/*! 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 36 A 9.75-m ladder with a mass of 2... [FREE SOLUTION] | 91Ó°ÊÓ

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A 9.75-m ladder with a mass of 23.2 kg lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of \(245 \mathrm{N}\). At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of \(1.80 \mathrm{rad} / \mathrm{s}^{2}\) about an axis passing through the bottom end of the ladder. The ladder's center of gravity lics halfway between the top and bottom ends. (a) What is the net torque acting on the ladder? (b) What is the ladder's moment of inertia?

Short Answer

Expert verified
The net torque is 2388.75 Nm, and the moment of inertia is 1327.08 kg·m².

Step by step solution

01

Identify Variables and Known Values

First, let's identify the information given in the problem:- Length of the ladder, \(L = 9.75\, \text{m}\).- Mass of the ladder, \(m = 23.2\, \text{kg}\).- Force exerted by the painter, \(F = 245\, \text{N}\).- Angular acceleration, \(\alpha = 1.80\, \text{rad/s}^2\).The ladder's center of gravity is at its midpoint, i.e., \(x_{cg} = \frac{L}{2} = \frac{9.75}{2} = 4.875\, \text{m}\).
02

Calculate Net Torque

The torque \(\tau\) due to a force is given by the formula: \(\tau = F \cdot r\), where \(r\) is the distance from the pivot point to the point where the force is applied.When the painter pulls the ladder, the distance \(r\) is the length of the ladder, 9.75 m.The net torque \(\tau_{\text{net}}\) is therefore: \[\tau_{\text{net}} = F \cdot L = 245 \times 9.75 = 2388.75 \, \text{Nm}.\]
03

Calculate the Moment of Inertia

The relationship between net torque, moment of inertia \(I\), and angular acceleration \(\alpha\) is given by: \[\tau_{\text{net}} = I \cdot \alpha.\]Rearrange to find the moment of inertia \(I\): \[I = \frac{\tau_{\text{net}}}{\alpha} = \frac{2388.75}{1.80} = 1327.08 \, \text{kg} \cdot \text{m}^2.\]
04

Evaluate Solution

The net torque calculated is \(2388.75 \, \text{Nm}\), and the moment of inertia is \(1327.08 \, \text{kg} \cdot \text{m}^2\). These values are based on the simplified assumptions that we've considered only the ladder and not any additional forces or resistances.

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Key Concepts

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

Moment of Inertia
Understanding the concept of **moment of inertia** is key in rotational dynamics. It describes how mass is distributed relative to a rotational axis and affects how easily an object can be rotated.
For a ladder being lifted with a rotational axis at the bottom end, its mass distribution determines the inertia. The moment of inertia (I) is calculated using:\[I = \frac{\tau_{\text{net}}}{\alpha}\]Where \( \tau_{\text{net}} \) is the net torque applied and \( \alpha \) is the angular acceleration.
A higher moment of inertia means more effort is needed to spin the object. Here, the calculated value is \( 1327.08 \, \text{kg} \cdot \text{m}^2 \), indicating the required effort to get the ladder rotating around its base.
Angular Acceleration
Angular acceleration is how quickly the rotational speed of an object changes, measured in radians per second squared (\text{rad/s}^2).
  • It represents how fast the ladder gains speed as it starts rotating.
  • This concept is similar to linear acceleration, but in circular motion.
In our ladder problem, the angular acceleration is given as \( 1.80 \, \text{rad/s}^2 \). It defines how aggressively the ladder spins from its resting position when the painter applies the lifting force. Understanding angular acceleration helps in predicting the rotation behavior and timing of objects when forces are applied.
Center of Gravity
The **center of gravity** is the average location of an object’s weight distribution. It's the point where gravity acts on an object, naturally balancing it.
For the ladder, the center of gravity is at its midpoint, given as half the ladder's length (\(4.875 \, \text{m}\)).Knowing where the center of gravity lies is vital when calculating torque and stability. It affects how an object will rotate or move when forces are applied.
In practical terms, the center of gravity guides you on where to apply force efficiently to lift or balance objects steadily.
Physics Problem Solving
**Solving physics problems** requires a systematic approach. Here's how you can tackle problems like the ladder one:
  • **Identify Variables:** List known values like mass, length, or force.
  • **Choose Formulas:** Pick the right formula for what you need to find, such as torque or inertia.
  • **Substitute Values:** Place your known variables into the formula to compute the unknowns.
  • **Evaluate:** Always check your computations and ensure they logically match with the scenario.
This method not only helps in understanding the problem but also strengthens analytical skills, allowing you to solve more complex scenarios with confidence.

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Most popular questions from this chapter

A stationary bicycle is raised off the ground, and its front wheel \((m=1.3 \mathrm{kg})\) is rotating at an angular velocity of 13.1 rad/s (see the drawing). The front brake is then applied for \(3.0 \mathrm{s},\) and the wheel slows down to \(3.7 \mathrm{rad} / \mathrm{s}\). Assume that all the mass of the wheel is concentrated in the rim, the radius of which is \(0.33 \mathrm{m}\). The coefficient of kinetic friction between each brake pad and the rim is \(\mu_{\mathrm{k}}=0.85 .\) What is the magnitude of the normal force that each brake pad applies to the rim?

One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of 6.00 N and acts at a $$30.0^{\circ}$$ angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned.

Just after a motorcycle rides off the end of a ramp and launches into the air, its engine is turning counterclockwise at 7700 rev/min. The motorcycle rider forgets to throttle back, so the engine"s angular speed increases to 12.500 rev/min. As a result, the rest of the motorcycle (including the rider) begins to rotate clockwise about the engine at 3.8 rev/min. Calculate the ratio \(I_{\mathrm{L}} / I_{\mathrm{M}}\) of the moment of inertia of the engine to the moment of inertia of the rest of the motorcycle (and the rider). Ignore torques due to gravity and air resistance.

A block (mass \(=2.0 \mathrm{kg}\) ) is hanging from a massless cord that is wrapped around a pulley (moment of inertia \(=1.1 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2}\) ), as the drawing shows. Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of \(0.040 \mathrm{m}\) during the block's descent. Find the angular acceleration of the pulley and the tension in the cord.

Two spheres are each rotating at an angular speed of \(24 \mathrm{rad} / \mathrm{s}\) about axes that pass through their centers. Each has a radius of \(0.20 \mathrm{m}\) and a mass of \(1.5 \mathrm{kg}\). However, as the figure shows, one is solid and the other is a thin-walled spherical shell. Suddenly, a net external torque due to friction (magnitude \(=0.12 \mathrm{N} \cdot \mathrm{m}\) ) begins to act on each sphere and slows the motion down. Concepts: (i) Which sphere has the greater moment of inertia and why? (ii) Which sphere has the angular acceleration (a deceleration) with the smaller magnitude? (iii) Which sphere takes a longer time to come to a halt? Calculations: How long does it take each sphere to come to a halt?

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