Problem 38 - Applied Physics

Question

A rotating flywheel of diameter \(40.0 \mathrm{~cm}\) uniformly accelerates from rest to \(25 \overline{0} \mathrm{rad} / \mathrm{s}\) in \(15.0 \mathrm{~s}\). (a) Find its angular acceleration. (b) Find the linear velocity of a point on the rim of the wheel after \(15.0 \mathrm{~s}\). (c) How many revolutions does the wheel make during the \(15.0 \mathrm{~s}\) ?

Step-by-Step Solution

Answer

(a) 16.67 rad/s², (b) 50 m/s, (c) 298.56 revolutions

Step 1: Calculating Angular Acceleration

To find the angular acceleration (\(\alpha\)), use the formula for angular acceleration:\[\alpha = \frac{\Delta \omega}{\Delta t}\]where \(\Delta \omega\) is the change in angular velocity and \(\Delta t\) is the time interval. Here, the initial angular velocity \(\omega_0\) is \(0\ \text{rad/s}\) (since it starts from rest) and the final angular velocity \(\omega\) is \(250\ \text{rad/s}\). The time is \(15\ \text{s}\).\[\alpha = \frac{250\ \text{rad/s} - 0\ \text{rad/s}}{15\ \text{s}} = \frac{250}{15} = 16.67\ \text{rad/s}^2\]

Step 2: Finding Linear Velocity on the Rim

The linear velocity (\(v\)) of a point on the rim can be found using the relationship between linear velocity and angular velocity:\[v = r \cdot \omega\]where \(r\) is the radius of the flywheel and \(\omega\) is the final angular velocity. The diameter given is \(40.0\ \text{cm}\), so the radius \(r\) is \(20.0\ \text{cm} = 0.20\ \text{m}\).Substituting the values:\[v = 0.20\ \text{m} \times 250\ \text{rad/s} = 50\ \text{m/s}\]

Step 3: Calculating Number of Revolutions

To find the number of revolutions, use the formula relating angular displacement (\(\theta\)) to initial velocity, time, and angular acceleration:\[\theta = \omega_0 \cdot t + \frac{1}{2} \alpha \cdot t^2\]Since \(\omega_0 = 0\ \text{rad/s}\), the equation simplifies to:\[\theta = \frac{1}{2} \times 16.67\ \text{rad/s}^2 \times (15\ \text{s})^2\]\[\theta = \frac{1}{2} \times 16.67 \times 225 = 1875.75\ \text{rad}\]To convert radians to revolutions, use:\[\text{Revolutions} = \frac{1875.75\ \text{rad}}{2\pi} \approx 298.56\ \text{revolutions}\]

Key Concepts

Angular Acceleration

Angular acceleration is a measure of how quickly an object changes its angular velocity. It is akin to linear acceleration but applies to rotational motion. The symbol for angular acceleration is \( \alpha \). When an object like a flywheel accelerates uniformly from rest, its angular acceleration can be calculated using:

• \( \alpha = \frac{\Delta \omega}{\Delta t} \)

This formula equates angular acceleration to the change in angular velocity \( \Delta \omega \) divided by the time \( \Delta t \) taken for that change. In our exercise, the flywheel starts from rest, so \( \omega_0 = 0 \). Over 15 seconds, it reaches an angular velocity of 250 rad/s:

• \( \alpha = \frac{250 \ \text{rad/s} - 0 \ \text{rad/s}}{15 \ \text{s}} = 16.67 \ \text{rad/s}^2 \)

This tells us the wheel's rate of speeding up in terms of its rotational motion.

Linear Velocity

Linear velocity is the speed at which a point on the outer edge of a rotating object moves through space. For circular motion, it relates to angular velocity \( \omega \) using the radius \( r \) of the motion path. The linear velocity \( v \) is given by:

• \( v = r \cdot \omega \)

Here, the flywheel has a radius of 0.20 meters, and at \( 250 \ \text{rad/s} \), the calculation becomes:

• \( v = 0.20 \ \text{m} \times 250 \ \text{rad/s} = 50 \ \text{m/s} \)

This velocity shows how fast a point on the rim moves linearly, as if tracing a straight path tangent to the circular motion.

Revolutions

Revolutions measure the number of complete circular turns an object makes. In rotational dynamics, this involves converting angular displacement (in radians) into complete turns. Using the formula for angular displacement \( \theta \):

• \( \theta = \omega_0 \cdot t + \frac{1}{2} \alpha \cdot t^2 \)

Since initial velocity \( \omega_0 = 0 \), this reduces to:

• \( \theta = \frac{1}{2} \times 16.67 \ \text{rad/s}^2 \times 225 \ \text{s}^2 = 1875.75 \ \text{rad} \)

To find revolutions, convert radians using:

• \( \text{Revolutions} = \frac{\theta}{2\pi} \approx 298.56 \ \text{revolutions} \)

This calculation shows how many times the wheel spins around completely.

Rotational Kinematics

Rotational kinematics deals with the motion of objects that rotate. It is parallel to linear kinematics, providing equations and relationships to describe angular motion. Key terms used are:

• Angular position (\( \theta \))
• Angular velocity (\( \omega \))
• Angular acceleration (\( \alpha \))

These are connected through specific kinematic equations, such as:

• \( \theta = \omega_0 \cdot t + \frac{1}{2} \alpha \cdot t^2 \)
• \( \omega = \omega_0 + \alpha \cdot t \)

These equations help predict future motion aspects like speed or the distance traveled rotationally, much like forecasting the future position and velocity of linear motion.