Problem 42 - University Physics with Modern Physics

Question

A light, flexible rope is wrapped several times around a \(hollow\) cylinder, with a weight of 40.0 N and a radius of 0.25 m, that rotates without friction about a fixed horizontal axis. The cylinder is attached to the axle by spokes of a negligible moment of inertia. The cylinder is initially at rest. The free end of the rope is pulled with a constant force \(P\) for a distance of 5.00 m, at which point the end of the rope is moving at 6.00 m/s. If the rope does not slip on the cylinder, what is \(P\)?

Step-by-Step Solution

Answer

The force P is 14.7 N.

Step 1: Understand the system

We have a hollow cylinder with a radius of 0.25 m and a weight of 40.0 N, which indicates a mass of approximately 4.08 kg (using gravity as 9.8 m/s²). The rope is wrapped around and pulls with a force P that we need to find, given that the cylinder starts from rest and reaches a speed of 6.00 m/s after 5.00 m.

Step 2: Calculate the moment of inertia

The moment of inertia for a hollow cylinder about its central axis is given by: \( I = M r^2 \).Where \( M \) is the mass (4.08 kg) and \( r \) is the radius (0.25 m). Thus, \( I = 4.08 \times 0.25^2 = 0.255 \text{ kg}\cdot\text{m}^2 \).

Step 3: Relate linear and angular quantities

Since there is no slipping, the linear speed \(v\) and angular speed \(\omega\) are related by: \( v = r\omega \). At the end of 5.00 m, \( v = 6.00 \text{ m/s} \) and \( r = 0.25 \text{ m} \), so \( \omega = \frac{v}{r} = \frac{6.00}{0.25} = 24.0 \text{ rad/s} \).

Step 4: Use the work-energy principle

The work done by the rope is equal to the rotational kinetic energy gained by the cylinder. The work done by the rope is \( W = P \times d \), where \( d = 5.00 \text{ m} \). The rotational kinetic energy is \( K \) = \( \frac{1}{2}I\omega^2 \).Thus, equate \( P \times 5.00 = \frac{1}{2} \times 0.255 \times 24.0^2 \).

Step 5: Solve for force P

Substitute \( I = 0.255 \text{ kg}\cdot\text{m}^2 \) and \( \omega = 24.0 \text{ rad/s} \) into the equation:\( P \times 5.00 = \frac{1}{2} \times 0.255 \times 24.0^2 \).Calculate: \( P = \frac{0.5 \times 0.255 \times 576.0}{5.00} = 14.7 \text{ N} \).

Key Concepts

Rotational Kinematics

Rotational kinematics describes the motion of rotating objects, much like linear kinematics describes the motion in a straight line. In rotational kinematics, we deal with quantities such as angular position, angular velocity, and angular acceleration. These concepts are closely related to their linear counterparts but apply to objects rotating about an axis.
When a force is applied to an object wrapped around a rotating cylinder, like in our example, it sets the object into a rotational motion. The angular speed (\( \omega \)) is a key variable in rotational motion, representing how fast the object spins. Just as in linear motion where speed is distance covered over time, angular speed is the angle covered in a given time. It's related to linear speed by the equation: \( v = r \omega \), where \( v \) is linear speed, and \( r \) is the radius of the cylinder.
This relationship is crucial because it shows that you can determine how fast something is moving along a path by knowing how it spins around its axis and the size of the path (the radius). Therefore, studying rotational kinematics broadens our understanding of motion and provides essential tools for solving mechanical problems.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotation. It plays a similar role to mass in linear dynamics, but specifically for rotational motion. The larger the moment of inertia, the harder it is to start or stop an object's rotation.
For a hollow cylinder, like the one in our problem, the moment of inertia (\(I\)) is calculated using the formula \( I = M r^2 \), where \(M\) is the mass, and \(r\) is the radius. This formula tells us that as either the mass or the radius increases, the moment of inertia will also increase, making it more difficult to change its rotational state.
In the exercise, we calculated the moment of inertia as \(0.255 \text{ kg}\cdot\text{m}^2\), which is used to determine how the cylinder reacts under the applied force. Understanding this concept not only enhances problem-solving skills but also provides insights into designing systems that involve rotational motion, such as gears and pulleys.

Work-Energy Principle

The work-energy principle in physics states that work done on an object results in a change in energy. In the case of rotational motion, this principle connects the work done by a force with the rotational kinetic energy gained by the object.
In simpler terms, when you apply a force to turn an object, you do work on it. This work gets converted into the energy of motion, or kinetic energy. For rotations, the formula for rotational kinetic energy is \( K = \frac{1}{2} I \omega^2 \), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.
In the cylinder problem, the force \(P\) pulls the rope, performing work calculated by \(W = P \times d\), where \(d\) is the distance the rope was pulled. By using the work-energy principle, we equated this work to the cylinder's gained kinetic energy, allowing us to solve for \(P\). This approach showcases how energy principles can simplify solving physical problems by focusing on energy conversion.