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If a \(32.0 \mathrm{~N} \cdot \mathrm{m}\) torque on a wheel causes angular acceleration \(25.0 \mathrm{rad} / \mathrm{s}^{2}\), what is the wheel's rotational inertia?

Short Answer

Expert verified
The wheel's rotational inertia is \(1.28 \, \mathrm{kg} \cdot \mathrm{m^2}\).

Step by step solution

01

Identify the Given Information

We are given the torque acting on the wheel, \( \tau = 32.0 \, \mathrm{N} \cdot \mathrm{m} \), and the angular acceleration, \( \alpha = 25.0 \, \mathrm{rad/s^2} \). We need to find the wheel's rotational inertia, which is denoted by \( I \).
02

Understand the Relationship Between Torque, Angular Acceleration, and Rotational Inertia

The relationship between torque (\( \tau \)), angular acceleration (\( \alpha \)), and rotational inertia (\( I \)) is given by the formula:\[ \tau = I \cdot \alpha \]
03

Rearrange the Formula to Solve for Rotational Inertia

We need to solve for \( I \), the rotational inertia, so we rearrange the formula:\[ I = \frac{\tau}{\alpha} \]
04

Substitute the Given Values into the Formula

Substitute \( \tau = 32.0 \, \mathrm{N} \cdot \mathrm{m} \) and \( \alpha = 25.0 \, \mathrm{rad/s^2} \) into the formula to find \( I \):\[ I = \frac{32.0}{25.0} \]
05

Calculate the Rotational Inertia

Perform the division to find the rotational inertia:\[ I = 1.28 \, \mathrm{kg} \cdot \mathrm{m^2} \]

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Key Concepts

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

Torque
Torque is a fundamental concept in the study of rotational motion. Think of torque as the rotational equivalent of linear force. It measures how much a force acting on an object causes that object to rotate. Imagine pushing a door open or tightening a screw: these actions involve applying torque. The amount of torque applied depends on two things:
  • The magnitude of the force applied.
  • The distance from the pivot point (or axis of rotation) where the force is applied. This distance is often referred to as the "lever arm" or "moment arm".
The mathematical formula for torque (\( \tau \)) is:\[ \tau = F \times r \times \sin(\theta) \]where \( F \) is the force applied, \( r \) is the distance from the rotation axis, and \( \theta \) is the angle between the force vector and the lever arm. In many cases, \( \sin(\theta) \) is 1 when the force is perpendicular to the lever arm, simplifying the equation to \( \tau = F \times r \). Understanding torque is crucial for solving rotational dynamics problems, especially when determining the resulting motion like what we see in this exercise.
Angular Acceleration
Angular acceleration is a measure of how quickly an object's angular velocity changes. While its linear counterpart, acceleration, describes changes in speed along a path, angular acceleration (\( \alpha \)) focuses on changes in rotational speed.It is calculated using the difference in angular velocity over time. Mathematically, angular acceleration is expressed as:\[ \alpha = \frac{\Delta \omega}{\Delta t} \]where \( \Delta \omega \) is the change in angular velocity and \( \Delta t \) is the time over which this change occurs. Like linear acceleration, angular acceleration can be positive (speeding up) or negative (slowing down), which is often called deceleration.In our exercise, when a wheel experiences a torque, it gains angular acceleration, showing how much faster it spins due to the applied torque. This relationship is a core aspect when looking at rotational motion dynamics and connects directly to the wheel's rotational inertia.
Physics Problem Solving
Physics problem solving is all about understanding concepts and applying them logically and methodically. To tackle physics problems, especially those involving rotation like this exercise, follow these key steps:
  • Identify and list known values from the problem (such as torque \( \tau \) and angular acceleration \( \alpha \)).
  • Understand the core physics principles at play (like torque and angular motion relationships).
  • Use relevant equations that connect these principles, such as \( \tau = I \cdot \alpha \) in this exercise.
  • Rearrange these equations if needed to solve for the unknown values, like the rotational inertia \( I \). This often means algebraic manipulation to find the desired quantity.
  • Substitute known values into your equations and perform the calculations carefully.
  • Always check your units and consider if your answer makes sense within the context of the problem.
By following these steps, you'll be better prepared to solve physics problems effectively. Practice regularly to become proficient at identifying the applicable principles for a wide range of scenarios.

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Most popular questions from this chapter

At \(t=0,\) a flywheel has an angular velocity of \(4.7 \mathrm{rad} / \mathrm{s}, \mathrm{a}\) constant angular acceleration of \(-0.25 \mathrm{rad} / \mathrm{s}^{2},\) and a reference line at \(\theta_{0}=0 .\) (a) Through what maximum angle \(\theta_{\max }\) will the reference line turn in the positive direction? What are the (b) first and (c) second times the reference line will be at \(\theta=\frac{1}{2} \theta_{\max } ?\) At what (d) negative time and (e) positive time will the reference line be at \(\theta=10.5\) rad? (f) Graph \(\theta\) versus \(t,\) and indicate your answers.

A uniform spherical shell of mass \(M=4.5 \mathrm{~kg}\) and radius \(R=8.5 \mathrm{~cm}\) can rotate about a vertical axis on frictionless bearings (Fig. \(10-47) .\) A massless cord passes around the equator of the shell, over a pulley of rotational inertia \(I=3.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) and radius \(r=5.0 \mathrm{~cm},\) and is attached to a small object of mass \(m=0.60 \mathrm{~kg}\) There is no friction on the pulley's axle; the cord does not slip on the pulley. What is the speed of the object when it has fallen \(82 \mathrm{~cm}\) after being released from rest? Use energy considerations.

During the launch from a board, a diver's angular speed about her center of mass changes from zero to \(6.20 \mathrm{rad} / \mathrm{s}\) in \(220 \mathrm{~ms}\). Her rotational inertia about her center of mass is \(12.0 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) During the launch, what are the magnitudes of (a) her average angular acceleration and (b) the average external torque on her from the board?

The angular position of a point on the rim of a rotating wheel is given by \(\theta=4.0 t-3.0 t^{2}+t^{3},\) where \(\theta\) is in radians and \(t\) is in seconds. What are the angular velocities at (a) \(t=2.0 \mathrm{~s}\) and \((\mathrm{b}) t=4.0 \mathrm{~s} ?\) (c) What is the average angular acceleration for the time interval that begins at \(t=2.0 \mathrm{~s}\) and ends at \(t=4.0 \mathrm{~s} ?\) What are the instantaneous angular accelerations at (d) the beginning and (e) the end of this time interval?

A flywheel with a diameter of \(1.20 \mathrm{~m}\) is rotating at an angular speed of 200 rev \(/\) min. (a) What is the angular speed of the flywheel in radians per second? (b) What is the linear speed of a point on the rim of the flywheel? (c) What constant angular acceleration (in revolutions per minute-squared) will increase the wheel's angular speed to 1000 rev/min in 60.0 s? (d) How many revolutions does the wheel make during that \(60.0 \mathrm{~s} ?\)

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