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Discuss qualitatively the motion of the atoms in a block of steel that falls onto another steel block. Why and how do large-scale vibrations damp out?

Short Answer

Expert verified

When the atoms in the steel block fall into another steel block, then the particles mainly behave in the waveform. The large-scale vibrations damp out through the inner friction which mainly translates into heat. A small amount of dissipation may turn into radiation and sound.

Step by step solution

01

Significance of the translational motion and the law of vibration

The atoms can easily move from one position to another position which is referred to as the translational motion of the atoms.

The law of vibration illustrates that every single particle in this universe is in a constant movement state.

The motion of the atoms can be predicted by the translational motion and the law of vibrations gives the reason for the damping out of the vibrations.

02

Determination of the atom’s motion and damping out the vibrations

As a steel block is treated as a rigid object, hence it is a multiparticle system. However, while collision, the particles mainly acquire kinetic energy and transfer the energy into a waveform that eventually creates vibration.

As materials have a type of damping effect, hence, the kinetic energy gets transformed into heat for the friction that exists between the materials. However, due to the loss of energy, the amplitude of the wave decreases with time. Hence, the property of the materials for which the oscillations are damped due to the inner friction is mainly called hysteretic damping.

Hysteretic damping is mainly used for dissipating energy. A small amount of energy may be dissipated in radiation and sound form.

Thus, when the atoms in the steel block fall into another steel block, then the particles mainly behave in the waveform. The large-scale vibrations damp out through the inner friction which mainly translates into heat. A small amount of dissipation may turn into radiation and sound.

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Most popular questions from this chapter

Show that moment of inertia of a disk of mass M and radius R is 12MR2. Divide the disk into narrow rings, each of radius r and width dr. The contribution I of by one of these rings is r2dm, where dm is amount of mass contained in that particular ring. The mass of any ring is the total mass times the fraction of the total area occupied by the area of the ring. The area of this ring is approximately 2Ï€°ù»å°ù. Use integral calculus to add up all the calculations.

String is wrapped around an object of mass 1.5kg and moment of inertia 0.0015kg-m2(the density of the object is not uniform). With your hand you pull the string straight up with some constant force F such that the center of the object does not move up or down, but the object spins faster and faster (Figure 9.62). This is like a yo-yo; nothing but the vertical string touches the object. When your hand is a height y0=0.25mabove the floor, the object has an angular speed Ó¬0=12rad/s. When your hand has risen to a height y=0.35m above the floor, what is the angular speed of the object? Your answer must be numeric and not contain the symbol F.

Can you give an example of a system that has no atoms located at its center of mass?

A rod of length Land negligible mass is attached to a uniform disk of mass Mand radius R (Figure 9.64). A string is wrapped around the disk, and you pull on the string with a constant force F . Two small balls each of mass mslide along the rod with negligible friction. The apparatus starts from rest, and when the center of the disk has moved a distance d, a length of string shas come off the disk, and the balls have collided with the ends of the rod and stuck there. The apparatus slides on a nearly frictionless table. Here is a view from above:

(a) At this instant, what is the speed vof the center of the disk? (b) At this instant the angular speed of the disk isÓ¬ . How much internal energy change has there been?

By calculating numerical quantities for a multiparticle system. One can get a concrete sense of the meaning of the relationships p→sys=Mtotv→CMand Ktot=Ktrans+Krel. Consider an object consisting of two balls connected by a spring, whose stiffness is 400 N/m. The object has been thrown through the air and is rotating and vibrating as it moves. At a particular instant, the spring is stretched 0.3m, and the two balls at the ends of the spring have the following masses and velocities: 1:5kg.(8,14,0)m/s2:3kg(-5,9,0)m/s

(a)For this system, calculate p→sys. (b) Calculate v→CM(c) Calculate Ktot3. (d) Calculate Ktrans. (e) Calculate Krel. (f) Here is a way to check your result for Krel. The velocity of a particle relative to the center of mass is calculated by subtracting v→CMfrom the particle’s velocity. To take a simple example, if you’re riding in a car that’s moving with v→CM,x=20m/sand you throw a ball with v→CM,x=35m/s, relative to the car, a bystander on the ground sees the ball moving with vx=55m/sSo v→=v→CM=v→reland therefore we have=v→relv→=v→CMfor each mass and calculate the correspondingKrel. Compare with the result you obtained in part (e).

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