Problem 8 - College Physics

Question

A cylinder with a diameter of \(20 \mathrm{~cm}\) rolls with an angular speed of \(0.050 \mathrm{rad} / \mathrm{s}\) on a level surface. If the cylinder experiences a uniform tangential acceleration of \(0.018 \mathrm{~m} / \mathrm{s}^{2}\) without slipping until its angular speed is \(1.2 \mathrm{rad} / \mathrm{s},\) through how many complete revolutions does the cylinder rotate during the time it accelerates?

Step-by-Step Solution

Answer

The cylinder rotates approximately 0.65 complete revolutions during acceleration.

Step 1: Determine the Initial and Final Angular Speed

The initial angular speed \( \omega_i \) of the cylinder is given as \( 0.050 \, \mathrm{rad/s} \) and the final angular speed \( \omega_f \) is \( 1.2 \, \mathrm{rad/s} \).

Step 2: Calculate the Angular Acceleration

The tangential acceleration is \( a_t = 0.018 \, \mathrm{m/s^2} \). The radius of the cylinder is half of the diameter: \( r = \frac{20 \, \mathrm{cm}}{2} = 10 \, \mathrm{cm} = 0.1 \, \mathrm{m} \). The angular acceleration is calculated using the formula \( \alpha = \frac{a_t}{r} \). Substituting in the values, we find: \[ \alpha = \frac{0.018}{0.1} = 0.18 \, \mathrm{rad/s^2}. \]

Step 3: Calculate Time to Reach Final Speed

Using the formula for angular acceleration \( \omega_f = \omega_i + \alpha \times t \), we can solve for the time \( t \):\[1.2 = 0.050 + 0.18 \times t \Rightarrow t = \frac{1.2 - 0.050}{0.18} \approx 6.39 \, \mathrm{s}.\]

Step 4: Calculate Angular Displacement in Radians

The angular displacement \( \theta \) can be calculated using the formula:\[ \theta = \omega_i \times t + \frac{1}{2} \times \alpha \times t^2 \]Substitute the known values:\[ \theta = 0.050 \times 6.39 + \frac{1}{2} \times 0.18 \times (6.39)^2 \approx 4.09 \, \mathrm{rad}. \]

Step 5: Convert Angular Displacement from Radians to Revolutions

To convert the angular displacement from radians to revolutions, use the conversion factor of \( 2\pi \) radians per revolution:\[ \text{Number of revolutions} = \frac{4.09}{2\pi} \approx 0.65 \, \mathrm{revolutions}. \]

Key Concepts

Cylindrical Motion

Cylindrical motion involves the movement of a cylindrical object, such as a roller or a wheel, as it rolls on a surface. Here, the cylinder showcases both rotation and translation. This dual nature makes cylindrical motion distinct and interesting.
As a cylinder rolls without slipping, every point on its surface that touches the ground momentarily comes to a stop. This means that the linear distance it covers is equal to the distance it rotates. In our exercise, the cylinder with a diameter of 20 cm has to convert its angular speed into linear speed as it moves forward.

• The cylinder' diameter determines its circumference, linking its rotation to translation on the surface.
• The rolling motion ensures equal progress both rotationally and linearly, adhering to principles of physics.

This is a classical problem used to understand concepts like angular displacement, angular velocity, and how they relate to real-world objects like wheels or barrels on a road.

Angular Acceleration

Angular acceleration (\(\alpha\)) is the rate of change of angular velocity over time. It tells us how quickly an object speeds up or slows down in its rotational motion. In our scenario, the cylinder's angular velocity increases due to this uniform acceleration.
The formula for calculating angular acceleration is: \(\alpha = \frac{a_t}{r}\), where \(a_t\) is tangential acceleration and \(r\) is the radius of the cylinder. This relationship is key to connecting linear and rotational motion.

• Tangential acceleration causes changes in the cylinder's angular speed.
• The radius plays a crucial role in determining how effectively a given tangential force can accelerate the cylinder rotationally.

By understanding angular acceleration, we can see how rotational activities lead to observable changes in velocity and movement.

Tangential Acceleration

Tangential acceleration is concerned with the linear acceleration that occurs along the edge of a rotating object. This can be thought of as how fast a point on the edge of the cylinder is gaining speed as the cylinder rolls. In the exercise, the cylinder experiences a tangential acceleration of 0.018 m/s².
To find out how this affects the rotation, it's essential to recognize that tangential acceleration is directly linked to angular acceleration through the cylinder's radius.

• The equation \(a_t = \alpha \times r\) defines the relationship between angular and tangential accelerations.
• Tangential acceleration is a key factor in determining how quickly an object reaches its final angular speed in rolling motion.

Simplifying this through observation and calculations helps sharpen intuition regarding rotational forces in everyday phenomena.

Kinematics in Rotation

Kinematics in rotation focuses on the study of rotational motion without considering the forces that cause it. This field covers properties like angular displacement, angular velocity, and angular acceleration, comparing them to their linear counterparts.
In this exercise, we are looking at how these properties allow us to calculate movements in terms of radians and later, revolutions—a complete cycle of motion for any rotating body.

• The angular displacement \(\theta\) offers a glance into how far the cylinder has rolled, revealing the total number of rotations it undergoes.
• Conversions between radians and revolutions (\(2\pi\) radians = 1 revolution) are foundational to understanding circular motion comprehensively.

Kinematics in rotational contexts make possible the exploration of complex phenomena like gears, wheels, and record players—each embodying rotational activities in a variety of ways.