Classical Mechanics - Acceleration

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = βx⁻ⁿ where β and n are constants and x is the position of the particle. What is the acceleration of the particle as a function of x?

A. −nβ²x⁻²ⁿ⁻¹
B. −nβ²x⁻ⁿ⁻¹
C. −nβ²x⁻ⁿ
D. −β²x⁻ⁿ⁺¹
E. −β²x⁻²ⁿ⁺¹

(GR9677 #44)

Solution:

a = dv/dt = (dv/dx) (dx/dt) =  (dv/dx) v

Given: v(x) = βx⁻ⁿ
dv/dx =  −nβx⁻ⁿ⁻¹

Thus, a = (−nβx⁻ⁿ⁻¹) βx⁻ⁿ = −nβ²x⁻²ⁿ⁻¹

Answer: A

Classical Mechanics - 1D Vertical Motion

A rock is thrown vertically upward with initial speed v₀. Assume a friction force proportional to –v, where v is the velocity of the rock, and neglect the buoyant force exerted by air. Which of the following is correct?

1. The acceleration of the rock is always equal to g.
2. The acceleration of the rock is equal to g only at the top of the flight.
3. The acceleration of the rock is always less than g.
4. The speed of the rock upon return to its starting point is v₀.
5. The rock can attain a terminal speed greater than v₀ before it returns to its starting point.

(GR8677 #01)

Solution:

There is a friction force:

• acceleration is not constant → (A) and (C) FALSE
• energy is not conserved so it's initial and final speed is not the same → (D) FALSE
• frictional force slows down the object, so its speed at time t has to be less than its initial speed  → (E) FALSE

Answer: B


Math analysis:

Friction force: F_f  = − kv
The equation of motion with friction force: ma = − mg − kv.

a = − g − (kv/m)
(A) FALSE

At the top of the flight, v = 0
a = − g − (k∙0/m) = − g
(B) TRUE

Moving up: v positive
a = − g − (kv/m) = − g − c
a − g
Moving Down: v negative
a = − g − [k(−v)/m] = − g + c
a − g
(C) FALSE

Classical Mechanics - Lagrangian

The Lagrangian for a mechanical system is where q is a generalized coordinate and a and b are constants. The equation of motion for this system is

(GR0177 #74)

Solution:

The Lagrangian equation of motion for the generalized coordinate q:





Answer: D