/*! 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 37 A wad of sticky clay with mass \... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 37

A wad of sticky clay with mass \(m\) and velocity \(\mathbf{v}_{i}\) is fired at a solid cylinder of mass \(M\) and radius \(R\) (Figure \(\mathrm{P} 11.37\) ). The cylinder is initially at rest and is mounted on a fixed horizontal axle that runs through its center of mass. The line of motion of the projectile is perpendicular to the axle and at a distance \(d

Short Answer

Expert verified
The angular speed of the system just after the clay strikes and sticks to the surface of the cylinder is \(\omega = \frac{m \cdot v_i \cdot d}{(M+m) \cdot R^{2}}\). The mechanical energy of the clay-cylinder system is not conserved in this process because it involves an inelastic collision.

Step by step solution

01

Identify Initial conditions

Initially, the system is at a rest with no angular velocity. The clay has a mass \(m\) and velocity \(\mathbf{v}_{i}\). The clay hits the cylinder at a distance \(d\) from its center of mass.
02

Conservation of Angular Momentum

Since there's no external torque acting on the system, total angular momentum before and after the collision is conserved. That is \(m \cdot v_i \cdot d = (M+m) \cdot \omega R^{2}\), where \(\omega\) is the angular speed of the system after the collision.
03

Solve for Angular Speed

To find the angular speed, rearrange the equation to solve for \(\omega\): \[\omega = \frac{m \cdot v_i \cdot d}{(M+m) \cdot R^{2}}\]
04

Analysis of Mechanical Energy Conservation

Mechanical energy in general can be preserved if the collision is completely elastic. In this case, however, the clay sticks to the surface of the cylinder, thus making it an inelastic collision. So, mechanical energy of the system is not preserved during this process.

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Key Concepts

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

Understanding Inelastic Collisions
In an inelastic collision, two objects collide and do not bounce off each other; they stick together or deform upon impact, which means that they move with a common velocity after the collision. Unlike elastic collisions, where kinetic energy is conserved, in inelastic collisions, some of the kinetic energy is transformed into other forms of energy, such as heat or sound.

The wad of clay-firing exercise illustrates this when the clay hits and sticks to the cylinder, coalescing into a single rotating mass. The conservation of momentum is a crucial principle used to solve such problems. Since this is an inelastic encounter, the kinetic energy before and after isn't the same as the collision is not perfectly elastic. This inelastic nature causes the kinetic energy to be partially converted to other forms, explaining why in terms of energy, not everything adds up as it did before the impact.
Angular Speed and its Relevance
Angular speed describes how fast an object rotates or revolves relative to another point, expressed as radians per second. The angular speed tells us the rate at which the angle changes as a body rotates around an axis.

In the given exercise, the angular speed \(\omega\) is the speed at which the combined mass of the clay and cylinder rotate after the collision. The crucial aspect here is understanding that angular momentum before the collision must equal that after the collision, assuming no outside torques act on it. By applying the conservation of angular momentum, we can solve for \(\omega\) and determine the new rotational state of the system.
Mechanical Energy Conservation
Energy conservation is a fundamental principle stating that the total mechanical energy in a closed system remains constant if only conservative forces are acting. Mechanical energy is the sum of potential and kinetic energy.

However, in the case of inelastic collisions, like the scenario with the clay and cylinder, the mechanical energy of the system isn't conserved. The kinetic energy before the collision is not equal to the kinetic energy after鈥攕ome of it gets transformed into other forms of energy due to the deformation and heat generated during the inelastic collision. Such an event is a prime example of where mechanical energy conservation does not apply because the energy is not simply transferred but changed in form.

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Most popular questions from this chapter

Global warming is a cause for concern because even small changes in the Earth's temperature can have significant consequences. For example, if the Earth's polar ice caps were to melt entirely, the resulting additional water in the oceans would flood many coastal cities. Would it appreciably change the length of a day? Calculate the resulting change in the duration of one day. Model the polar ice as having mass \(2.30 \times 10^{19} \mathrm{kg}\) and forming two flat disks of radius \(6.00 \times 10^{5} \mathrm{m} .\) Assume the water spreads into an unbroken thin spherical shell after it melts.

A puck of mass \(m\) is attached to a cord passing through a small hole in a frictionless, horizontal surface (Fig. P11.49). The puck is initially orbiting with speed \(v_{i}\) in a circle of \(\mathrm{ra}-\) dius \(r_{i}\). The cord is then slowly pulled from below, decreasing the radius of the circle to \(r\). (a) What is the speed of the puck when the radius is \(r ?\) (b) Find the tension in the cord as a function of \(r .\) (c) How much work \(W\) is done in moving \(m\) from \(r_{i}\) to \(r ?\) (Note: The tension depends on \(r\) ) (d) Obtain numerical values for \(v, T,\) and \(W\) when \(r=0.100 \mathrm{m}\) \(m=50.0 \mathrm{g}, r_{i}=0.300 \mathrm{m},\) and \(v_{i}=1.50 \mathrm{m} / \mathrm{s}.\)

In the Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius \(0.529 \times 10^{-10} \mathrm{m}\) around the proton. Assuming the orbital angular momentum of the electron is equal to \(h / 2 \pi,\) calculate (a) the orbital speed of the electron, (b) the kinetic energy of the electron, and (c) the angular frequency of the electron's motion.

A 60.0-kg woman stands at the rim of a horizontal turntable having a moment of inertia of \(500 \mathrm{kg} \cdot \mathrm{m}^{2}\) and a radius of \(2.00 \mathrm{m} .\) The turntable is initially at rest and is free to rotate about a frictionless, vertical axle through its center. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of \(1.50 \mathrm{m} / \mathrm{s}\) relative to the Earth. (a) In what direction and with what angular speed does the turntable rotate? (b) How much work does the woman do to set herself and the turntable into motion?

A space station is constructed in the shape of a hollow ring of mass \(5.00 \times 10^{4} \mathrm{kg} .\) Members of the crew walk on a deck formed by the inner surface of the outer cylindrical wall of the ring, with radius 100 m. At rest when constructed, the ring is set rotating about its axis so that the people inside experience an effective free-fall acceleration equal to \(g .\) (Figure \(\mathrm{P} 11.27\) shows the ring together with some other parts that make a negligible contribution to the total moment of inertia.) The rotation is achieved by firing two small rockets attached tangentially to opposite points on the outside of the ring. (a) What angular momentum does the space station acquire? (b) How long must the rockets be fired if each exerts a thrust of \(125 \mathrm{N} ?\) (c) Prove that the total torque on the ring, multiplied by the time interval found in part (b), is equal to the change in angular momentum, found in part (a). This equality represents the angular impulse-angular momentum theorem.
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