91Ó°ÊÓ

A wheel changes its angular velocity with a constant angular acceleration while rotating about a fixed axis through its center. (a) Show that the change in the magnitude of the radial acceleration during any time interval of a point on the wheel is twice the product of the angular acceleration, the angular displacement, and the perpendicular distance of the point from the axis. (b) The radial acceleration of a point on the wheel that is 0.250 \(\mathrm{m}\) from the axis changes from 25.0 \(\mathrm{m} / \mathrm{s}^{2}\) to 85.0 \(\mathrm{m} / \mathrm{s}^{2}\) as the wheel rotates through 20.0 rad. Calculate the tangential acceleration of this point. (c) Show that the change in the wheel's kinetic energy during any time interval is the product of the moment of inertia about the axis, the angular acceleration, and the angular displacement. (d) During the 20.0 -rad angular displacement of part (b), the kinetic energy of the wheel increases from 20.0 \(\mathrm{J}\) to 45.0 \(\mathrm{J} .\) What is the moment of inertia of the wheel about the rotation axis?

Step by step solution

01

Understanding Radial Acceleration

The radial acceleration of a point on a rotating wheel is given by \( a_r = \omega^2 r \), where \( \omega \) is the angular velocity and \( r \) is the distance from the axis. Thus, any change in radial acceleration is related to the change in \( \omega^2 \).

02

Relating Radial Acceleration Change to Angular Quantities

The change in radial acceleration is \( \Delta a_r = (\omega_f^2 - \omega_i^2)r \), where \( \omega_f \) and \( \omega_i \) are final and initial angular velocities. Using \( \omega_f^2 - \omega_i^2 = 2\alpha\theta \) (from rotational kinematics), \( \Delta a_r = 2\alpha\theta r \), proving our first part.

03

Calculate Tangential Acceleration from Radial Change

Given initial and final radial accelerations (25.0 m/s² and 85.0 m/s²) and angular displacement of 20.0 rad, use \( \Delta a_r = 2\alpha\theta r \). With \( r = 0.250 \) m, solve for \( \alpha \): \( 85 - 25 = 2\times\alpha\times 20 \times 0.250 \rightarrow \alpha = 6 \text{ rad/s}^2 \). Then, \( a_t = \alpha r = 6 \times 0.250 = 1.5 \text{ m/s}^2 \).

04

Relating Kinetic Energy Change to Angular Quantities

Rotational kinetic energy is \( K = \frac{1}{2}I\omega^2 \). Changes in kinetic energy are \( \Delta K = \frac{1}{2}I(\omega_f^2 - \omega_i^2) \). Using \( \omega_f^2 - \omega_i^2 = 2\alpha\theta \), \( \Delta K = I\alpha\theta \), proving our second part.

05

Calculate Moment of Inertia from Kinetic Energy Change

Given \( \Delta K = 45.0 - 20.0 = 25.0 \text{ J} \) and using \( \Delta K = I\alpha\theta \). Substitute \( \alpha = 6 \text{ rad/s}^2 \) and \( \theta = 20 \text{ rad} \): \( 25 = I \cdot 6 \cdot 20 \). Solve \( I = \frac{25}{120} = 0.208 \text{ kg} \cdot \text{m}^2 \).

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Key Concepts

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

Angular Acceleration

Angular acceleration is a key concept in understanding rotational motion. It describes how quickly the angular velocity of an object changes over time. Angular velocity, often denoted by the symbol \( \omega \), is the rate of change of the angular position of an object, usually measured in radians per second. Angular acceleration, on the other hand, is the rate at which this angular velocity changes, and it is denoted by the symbol \( \alpha \). When an object experiences a constant angular acceleration, its angular velocity \( \omega \) changes linearly over time.

• Calculating Angular Acceleration: In the exercise, angular acceleration is calculated using the change in radial acceleration formula: \( \Delta a_r = 2\alpha\theta r \), relating to the angular displacement \( \theta \) and distance \( r \) from the axis.
• Example: Given the radial acceleration changes from 25 m/s² to 85 m/s² over an angular displacement \( \theta \) of 20 radians for a point 0.250 m from the axis, \( \alpha \) is found to be 6 rad/s².

Understanding angular acceleration helps predict how the speed of rotation increases or decreases over time.

Kinetic Energy

In rotational motion, kinetic energy also plays a crucial role, just like in linear motion. Rotational kinetic energy refers to the energy of an object due to its rotation. It is expressed with the formula \( K = \frac{1}{2}I\omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

• Relationship between Kinetic Energy and Motion: The change in kinetic energy as the wheel rotates can be calculated using the equation: \( \Delta K = I\alpha\theta \). This shows the direct relationship between kinetic energy change and angular quantities.
• Practical Understanding: In the problem, the wheel's kinetic energy increases from 20 J to 45 J as it undergoes a 20 rad angular displacement. This increase directly relates to the energy required to spin the wheel faster.

It helps us understand how energy transforms when an object experiences rotational motion, affecting speed and motion characteristics.

Moment of Inertia

Moment of inertia is an essential concept when studying rotational motion, akin to mass in linear motion. It measures an object's resistance to changes in its angular motion. Basically, the moment of inertia indicates how difficult it is to rotate an object around a given axis. The formula for moment of inertia varies depending on the object's shape and the axis about which it is rotating, generally expressed as \( I \cdot \alpha \cdot \theta \) in energy equations.

• Role in Rotation: The moment of inertia affects how much torque is needed to rotate an object, influencing how quickly it accelerates for a given torque.
• Calculation Example: In the exercise, the moment of inertia is calculated using the kinetic energy change formula. With a known \( \Delta K = 25 \, \text{J} \) and angular terms \( \alpha = 6 \, \text{rad/s}^2 \) and \( \theta = 20 \, \text{rad} \), solving gives \( I = 0.208 \, \text{kg}\cdot\text{m}^2 \).

Understanding moment of inertia is crucial for designing and analyzing mechanical systems, ensuring efficient and effective motion control.

Tangential Acceleration

Tangential acceleration is another vital concept that refers to the rate of change of velocity experienced by an object moving along a circular path. Specifically, it is the component of linear acceleration that acts along the direction of the tangential to the motion, as opposed to the radial component. It is related directly to angular acceleration via the formula \( a_t = \alpha \times r \), where \( \alpha \) is angular acceleration, and \( r \) is the radius or distance from the axis.

• Relation to Circular Motion: Tangential acceleration suggests how fast the object's speed changes along its path.
• Application from Exercise: With an angular acceleration \( \alpha = 6 \, \text{rad/s}^2 \) and \( r = 0.250 \, \text{m} \), the tangential acceleration is \( a_t = 1.5 \, \text{m/s}^2 \).

This concept is fundamental in understanding how rotational and linear dynamics interplay, especially in systems where rotation translates to linear movement.