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A \(4.00-\mathrm{m}\) length of light nylon cord is wound around a uniform cylindrical spool of radius \(0.500 \mathrm{m}\) and mass 1.00 kg. The spool is mounted on a frictionless axle and is initially at rest. The cord is pulled from the spool with a constant acceleration of magnitude \(2.50 \mathrm{m} / \mathrm{s}^{2}\). (a) How much work has been done on the spool when it reaches an angular speed of \(8.00 \mathrm{rad} / \mathrm{s} ?\) (b) Assuming there is enough cord on the spool, how long does it take the spool to reach this angular speed? (c) Is there enough cord on the spool?

Short Answer

Expert verified
The work done on the spool is 16 J. The time required for the spool to reach the angular speed of \( 8.00 rad/s \) is 1.60 s. There is enough cord on the spool as the required length is less than the available length of the cord.

Step by step solution

01

Calculate the Final Rotational Kinetic Energy

The final rotational kinetic energy of the spool, when it is rotating with an angular speed \( \omega = 8.00 \, rad/s \), can be calculated using the formula for rotational kinetic energy: \( K = 0.5 * I * \omega^2 \). The moment of inertia 'I' of a uniform cylindrical spool can be calculated as \( I = 0.5 * m * R^2 \) where 'm' is the mass of the spool and 'R' is the radius.
02

Determine the Work Done

According to the Work-Energy theorem, the work done on the spool is equal to the change in its kinetic energy. Since the spool starts from rest, its initial kinetic energy is zero. Therefore, the work done \( W \) is equal to its final kinetic energy, or \( W = K \).
03

Calculate the Time to Reach Angular Speed

Using the relation between linear and angular accelerations \( \alpha = a/R \), the angular acceleration 'a' is given by \( a = 2.50 m/s^2 / 0.500 m = 5.00 rad/s^2 \). Using the equation of motion \( \omega = \omega_0 + \alpha * t \), where \( \omega_0 \) is the initial angular velocity (which is zero as the spool starts from rest) and 't' is time, solve for 't'.
04

Verify if Enough Cord is Available

The total angular displacement \( \theta \) while accelerating to final angular speed can be calculated using the equation \( \theta = \omega_0 * t + 0.5 * \alpha * t^2 \). After obtaining \( \theta \), the total length of the cord required can be calculated as \( R * \theta \). Using the given length of the cord, check if the cord is long enough for the spool to reach the final angular speed.

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Key Concepts

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

Rotational Kinetic Energy
Rotational kinetic energy is a crucial concept in rotational dynamics that describes the energy an object has due to its rotation. It is similar to the kinetic energy in linear motion, but instead of dealing with mass and linear velocity, it deals with the moment of inertia and angular velocity.
The formula for rotational kinetic energy is given by:
\[ K = \frac{1}{2} I \omega^2 \]where:
  • \( K \) is the rotational kinetic energy.
  • \( I \) is the moment of inertia.
  • \( \omega \) is the angular speed.
In the original exercise, for a spool of radius \( 0.500 \) m and mass \( 1.00 \) kg, the moment of inertia needs to be calculated first to find its rotational kinetic energy when it's spinning at the given angular speed. The rotational kinetic energy becomes essential for understanding how much work has been performed on the system.
Moment of Inertia
The moment of inertia is a measure of how difficult it is to change the rotational speed of an object. It depends on the distribution of mass around the axis of rotation and plays a role analogous to mass in linear motion.
For a cylindrical spool, like in the exercise, the moment of inertia is calculated using:
\[ I = \frac{1}{2} m R^2 \]where:
  • \( m \) is the mass of the spool.
  • \( R \) is the radius.
The spool's mass and radius are given, allowing calculation of its moment of inertia. This value is critical when calculating the rotational kinetic energy or determining the work done on the system, linking directly into the problems tackled in rotational dynamics.
Work-Energy Theorem
The work-energy theorem is a fundamental principle that relates the work done on an object to its change in kinetic energy. In the context of rotational dynamics, this theorem shows how the work required to spin an object is directly connected to its rotational kinetic energy.
According to the work-energy theorem:
\[ W = \Delta K \]where \( W \) is the work done, and \( \Delta K \) is the change in kinetic energy.
For a spool starting from rest, the initial kinetic energy is zero, making the work done equal to its final rotational kinetic energy. Understanding this principle enables the determination of how much work is needed to achieve a certain angular speed in the spool, as calculated in part of the exercise. This theorem provides a straightforward way to connect rotational motion and energy input.
Angular Acceleration
Angular acceleration describes how quickly an object's rotational speed changes and is analogous to linear acceleration in translational motion. It helps understand how fast an object accelerates rotationally when a force is applied.
In rotational dynamics, angular acceleration \( \alpha \) can be derived from linear acceleration \( a \) with the equation:
\[ \alpha = \frac{a}{R} \]where:
  • \( a \) is the linear acceleration of the cord being pulled.
  • \( R \) is the radius.
For the spool in the exercise, the angular acceleration is calculated as part of finding out how long it takes to reach a certain angular speed. This leads to solving important equations of motion that describe the system's dynamics in the rotational context.

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Most popular questions from this chapter

(a) Determine the acceleration of the center of mass of a uniform solid disk rolling down an incline making angle \(\theta\) with the horizontal. Compare this acceleration with that of a uniform hoop. (b) What is the minimum coefficient of friction required to maintain pure rolling motion for the disk?

(a) Without the wheels, a bicycle frame has a mass of \(8.44 \mathrm{kg} .\) Each of the wheels can be roughly modeled as a uniform solid disk with a mass of \(0.820 \mathrm{kg}\) and a radius of \(0.343 \mathrm{m} .\) Find the kinetic energy of the whole bicycle when it is moving forward at \(3.35 \mathrm{m} / \mathrm{s}\). (b) Before the invention of a wheel turning on an axle, ancient people moved heavy loads by placing rollers under them. (Modern people use rollers too. Any hardware store will sell you a roller bearing for a lazy susan.) A stone block of mass 844 kg moves forward at \(0.335 \mathrm{m} / \mathrm{s}\), supported by two uniform cylindrical tree trunks, each of mass \(82.0 \mathrm{kg}\) and radius \(0.343 \mathrm{m}\) No slipping occurs between the block and the rollers or between the rollers and the ground. Find the total kinetic energy of the moving objects.

The density of the Earth, at any distance \(r\) from its center, is approximately $$ \rho=[14.2-11.6(r / R)] \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} $$ where \(R\) is the radius of the Earth. Show that this density leads to a moment of inertia \(I=0.330 M R^{2}\) about an axis through the center, where \(M\) is the mass of the Earth.

A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of \(12.0 \mathrm{rad} / \mathrm{s}\) in 3.00 s. Find (a) the magnitude of the angular acceleration of the wheel and (b) the angle in radians through which it rotates in this time.

A uniform solid disk and a uniform hoop are placed side by side at the top of an incline of height \(h\). If they are released from rest and roll without slipping, which object reaches the bottom first? Verify your answer by calculating their speeds when they reach the bottom in terms of \(h\)

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