**Problem Statement:**A constant torque is applied to a rigid body whose moment of inertia is 4.40 kg·m² around the axis of rotation. If the wheel starts from rest and attains an angular velocity of 20.0 rad/s in 10.4 s, what is the applied torque?**Explanation:**This problem involves the concepts of torque, moment of inertia, and angular motion. Given the moment of inertia and the angular acceleration needed to achieve the final angular velocity, the applied torque can be calculated using the formula:\[ \tau = I \cdot \alpha \]Where:- \( \tau \) is the torque,- \( I \) is the moment of inertia,- \( \alpha \) is the angular acceleration.**Steps to Solve:**1. **Calculate Angular Acceleration** (\( \alpha \)): - Use the formula for angular acceleration: \[ \alpha = \frac{\Delta \omega}{\Delta t} \] - Given: - Initial angular velocity (\( \omega_0 \)) = 0 rad/s (starts from rest) - Final angular velocity (\( \omega \)) = 20.0 rad/s - Time (\( \Delta t \)) = 10.4 s - Now, calculate: \[ \alpha = \frac{20.0 \, \text{rad/s} - 0 \, \text{rad/s}}{10.4 \, \text{s}} \] 2. **Calculate Torque** (\( \tau \)): - Substitute the values of \( I \) and \( \alpha \) into the torque formula: - Given moment of inertia, \( I = 4.40 \, \text{kg·m}^2 \) - Calculate: \[ \tau = 4.40 \, \text{kg·m}^2 \times \alpha \]This exercise helps understand how to relate angular motion with torque using the principles of rotational dynamics. **Problem Statement:**When opening a door, you push on it perpendicularly with a force of 62.6 N at a distance of 0.792 m from the hinges. What torque are you exerting relative to the hinges? (Does it matter if you push at the same height as the hinges?)---**Explanation:**This question involves calculating the torque exerted on a door. Torque (\(\tau\)) is the rotational equivalent of force and is calculated using the formula:\[\tau = F \times r \times \sin(\theta)\]Where:- \(\tau\) = torque- \(F\) = force applied (62.6 N)- \(r\) = distance from the pivot point (0.792 m)- \(\theta\) = angle between force direction and the lever arm (90 degrees since it's perpendicular)Since the force is applied perpendicularly, \(\sin(90^{\circ}) = 1\).Therefore, the torque is:\[\tau = 62.6 \, \text{N} \times 0.792 \, \text{m} = 49.5552 \, \text{N}\cdot\text{m}\]**Key Concept:**- Torque is greater when the force is applied further from the pivot point (hinges) or when the angle is closer to 90 degrees.- It doesn't matter if you push at the same height as the hinges as long as the force is applied perpendicularly.

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A constant torque is applied to a rigid body whose moment of inertia is 4.40 kg-m2 around the axis of rotation. If the wheel starts from rest and attains an angular velocity of 20.0 rad/s in 10.4 s, what is the applied torque?

A constant torque is applied to a rigid body whose moment of inertia is 4.40 kg-m2 around the axis of rotation. If the wheel starts from rest and attains an angular velocity of 20.0 rad/s in 10.4 s, what is the applied torque?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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