Problem 21 - University Physics with Modern Physics

Question

A wheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of 3.00 rad/s\(^2\). Compute the radial acceleration of a point on the rim for the instant the wheel completes its second revolution from the relationship (a) \(a_{rad} = \omega^2r\) and (b) \(a_{rad} = v^2/r\)

Step-by-Step Solution

Answer

The radial acceleration is approximately 15.08 m/s² for both (a) and (b).

Step 1: Find Angular Displacement Required

To find when the wheel completes its second revolution, calculate the total angular displacement. Since one revolution is \(2\pi\) rad, two revolutions will be \(4\pi\) rad.

Step 2: Calculate Final Angular Velocity

Use the formula for angular motion: \(\omega^2 = \omega_0^2 + 2\alpha \theta\), where \(\omega_0 = 0\) (starts from rest), \(\alpha = 3.00\,\text{rad/s}^2\), and \(\theta = 4\pi\). Solving for \(\omega\), we get:\[\omega^2 = 0 + 2 \times 3.00 \times 4\pi\]\[\omega^2 = 24\pi\]\.\[\omega \approx \sqrt{24\pi} \approx 8.68 \, \text{rad/s}\].

Step 3: Find Radius of the Wheel

The diameter of the wheel is given as 40.0 cm. Thus, the radius \(r\) is half of the diameter: \(r = 20.0\, \text{cm} = 0.20\, \text{m}\).

Step 4: Calculate Radial Acceleration Using \(a_{rad} = \omega^2 r\)

Using the formula \(a_{rad} = \omega^2 r\) and \(\omega = 8.68 \, \text{rad/s}\), \(r = 0.20 \, \text{m}\):\[a_{rad} = (8.68)^2 \times 0.20 \approx 15.08 \, \text{m/s}^2\].

Step 5: Calculate Linear Velocity

The linear velocity \(v\) is related to the angular velocity by \(v = \omega r\). Thus,:\[v = 8.68 \times 0.20 \approx 1.74 \, \text{m/s}\].

Step 6: Calculate Radial Acceleration Using \(a_{rad} = v^2/r\)

Using the formula \(a_{rad} = v^2 / r\) with \(v = 1.74 \, \text{m/s}\) and \(r = 0.20 \, \text{m}\), calculate:\[a_{rad} = \frac{(1.74)^2}{0.20} \approx 15.08 \, \text{m/s}^2\].

Key Concepts

Angular Acceleration

Angular acceleration is a measure of how quickly an object's rotational speed changes. It is represented by the symbol \(\alpha \), and is typically measured in radians per second squared (rad/s\(^2\)). In the context of the rotating wheel starting from rest with a constant angular acceleration of 3.00 rad/s\(^2\), the wheel's speed increases gradually until it reaches the desired rotational speed. Angular acceleration can be calculated using the formula:

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

This formula denotes that angular acceleration is the change in angular velocity (\( \Delta \omega \)) over the change in time (\( \Delta t \)). In this exercise, since the wheel starts from rest, the initial angular velocity (\( \omega_0 \)) is zero. Thus, the formula effectively shows us how the wheel spins faster, resulting in an increasing angular velocity.

Angular Displacement

Angular displacement refers to the angle through which an object rotates about a fixed point. In our exercise, the wheel undergoes an angular displacement when it turns completely around. Measured in radians, angular displacement is calculated by multiplying the number of complete revolutions by \( 2\pi \) radians per revolution. As given in the step by step solution, to find the angular displacement of a wheel that completes its second revolution, we compute:

• Total angular displacement \( \theta = 2 \times 2\pi \text{ rad} = 4\pi \text{ rad} \)

This measurement tells us how far around the circle the wheel has moved. Understanding angular displacement is crucial when linking how linear motion translates to circular motion, as each revolution brings the wheel closer to a full circle.

Angular Velocity

Angular velocity defines how fast an object rotates or revolves relative to another point, measured in radians per second (rad/s). Think of it as the rate at which the angle changes with time. In the case of the wheel in the problem, the angular velocity was initially zero, as it started from rest. It then increased due to the wheel's constant angular acceleration. To calculate the final angular velocity (\( \omega \)) when the wheel finishes its second rotation, we use the formula:

• \( \omega^2 = \omega_0^2 + 2\alpha \theta \)

With the values provided, solving gives \( \omega \approx 8.68 \, \text{rad/s} \), indicating how fast the wheel is spinning relative to its center. This angular velocity also helps derive the radial acceleration, essential to analyze the motion of any point on the wheel's rim.

Linear Velocity

Linear velocity is the rate at which an object moves along a path. In circular motion, it's derived from the angular velocity and connects rotational and linear movement. To find the linear velocity (\( v \)) of a point on the rim of our wheel, we use the relationship with the angular velocity:

• \( v = \omega r \)

Given that \( \omega = 8.68 \, \text{rad/s} \) and the radius \( r = 0.20 \, \text{m} \), we calculate \( v \approx 1.74 \, \text{m/s} \). This value reveals the speed at which a point on the edge of the wheel is moving in a straight line. Linear velocity is crucial in understanding the mechanics of objects in circular path movements. It exemplifies how angular metrics convert into practical speeds on physical paths, aiding in computing radial acceleration using \( a_{rad} = \frac{v^2}{r} \). This seamless conversion highlights the interconnected nature of rotational and linear motion.