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Chapter 10: Problem 56

A uniform hollow disk has two pieces of thin light wire wrapped around its outer rim and is supported from the ceiling (Fig. 10.51 ). Suddenly one of the wires breaks, and the remaining wire does not slip as the disk rolls down. Use energy conservation to find the speed of the center of this disk after it has fallen a distance of 1.20 \(\mathrm{m} .\)

Short Answer

Expert verified
The speed of the disk's center after falling 1.20 m is approximately 3.43 m/s.

Step by step solution

01

Identify the Given Values and Constants

We begin by identifying the given values from the problem. The disk falls a distance of \( h = 1.20 \) meters. The mass of the hollow disk (\( m \)), radius (\( R \)), and gravitational acceleration (\( g = 9.81 \text{ m/s}^2 \)) will be used. Additionally, the moment of inertia for a hollow disk is \( I = mR^2 \).
02

Setup the Energy Conservation Equation

The conservation of mechanical energy states that the total mechanical energy (potential plus kinetic) at the start will be equal to the total mechanical energy after falling 1.20 m. Initially, all energy is gravitational potential: \( U_i = mgh \). When the disk has fallen the distance, its energy is kinetic: \( K_f = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \).
03

Relate Linear and Angular Speeds

Since the disk rolls without slipping, the linear speed \( v \) is related to the angular speed \( \omega \) by \( \omega = \frac{v}{R} \). Substitute this relationship into the kinetic energy \( I \omega^2 \) term: \( K_f = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)\left(\frac{v}{R}\right)^2 \).
04

Simplify the Energy Equation and Solve for \( v \)

Substitute \( \omega = \frac{v}{R} \) into the kinetic energy equation to get: \( K_f = \frac{1}{2}mv^2 + \frac{1}{2}m v^2 = mv^2 \). Set \( mv^2 = mgh \), thus \( v^2 = gh \). Solve for \( v \): \( v = \sqrt{gh} \).
05

Calculate the Speed

Plug in the given numerical values: \( v = \sqrt{9.81 \text{ m/s}^2 \times 1.20 \text{ m}} \). This equals \( v = \sqrt{11.772} \approx 3.43 \text{ m/s} \).

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Key Concepts

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

Moment of Inertia
The moment of inertia is a crucial concept when discussing rotational dynamics. It acts as the rotational equivalent of mass in linear motion. Specifically, it measures how difficult it is to change the rotational state of an object. For a hollow disk, the moment of inertia is given by the formula:
  • \( I = mR^2 \),
where \( m \) is the mass of the disk and \( R \) is its radius.

This formula shows that the further the mass is distributed from the axis of rotation, the higher the inertia. Hence, hollow disks have a different moment of inertia compared to solid ones, even if they have the same mass and radius.

Understanding moment of inertia helps explain why different shaped objects might roll down a hill at different speeds. A hollow disk, with its unique inertia, primarily resists rotational changes due to its mass being distributed towards the edge.
Rotational Motion
Rotational motion occurs when an object spins around an internal or external axis. In the context of our problem, the hollow disk displays rotational motion as it rolls down without slipping.

In such scenarios, linear speed \( v \) is intricately linked to angular speed \( \omega \). The relationship is given by:
  • \( \omega = \frac{v}{R} \),
where \( R \) is the radius of the disk.

This equation illustrates that faster linear motion directly translates to faster rotational motion, provided the disk doesn't slip. As the disk begins its descent, both linear and rotational kinetic energies factor into the energy conservation equation, underscoring the dual nature of the disk's movement.
Kinetic and Potential Energy
Energy conservation principles are key to solving the problem of the disk's falling speed. Initially, the disk possesses potential energy thanks to its height. This energy is expressed by:
  • \( U_i = mgh \),
where \( m \) is mass, \( g \) is gravitational acceleration, and \( h \) is height. As the disk moves, gravitational potential energy transforms into kinetic energy.

The kinetic energy of a rolling object like our disk comprises two components:
  • Translational kinetic energy: \( \frac{1}{2}mv^2 \),
  • Rotational kinetic energy: \( \frac{1}{2}I\omega^2 \).
This comprehensive view explains how gravitational potential energy is entirely converted into these two forms of kinetic energy during its descent. Ultimately, all potential energy converts to these kinetic forms, allowing us to derive the disk's speed after falling a defined distance.

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Most popular questions from this chapter

Under some circumstances, a star callapse into an extremely dense object made mostly of neurrons and called a neutron star. The density of a neutron star is roughly \(10^{14}\) times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was \(7.0 \times 10^{5} \mathrm{km}\) (comparable to our sun); its final radius is 16 \(\mathrm{km}\) . If the original star rotated once in 30 days, find the angular speed of the neutron star.

Two metal disks, one with radius \(R_{1}=2.50 \mathrm{cm}\) and mass \(M_{1}=0.80 \mathrm{kg}\) and the other with radius \(R_{2}=5.00 \mathrm{cm}\) and mass \(M_{2}=1.60 \mathrm{kg}\) , are welded together and mounted on a frictionless axis through their common center, as in Problem \(9.89 .\) (a) A light string is wrapped around the edge of the smaller disk, and a 1.50 \(\mathrm{kg}\) block is suspended from the free end of the string. What is the magnitude of the downward acceleration of the block after it is released? (b) Repeat the calculation of part (a), this time with the string wrapped around the edge of the larger disk. In which case is the acceleration of the block greater? Does your answer make sense?

A \(15.0-\mathrm{kg}\) bucker of water is suspended by a very light rope wrapped around a solid uniform cylinder 0.300 \(\mathrm{m}\) in diameter with mass 12.0 \(\mathrm{kg}\) . The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 10.0 \(\mathrm{m}\) to the water. (a) What is the tension in the rope while the bucket is falling? (b) With what speed does the bucket strike the water? (c) What is the time of fall? (d) While the bucket is falling, what is the force exerted on the cylinder by the axle?

A cord is wrapped around the rim of a solid uniform wheel 0.250 \(\mathrm{m}\) in radius and of mass 9.20 \(\mathrm{kg}\) . A steady horizontal pull of 40.0 \(\mathrm{N}\) to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. (a) Compute the angular acceleration of the wheel and the acceleration of the part of the cord that has already been pulled off the wheel. (b) Find the magnitude and direction of the force that the axe exerts on the wheel. (c) Which of the answers in parts (a) and (b) would change if the pull were upward instead of horizontal?

A block with mass \(m=\) 5.00 \(\mathrm{kg}\) slides down a surface inclined \(36.9^{\circ}\) to the horizontal (Fig. 10.55 ). The coefficient of kinetic friction is 0.25 . A string attached to the block is wrapped around a flywheel on a fixed axis at \(O\) . The flywheel has mass 25.0 \(\mathrm{kg}\) and moment of inertia 0.500 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.200 \(\mathrm{m}\) from that axis. (a) What is the acceleration of the block down the plane? (b) What is the tension in the string?
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