Classical Mechanics - Conservative Force

Suppose that the gravitational force law between two massive objects were

F₁₂ = r̂₁₂ Gm₁m₂/r₁₂^(2+ɛ) 

where ɛ is a small positive number. Which of the following statements would be FALSE?

1. The total mechanical energy of the planet-Sun system would be conserved.
2. The angular momentum of a single planet moving about the Sun would be conserved.
3. The periods of planets in circular orbits would be proportional to the (3+ɛ)/2 power of their respective orbital radii.
4. A single planet could move in a stationary non circular elliptical orbit about the Sun.
5. A single planet could move in a stationary circular orbit about the Sun.

(GR9677 #23)

Solution:

(A) TRUE.
Gravitational force is a conservative force.
In conservative field, the total mechanical energy is conserved.

(B) TRUE
In conservative field, angular momentum, L is conserved.

(C) TRUE
F_g = F_c
GMm/r^(2+ɛ) = mrω²
GMm/r^(2+ɛ) = mr(2π/T)²
GM/r^(3+ɛ) = 4π²/T²
T² = 4π²r^(3+ɛ)/GM
T ∝ r^(3+ɛ)/2 

(D) FALSE
Central force = centripetal force (F_g = F_c) produces circular orbit.
Non central forces do not produce circular orbit.

(E) TRUE
See (D)

Answer: D

Notes:

Central force:

1. It is a force whose magnitude depends only on the distance between the object and the origin.
2. It is a conservative field, can be expressed as F = − ∇V (the negative gradient of a potential energy).
3. Gravitational force, Coulomb force, and Elastic Force (Harmonic Oscillator) are examples of central (conservative) forces.
4. In conservative field, the net work done by the force is zero, W = ∮_c F ∙ dr = 0 → the total mechanical energy is conserved.
5. Conservative force is irrotional (torque = 0), since curl ∇V or ∇ × ∇V = 0.
6. Torque, τ = dL/dT = 0 → angular momentum, L is conserved
7. Central force = centripetal force (F_g = F_c) produces circular orbit.

Classical Mechanics - Circular Motion

A particle of mass M moves in a circular orbit of radius r around a fixed point under the influence of an attractive force F = ^K⁄_r³, where K is a constant. If the potential energy of the particle is zero at an infinite distance from the force center, the total energy of the particle in the circular orbit is

A. − ^K⁄_r² 
B. − ^K⁄_2r²
C. 0
D. ^K⁄_2r²
E. ^K⁄_r²

(GR9277 #87)

Solution:

Attractive force = Centripetal Force
 ^K⁄_r³ = ^mv²⁄_r
mv² = ^K⁄_r²

Kinetic energy:  T = ½mv² = ^K⁄_2r²
Potential energy: V(r) = − ∫ F dr = −K ∫ ¹⁄_r³ dr = ^K⁄_2r²
Attractive force → negative potential energy, V(r) = − ^K⁄_2r²
Total energy:  T + V = ^K⁄_2r²  −  ^K⁄_2r² = 0

Answer: C

Classical Mechanics - Conservative Force

Question 34-36

The potential energy of a body constrained to move on a straight line is kx⁴ where k is a constant. The position of the body is x, its speed v, its linear momentum p, and its mass m.

The force on the body is

A. ½mv²
B. −4kx³
C. kx⁴
D. −kx⁵/5
E. mg

(GR8677 #34)

Solution:



Answer: B

Classical Mechanics - Harmonic Oscillator

A particle of mass m that moves along the x-axis has potential energy V(x) = a + bx² , where a and b are positive constants. Its initial velocity is v₀ at x = 0. It will execute simple harmonic motion with a frequency determined by the value of

A. b alone
B. b and a alone
C. b and m alone
D. b, a and m alone
E. b, a, m and v₀

(GR8677 #60)

Solution:



Answer: C

Classical Mechanics - Potential Energy

A Particle of mass m moves in a one-dimensional potential V(x) = −ax² + bx⁴, where a and b are positive constants. The angular frequency of small oscillations about the minima of the potential is equal to

A. π(a/2b)^1/2
B. π(a/m)^1/2
C. (a/mb)^1/2
D. 2(a/m)^1/2
E. (a/2m)^1/2

(GR9677 #92)

Solution:

The minima of the potential (most probable value x or the equilibrium position of the mass):





Angular Frequency:

Conservative force:




Thus, the angular frequency about the minima of the potential:



Answer: D