Problem 32 - University Physics with Modern Physics

Question

You are to design a rotating cylindrical axle to lift \(800-\mathrm{N}\) buckets of cement from the ground to a rooftop 78.0 \(\mathrm{m}\) above the ground. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle; as the axle turms, the buckets will rise. (a) What should the diameter of the axle be in order to raise the buckets at a steady 2.00 \(\mathrm{cm} / \mathrm{s}\) when it is turning at 7.5 \(\mathrm{rpm} ?\) (b) If instead the axle must give the buckets an upward acceleration of \(0.400 \mathrm{m} / \mathrm{s}^{2},\) what should the angular acceleration of the axle be?

Step-by-Step Solution

Answer

(a) Diameter ≈ 5.1 cm; (b) Angular acceleration ≈ 15.7 rad/s².

Step 1: Understanding the Problem

We need to determine two things: (a) the diameter of the rotating axle that will allow a bucket to rise at a steady speed of 2.00 cm/s when turning at 7.5 rpm, and (b) the angular acceleration of the axle when the bucket should rise with an upward acceleration of 0.400 m/s².

Step 2: Convert Units

For part (a), convert the linear speed from cm/s to m/s: \(v = 2.00 \text{ cm/s} = 0.020 \text{ m/s}\). Also convert the rotational speed from rpm to radians per second: \(7.5 \text{ rpm} = \frac{7.5 \times 2\pi}{60} \text{ rad/s} = 0.7854 \text{ rad/s}\).

Step 3: Diameter from Steady Rise (Part a)

The relationship between linear speed \(v\) and angular speed \(\omega\) is \(v = r \omega\), where \(r\) is the radius of the axle. Solving for \(r\), we have \(r = \frac{v}{\omega} = \frac{0.020}{0.7854} = 0.0255 \text{ m}\). Therefore, the diameter \(D\) is \(2r = 0.0510 \text{ m}\) or 5.10 cm.

Step 4: Determine Angular Acceleration (Part b)

For part (b), relate the linear acceleration \(a = 0.400 \text{ m/s}^2\) to angular acceleration \(\alpha\) using \(a = r \alpha\). Using \(r = 0.0255 \text{ m}\) from step 3, \(\alpha = \frac{a}{r} = \frac{0.400}{0.0255} \approx 15.69 \text{ rad/s}^2\).

Key Concepts

Rotational Motion

Rotational motion describes the motion of objects that spin or rotate about an axis. In many real-world applications like engines, wheels, and our exercise's rotating cylindrical axle, rotational motion is central to their function. Key aspects include:

- **Rotational Speed**: This is how fast an object turns, often measured in revolutions per minute (rpm) or radians per second (rad/s). - **Radius and Diameter**: These define the size of the rotating object. The diameter is twice the radius, and in our problem, finding the correct radius is essential to ensure the axle lifts the cement buckets correctly.
Understanding these concepts helps predict how the axle will behave as it turns. It ensures that the rotational speed aligns with the linear requirements of the task—raising a bucket at a specified velocity. Grasping these basics can provide a deeper insight into how rotations facilitate mechanical lifting tasks.

Angular Acceleration

Angular acceleration refers to how quickly the rotational speed of an object changes. In simpler terms, it's similar to the concept of acceleration in linear motion, but it applies to rotation. In our axle lifting problem, angular acceleration comes into play when we want to lift the bucket not just at a constant speed but with an upward acceleration. The formula connecting linear acceleration (\( a \)) and angular acceleration (\( \alpha \)) is: \[ a = r \alpha \] where \( r \) is the radius.

This relationship ensures mechanical systems coordinate their linear and rotational parts effectively. An increased angular acceleration means the axle spins faster over time, increasing the upward acceleration of the bucket. The calculated angular acceleration of approximately 15.69 rad/s² ensures the bucket accelerates upwards as required. This concept is crucial in many engineering and design applications across machines and vehicles.

Unit Conversion

Unit conversion is essential for solving physics problems accurately, as it ensures all measurements are in compatible units. In our exercise, converting units was necessary to reach the solution.

- **Linear Speed Conversion**: From cm/s to m/s by dividing the speed in cm/s by 100 (e.g., 2.00 cm/s = 0.020 m/s). - **Rotational Speed Conversion**: From rpm to rad/s involves using the formula:\[ \text{rad/s} = \frac{\text{rpm} \times 2\pi}{60} \] This accounts for converting revolutions (a full circle of \( 2\pi \) radians) to per minute, then to per second.
Accurate unit conversion prevents errors that could lead to incorrect results. Units must be consistent across equations so that calculations, such as finding radius or angular acceleration, yield the expected results. Mastering unit conversion is a valuable skill for tackling a wide range of physics problems.