{2}\). Assume the ladder can slide freely on the horizontal ground and on the vertical wall.

Short Answer

Expert verified
The angular velocity \(\omega\) is \(\sqrt{\frac{2mg \frac{l}{2} (\sin(\theta_0) - \sin(\theta_2))}{\bar{I}}}\).

Step by step solution

01

Understand the Problem

We need to find the angular velocity of the ladder when it swings from an angle \(\theta_0\) to \(\theta_2\) after being released from rest. The ladder slides on both the ground and the wall without friction.
02

Identify Relevant Physical Principles

The primary principle here is the conservation of energy. Initially, the ladder has potential energy at \(\theta_0\), and when it swings down to \(\theta_2\), some of this energy is converted to kinetic energy, specifically rotational kinetic energy.
03

Establish the Initial Potential Energy

At \(\theta_0\), the potential energy \(PE_1\) can be calculated based on the vertical position of the center of mass. So, \[ PE_1 = mgh = mg \left( \frac{l}{2} \sin(\theta_0) \right) \] where \(h = \frac{l}{2} \sin(\theta_0)\) is the height of the center of mass above the ground.
04

Determine Final Potential Energy at \(\theta_2\)

Similarly, calculate the potential energy \(PE_2\) when the ladder is at \(\theta_2\): \[ PE_2 = mg \left( \frac{l}{2} \sin(\theta_2) \right) \]
05

Apply Conservation of Energy

Since energy is conserved, the decrease in potential energy equals the increase in kinetic energy. Therefore, \( PE_1 - PE_2 = KE\). The kinetic energy \(KE\) when the ladder is at \(\theta_2\) is purely rotational, \[ KE = \frac{1}{2} \bar{I} \omega^2 \] where \(\omega\) is the angular velocity.
06

Solve for Angular Velocity \(\omega\)

Substitute the expressions for \(PE_1\), \(PE_2\), and \(KE\) into the equation from Step 5: \[ mg \left( \frac{l}{2} \sin(\theta_0) \right) - mg \left( \frac{l}{2} \sin(\theta_2) \right) = \frac{1}{2} \bar{I} \omega^2 \] Simplify to get: \[ mg \frac{l}{2} (\sin(\theta_0) - \sin(\theta_2)) = \frac{1}{2} \bar{I} \omega^2 \] Solve for \(\omega\): \[ \omega = \sqrt{\frac{2mg \frac{l}{2} (\sin(\theta_0) - \sin(\theta_2))}{\bar{I}}} \]
07

Verify Dimensions and Solution

Ensure that the units are consistent throughout the calculation. Angular velocity has units of radians per second, which are consistent with the provided variables and constants.

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.

The Concept of Rotational Kinetic Energy
Rotational kinetic energy is a form of energy that an object possesses due to its rotation. When we think about kinetic energy, we usually consider the motion of the object in a straight line, known as translational kinetic energy. However, for objects that spin or rotate, we need to consider another component of kinetic energy - the rotational aspect.

The formula for rotational kinetic energy is given by \[ KE_{rot} = \frac{1}{2} I \omega^2 \]where:
  • \( I \) is the moment of inertia, which depends on the mass distribution relative to the axis of rotation
  • \( \omega \) is the angular velocity, which is the rate of rotation.
As the ladder in our original problem swings from its initial angle \( \theta_0 \) to \( \theta_2 \), its potential energy is converted into rotational kinetic energy.
Understanding how this energy transforms helps us calculate how quickly the ladder is rotating at any angle, using principles of energy conservation. The change in potential energy due to movement creates the rotational energy needed for determining angular velocity.
Centroidal Mass Moment of Inertia
The centroidal mass moment of inertia is an essential concept when analyzing rotational motion. It quantifies how difficult it is to change an object’s rotational motion about its center of mass.

The moment of inertia is determined by both the object's mass and how that mass is distributed with respect to the axis of rotation. A larger moment of inertia means more energy is required to achieve the same angular velocity, as rotation is more difficult to start or stop.
For a uniform rod, or a ladder in this case, rotating about its center, the formula to calculate \( I \) is:\[ I = \frac{1}{3}ml^2 \]
This formula gives us the inertia relative to the center of the rod, assuming a uniform mass distribution. In the context of our ladder problem, \( \bar{I} \) plays a crucial role in determining how fast the ladder is rotating at angle \( \theta_2 \), after its potential energy has been converted into rotational kinetic energy.
Understanding Angular Velocity
Angular velocity is a measure of how fast an object rotates or spins around a specific point or axis. It's an integral part of understanding rotational motion and is represented in units of radians per second (rad/s).

Unlike linear velocity, which measures how fast something moves through space, angular velocity tells us how quickly the angle of an object is changing over time. It provides an insight into how the speed of rotation varies at different points throughout an object's pathway.In the context of our ladder problem, angular velocity \( \omega \) is determined once we apply the principle of conservation of energy. The original potential energy stored when the ladder is at \( \theta_0 \) is transformed into rotational kinetic energy at \( \theta_2 \). Knowing the ladder's moment of inertia and initial and final positions allows us to find the precise angular velocity, giving a better understanding of the ladder's rotational speed at the given angle.

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

The \(7.5-\) -lb disk \(A\) has a radius \(r_{A}=6\) in, and is initially at rest. The 10 -lb disk \(B\) has a radius \(r_{B}=8\) in. and an angular velocity \(\omega_{0}\) of 900 rpm when it is brought into contact with disk \(A\). Neglecting friction in the bearings, determine \((a)\) the final angular velocity of each disk, \((b)\) the total impulse of the friction force exerted on disk \(A .\)

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 uniform slender rod AB is at rest on a frictionless horizontal table when end A of the rod is struck by a hammer that delivers an impulse that is perpendicular to the rod. In the subsequent motion, determine the distance b through which the rod will move each time it completes a full revolution.

A flywheel is rigidly attached to a 1.5-in.-radius shaft that rolls without sliding along parallel rails. Knowing that after being released from rest the system attains a speed of 6 in./s in 30 s, determine the centroidal radius of gyration of the system.

Sphere \(A\) of mass \(m\) and radius \(r\) rolls without slipping with a velocity \(\bar{v}_{1}\) on a horizontal surface when it hits squarely an identical sphere \(B\) that is at rest. Denoting by \(\mu_{k}\) the coefficient of kinetic friction between the spheres and the surface, neglecting friction between the spheres, and assuming perfectly elastic impact, determine \((a)\) the linear and angular velocities of each sphere immediately after the impact, (b) the velocity of each sphere after it has started rolling uniformly.

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.