Problem 24 - College Physics

Question

II A flywheel with a radius of \(0.300 \mathrm{~m}\) starts from rest and accelerates with a constant angular acceleration of \(0.600 \mathrm{rad} / \mathrm{s}^{2}\). Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim (a) at the start, (b) after it has turned through \(60.0^{\circ},\) and \((\mathrm{c})\) after it has turned through \(120.0^{\circ} .\)

Step-by-Step Solution

Answer

(a) At start: 0.180 m/s², (b) 60°: 0.627 m/s², and (c) 120°: 1.212 m/s².

Step 1: Understanding the problem

We have a flywheel with a radius of 0.300 m, starting from rest and accelerating with a constant angular acceleration of 0.600 rad/s². Our objective is to compute the tangential acceleration, radial acceleration, and resultant acceleration of a point on its rim at three positions: at the start, after turning 60°, and after turning 120°.

Step 1: Tangential Acceleration Calculation

Tangential acceleration for a point on the rim is given by the formula \( a_t = \alpha \cdot r \), where \( \alpha = 0.600 \text{ rad/s}^2 \) is the angular acceleration and \( r = 0.300 \text{ m} \) is the radius of the flywheel. Therefore, \( a_t = 0.600 \cdot 0.300 = 0.180 \text{ m/s}^2 \). This tangential acceleration is constant.

Step 2: Radial Acceleration at the Start

At the start, the angular velocity \( \omega = 0 \text{ rad/s} \) (since it starts from rest). Hence, the radial acceleration, given by \( a_r = \omega^2 \cdot r \), is \( a_r = 0^2 \cdot 0.300 = 0 \text{ m/s}^2 \).

Step 3: Resultant Acceleration at the Start

Since both the radial acceleration and the tangential acceleration are known, the magnitude of the resultant acceleration is given by \( a = \sqrt{a_t^2 + a_r^2} \). At the start, \( a = \sqrt{0.180^2 + 0^2} = 0.180 \text{ m/s}^2 \).

Step 4: Radial Acceleration after Turning 60°

First, calculate the angular displacement in radians: \( 60° = \pi/3 \text{ rad} \). Using \( \omega^2 = \omega_0^2 + 2\alpha\theta \) and starting from rest (\(\omega_0 = 0\)), \( \omega^2 = 2 \cdot 0.600 \cdot \pi/3 \). This gives \( \omega = \sqrt{1.2 \cdot \pi/3} \approx 1.416 \text{ rad/s} \). Then, \( a_r = \omega^2 \cdot r = 1.416^2 \cdot 0.300 \approx 0.601 \text{ m/s}^2 \).

Step 5: Resultant Acceleration after Turning 60°

Using the tangential and radial accelerations, \( a = \sqrt{a_t^2 + a_r^2} = \sqrt{0.180^2 + 0.601^2} \approx 0.627 \text{ m/s}^2 \).

Step 6: Radial Acceleration after Turning 120°

Convert the angle to radians: \( 120° = 2\pi/3 \). Again using \( \omega^2 = 0 + 2\cdot0.600\cdot 2\pi/3 \), we find \( \omega = \sqrt{2.4 \cdot \pi/3} \approx 1.999 \text{ rad/s} \). So the radial acceleration is \( a_r = 1.999^2 \cdot 0.300 \approx 1.199 \text{ m/s}^2 \).

Step 7: Resultant Acceleration after Turning 120°

With \( a_t = 0.180 \text{ m/s}^2 \) and \( a_r = 1.199 \text{ m/s}^2 \), the resultant acceleration is \( a = \sqrt{0.180^2 + 1.199^2} \approx 1.212 \text{ m/s}^2 \).

Key Concepts

Tangential Acceleration

Tangential acceleration is a key component when dealing with rotating bodies like a flywheel. It represents how fast a point on the rim of the wheel speeds up as the wheel rotates. Unlike linear acceleration, which refers to straight-line motion, tangential acceleration focuses on the speed increase along the edge of a circle.

The formula to calculate tangential acceleration (\( a_t \)) is derived from the relationship between angular acceleration (\( \alpha \)) and the radius (\( r \)) of the wheel. It is given by:

• \( a_t = \alpha \cdot r \)

Here, \( \alpha \) is the angular acceleration in radians per second squared, and \( r \) is the radius in meters. This relationship shows that the tangential acceleration is directly proportional to both the angular acceleration and the radius.

For the flywheel in our exercise, the angular acceleration is constant at \( 0.600 \text{ rad/s}^2 \), and the radius is \( 0.300 \text{ m} \). Therefore, the tangential acceleration remains constant at \( 0.180 \text{ m/s}^2 \), regardless of how far the wheel rotates.

Radial Acceleration

Radial acceleration, also known as centripetal acceleration, is the acceleration that keeps an object moving in a circular path. It points towards the center of the circle, acting like an invisible force that pulls the spinning point towards the middle.

Radial acceleration (\( a_r \)) is calculated using the formula:

• \( a_r = \omega^2 \cdot r \)

Here, \( \omega \) is the angular velocity in radians per second, and \( r \) is the radius in meters. Note that at the start, when the wheel begins from rest, the angular velocity (\( \omega \)) is zero, leading to zero radial acceleration.

As the flywheel turns through a certain angle, the radial acceleration changes. For instance, after a 60-degree turn, the flywheel gains some angular velocity because of the continuing angular acceleration. At this point, the radial acceleration becomes significant because the angular velocity is no longer zero.

Thus, radial acceleration is dependent on how fast the wheel is spinning, and it varies as the wheel’s angular velocity changes.

Resultant Acceleration

Resultant acceleration is the combination of both tangential and radial components. It's like finding the overall effect of two forces acting concurrently at a point at the edge of a rotating object. This combined acceleration gives a complete picture of how a point on the flywheel is moving.

The magnitude of resultant acceleration (\( a \)) can be calculated using the Pythagorean theorem, since tangential and radial accelerations are perpendicular:

• \( a = \sqrt{a_t^2 + a_r^2} \)

At the start, since the radial acceleration is zero due to the initial zero velocity, the resultant acceleration equals the tangential acceleration. As the flywheel turns through angles like 60° and 120°, the radial acceleration grows, altering the resultant acceleration.

For example, after a 60-degree turn, a point on the rim experiences both types of accelerations, and the resultant acceleration calculates to about \( 0.627 \text{ m/s}^2 \). After a 120-degree turn, this value increases further as the point on the rim speeds up more, leading to a resultant acceleration of around \( 1.212 \text{ m/s}^2 \).

Understanding resultant acceleration helps in grasping how various forces work together to influence a rotating system.