Problem 37 - Applied Physics

Question

A circular disk \(30.0 \mathrm{~cm}\) in diameter is rotating at \(275 \mathrm{rpm}\) and then uniformly stops within \(8.00 \mathrm{~s}\). (a) Find its angular acceleration. (b) Find the initial linear speed of a point on its rim. (c) How many revolutions does the disk make before it stops?

Step-by-Step Solution

Answer

(a) \(-3.60 \mathrm{~rad/s^2}\); (b) \(4.32 \mathrm{~m/s}\); (c) \(18.37\) revolutions.

Step 1: Convert Rotation Rate to Radians per Second

The disk is rotating at \( 275 \mathrm{~rpm} \). First, we need to convert this rate to radians per second. Use the conversion factor: \( 1 \mathrm{~rpm} = \frac{2\pi}{60} \mathrm{~rad/s} \). \[ \omega_i = 275 \times \frac{2\pi}{60} = \frac{275 \times 2\pi}{60} \approx 28.83 \mathrm{~rad/s} \] This is the initial angular velocity of the disk.

Step 2: Calculate Angular Acceleration

Given that the disk uniformly stops, we use the formula for angular acceleration \( \alpha \) when the final angular velocity \( \omega_f = 0 \). The formula is:\[ \alpha = \frac{\omega_f - \omega_i}{t} = \frac{0 - 28.83}{8} \approx -3.60 \mathrm{~rad/s^2} \] This gives the angular acceleration of the disk.

Step 3: Find Initial Linear Speed

The initial linear speed \( v \) of a point on the rim can be found using the relation \( v = r \times \omega_i \), where \( r \) is the radius of the disk. The radius is half the diameter, so:\[ r = \frac{30.0}{2} = 15.0 \mathrm{~cm} = 0.15 \mathrm{~m} \] \[ v = 0.15 \times 28.83 \approx 4.32 \mathrm{~m/s} \] This is the initial linear speed of a point on the disk's rim.

Step 4: Calculate Total Revolutions

Using the formula for total revolutions \( \theta \) in terms of initial angular velocity and angular acceleration, we have:\[ \theta = \frac{\omega_i^2 - \omega_f^2}{2\alpha} = \frac{28.83^2 - 0}{2 \times -3.60} \approx 115.4 \mathrm{~radians} \] To convert radians to revolutions, use the relation \( 1 \mathrm{~revolution} = 2\pi \mathrm{~radians} \):\[ \text{Revolutions} = \frac{115.4}{2\pi} \approx 18.37 \text{ revolutions} \] This is the number of revolutions before the disk stops.

Key Concepts

Angular Acceleration

When an object is rotating, its change in rotational speed over time is what we call angular acceleration. It helps to think of it similar to how we understand acceleration in terms of speed when a car speeds up or slows down. The formula for angular acceleration \( \alpha \) is:

• \( \alpha = \frac{\omega_f - \omega_i}{t} \)

Where:

• \( \omega_f \) is the final angular velocity,
• \( \omega_i \) is the initial angular velocity, and
• \( t \) is the time over which the change occurs.

In our exercise, the disk's final angular velocity is 0 because it comes to a stop. Starting from an initial angular speed of \( 28.83 \text{ rad/s} \) and stopping over 8 seconds, the angular acceleration is calculated to be approximately \( -3.60 \text{ rad/s}^2 \). The negative sign indicates a decrease in speed, meaning the disk is slowing down.
Understanding this helps in analyzing rotational movements and dynamics, especially when objects slow or speed up noticeably.

Linear Speed

Linear speed is a concept often used in connection with circular motion. It refers to how fast a point on the surface of a rotating object is moving in a straight line. The formula for linear speed \( v \) linked to angular velocity \( \omega \) is:

• \( v = r \times \omega \)

Where:

• \( r \) is the radius of the object, and
• \( \omega \) is the angular velocity.

For example, in the case of the rotating disk, with a radius of \( 0.15 \text{ m} \) and an initial angular velocity of \( 28.83 \text{ rad/s} \), the initial linear speed is calculated to be approximately \( 4.32 \text{ m/s} \).
Grasping linear speed is crucial because it relates how rotational motion translates into straightforward linear motion, which is especially relevant in gears, wheels, and circular processes. It can offer insights into kinetic energy and movement dynamics in diverse physics applications.

Conversion of Units

Conversions are an essential part of solving problems in physics, particularly when dealing with angular motion. Units need to be consistent for calculations to be accurate. In this exercise, we converted the disk’s rotation from revolutions per minute (rpm) to radians per second (rad/s).

• Conversion Factor: \( 1 \text{ rpm} = \frac{2\pi}{60} \text{ rad/s} \)

For our exercise:

• Initial rotation: \( 275 \text{ rpm} \)
• Converted to angular velocity: \( 28.83 \text{ rad/s} \)

Thus, utilizing the conversion factor, we convert the revolutions per unit time into an angular displacement which is easier to work with during problem-solving.
Always keep an eye on units to ensure that every part of your computation is coherent. Mistaking or overlooking units can lead to errors that affect the whole solution.

Revolutions

Revolutions describe how many full circles or cycles an object completes. It's a straightforward yet crucial aspect of circular motionIn rotational dynamics problems, knowing how many revolutions have been completed is important for understanding overall movement and distance around a circle. To find the number of revolutions when you have angular displacement in radians, simply divide by \( 2\pi \) since \( 2\pi \) radians equals one complete revolution:

• \( \text{Revolutions} = \frac{\text{Total radians}}{2\pi} \)

In our disk example, during the slowing down phase, it covered a total of \( 115.4 \text{ radians} \). Dividing by \( 2\pi \), we find it made approximately \( 18.37 \) revolutions.
Tracking revolutions helps us understand the completeness of cycles and the extent of rotational travel in many real-life applications like machinery, motors, and even Earth's rotation.