/*! 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

A projectile of mass \(m\) moves to the right with a speed \(v_{i}\) (Fig. \(P 11.50 a) .\) The projectile strikes and sticks to the end of a stationary rod of mass \(M,\) length \(d,\) pivoted about a frictionless axle through its center (Fig. P11.50b). (a) Find the angular speed of the system right after the collision. (b) Determine the fractional loss in mechanical energy due to the collision.

Two astronauts (Fig. P11.51), each having a mass \(M\), are connected by a rope of length \(d\) having negligible mass. They are isolated in space, orbiting their center of mass at speeds \(v.\) Treating the astronauts as particles, calculate (a) the magnitude of the angular momentum of the system and (b) the rotational energy of the system. By pulling on the rope, one of the astronauts shortens the distance between them to \(d / 2 .\) (c) What is the new angular momentum of the system? (d) What are the astronauts' new speeds? (e) What is the new rotational energy of the system? (f) How much work does the astronaut do in shortening the rope?

A uniform solid disk of mass \(3.00 \mathrm{kg}\) and radius \(0.200 \mathrm{m}\) rotates about a fixed axis perpendicular to its face. If the angular frequency of rotation is \(6.00 \mathrm{rad} / \mathrm{s},\) calculate the angular momentum of the disk when the axis of rotation \((a)\) passes through its center of mass and (b) passes through a point midway between the center and the rim.

A 100 -kg uniform horizontal disk of radius \(5.50 \mathrm{m}\) turns without friction at 2.50 rev/s on a vertical axis through its center, as in Figure \(\mathrm{P} 11.46 .\) A feedback mechanism senses the angular speed of the disk, and a drive motor at \(A\) maintains the angular speed constant while a 1.20 kg block on top of the disk slides outward in a radial slot. The \(1.20-\mathrm{kg}\) block starts at the center of the disk at time \(t=0\) and moves outward with constant speed \(1.25 \mathrm{cm} / \mathrm{s}\) relative to the disk until it reaches the edge at \(t=440\) s. The sliding block feels no friction. Its motion is constrained to have constant radial speed by a brake at \(B\), producing tension in a light string tied to the block. (a) Find the torque that the drive motor must provide as a function of time, while the block is sliding. (b) Find the value of this torque at \(t=440 \mathrm{s}\), just before the sliding block finishes its motion. (c) Find the power that the drive motor must deliver as a function of time. (d) Find the value of the power when the sliding block is just reaching the end of the slot. (e) Find the string tension as a function of time. (f) Find the work done by the drive motor during the 440 -s motion. (g) Find the work done by the string brake on the sliding block. (h) Find the total work on the system consisting of the disk and the sliding block.

The position vector of a particle of mass \(2.00 \mathrm{kg}\) is given as a function of time by \(\mathbf{r}=(6.00 \mathbf{i}+5.00 t \mathbf{j}) \mathrm{m} .\) Determine the angular momentum of the particle about the origin, as a function of time.
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