Consider the ring-shaped body of Fig. E13.35. A particle with mass m is placed a distance x from the center of the ring, along the line through the center of the ring and perpendicular to its plane. (a) Calculate the gravitational potential energy U of this system. Take the potential energy to be zero when the two objects are far apart. (b) Show that your answer to part (a) reduces to the expected result when x is much larger than the radius a of the ring. (c) Useto find the magnitude and direction of the force on the particle (see Section 7.4). (d) Show that your answer to part (c) reduces to the expected result when x is much larger than a. (e) What are the values of U andwhen? Explain why these results make sense.

Short Answer

Expert verified

a) The potential energy of the system is,

b) The potential energy becomes ,

c) The magnitude of the force on the particle isand the direction is negative as it is an attractive force character,

d) The magnitude of the force becomesand it is in the negative direction and

e) The values of when are and 0, respectively.

Step by step solution

01

Identification of the given data

The given data can be listed below as,

  • The mass of the particle is m.
  • The mass of the thin ring is M.
  • The particle is placed at a distance x.
  • The radius of the ring is a.
02

Significance of Newton’s gravitational law in evaluating the potential energy, magnitude and direction of the force and their values

Newton’s gravitational law states the force exerted is inversely proportional to the distances’ square and directly proportional to the products of the masses.

The equation of the force gives the gravitational potential energy, direction and the magnitude of the force, and the values of the potential energy.

03

Determination of the gravitational potential energy, magnitude and the direction of the force, and the values of potential energy and force

a)

Considering a mass segment on the ring that is expressed as:

Here, is the integrated angle.

From the Newton’s gravitational law, the differential potential energy between the mass m and the segment of the ring is expressed as:

Hence, the potential energy of this system is-

Thus, the potential energy of the system is.

b)

As , then becomes x.

Thus, the potential energy becomes .

c)

We know that,

Hence, using the value of the potential energy, the above equation becomes,

Thus, the magnitude of the force on the particle isand the direction is negative as it is an attractive force character.

d)

As , thenbecomes ,

Hence, the equation of the force becomes .

Thus, the magnitude of the force becomesand it is in the negative direction.

e)

At , the potential energy becomes-

At , the force becomes-

Thus, the values of when are and 0, respectively.

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