/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 A light, flexible rope is wrappe... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(P\)?

Short Answer

Expert verified
The force P is 14.7 N.

Step by step solution

01

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.
02

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, \( I = 4.08 \times 0.25^2 = 0.255 \text{ kg}\cdot\text{m}^2 \).
03

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 \( \omega = \frac{v}{r} = \frac{6.00}{0.25} = 24.0 \text{ rad/s} \).
04

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 \( P \times 5.00 = \frac{1}{2} \times 0.255 \times 24.0^2 \).
05

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: \( P = \frac{0.5 \times 0.255 \times 576.0}{5.00} = 14.7 \text{ N} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

The flywheel of a gasoline engine is required to give up 500 J of kinetic energy while its angular velocity decreases from 650 rev/min to 520 rev/min. What moment of inertia is required?

On a compact disc (CD), music is coded in a pattern of tiny pits arranged in a track that spirals outward toward the rim of the disc. As the disc spins inside a CD player, the track is scanned at a constant \(linear\) speed of \(v =\) 1.25m/s. Because the radius of the track varies as it spirals outward, the angular speed of the disc must change as the CD is played. (See Exercise 9.20.) Let's see what angular acceleration is required to keep \(v\) constant. The equation of a spiral is \(r(\theta) = r_0 + \beta\theta\), where \(r_0\) is the radius of the spiral at \(\theta =\) 0 and \(\beta\) is a constant. On a CD, \(r_0\) is the inner radius of the spiral track. If we take the rotation direction of the CD to be positive, \(\beta\) must be positive so that \(r\) increases as the disc turns and \(\theta\) increases. (a) When the disc rotates through a small angle \(d\theta\), the distance scanned along the track is \(ds = rd\theta\). Using the above expression for \(r(\theta)\), integrate \(ds\) to find the total distance \(s\) scanned along the track as a function of the total angle \(\theta\) through which the disc has rotated. (b) Since the track is scanned at a constant linear speed \(v\), the distance s found in part (a) is equal to \(vt\). Use this to find \(\theta\) as a function of time. There will be two solutions for \(\theta\) ; choose the positive one, and explain why this is the solution to choose. (c) Use your expression for \(\theta(t)\) to find the angular velocity \(\omega_z\) and the angular acceleration \(\alpha_z\) as functions of time. Is \(\alpha_z\) constant? (d) On a CD, the inner radius of the track is 25.0 mm, the track radius increases by 1.55 mm per revolution, and the playing time is 74.0 min. Find \(r_0, \beta,\) and the total number of revolutions made during the playing time. (e) Using your results from parts (c) and (d), make graphs of \(\omega_z\) (in rad/s) versus \(t\) and \(\alpha_z\) (in rad/s\(^2\)) versus \(t\) between \(t =\) 0 and \(t =\) 74.0 min. \(\textbf{The Spinning eel.}\) American eels (\(Anguilla\) \(rostrata\)) are freshwater fish with long, slender bodies that we can treat as uniform cylinders 1.0 m long and 10 cm in diameter. An eel compensates for its small jaw and teeth by holding onto prey with its mouth and then rapidly spinning its body around its long axis to tear off a piece of flesh. Eels have been recorded to spin at up to 14 revolutions per second when feeding in this way. Although this feeding method is costly in terms of energy, it allows the eel to feed on larger prey than it otherwise could.

A uniform bar has two small balls glued to its ends. The bar is 2.00 m long and has mass 4.00 kg, while the balls each have mass 0.300 kg and can be treated as point masses. Find the moment of inertia of this combination about an axis (a) perpendicular to the bar through its center; (b) perpendicular to the bar through one of the balls; (c) parallel to the bar through both balls; and (d) parallel to the bar and 0.500 m from it.

A circular saw blade with radius 0.120 m starts from rest and turns in a vertical plane with a constant angular acceleration of 2.00 rev/s\(^2\). After the blade has turned through 155 rev, a small piece of the blade breaks loose from the top of the blade. After the piece breaks loose, it travels with a velocity that is initially horizontal and equal to the tangential velocity of the rim of the blade. The piece travels a vertical distance of 0.820 m to the floor. How far does the piece travel horizontally, from where it broke off the blade until it strikes the floor?

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