Question

In outer space, a rod is pushed to the right by a constant force F (Figure 4.54). Describe the pattern of interatomic distances along the rod. Include a specific comparison of the situation at locations A, B, and C. Explain briefly in terms of fundamental principles.

Figure 4.54

Hint: Consider the motion of an individual atom inside the rod, and various locations along the rod.

(b) After the rod in part (a) reaches a speed v, the object that had been exerting the force on the rod is removed. Describe the subsequent motion of the rod and the pattern of interatomic distances inside the rod. Include a specific comparison of the situation at locations A, B, and C. Explain briefly.

Short answer

a) The interatomic bonds are more compressive at point A than they are at B and C.

b) Point A particles will vibrate more because have more displacement than other particles. Point C particle will have an amplitude smaller than point A and B particle.

Step-by-step solution

Constant force applies to the rod

Part a)

When the force applying (pushing) on the left end of the rod, It somewhat compresses the interatomic bond between adjacent atoms in the rod's adjoining region. The compress atoms of the bonds of their adjoining, the new interatomic bond compression spreads fast to the right, from the left end of the rod to right, and the rod goes into tension. The figure showing pattern of interatomic distance as force is applied on the left face so the interatomic distance is small as we going to along the right face the interatomic distance is large.

Force on the rod removed

Part b)

When force F is removed the particles start to go to the old or initial position and start vibrating. Vibrating only those atoms or particles who’s in atomic interactions. Point A particles will vibrate more because have more displacement than other particles. Point C particle will have an amplitude smaller than point A and B particle.