Mathematical Physics Vol 1

5.10 Examples

301

which yields the differential equation of motion 21

˙ ρ 2 + ρ 2 ˙ θ 2 + ˙ z 2 = 2 gz + h , h = const . (5.287) Thus, three unknowns need to be determined: ρ = ρ ( t ) , θ = θ ( t ) , z = z ( t ) . As we have one connecting equation, the degree of freedom is 2 (number of independent coordinates), and we thus need one more equation. Observe the law on motion momentum d L 0 d t = M 0 , where L 0 = r × m v - is the motion momentum of point O , and M 0 the force momentum of point O . In this case the following forces are acting: gravity G = mg k and the base reaction force R , which has the direction of the normal to the sphere and passes through the coordinate origin. As this force passes through the moment point, it follows that M R O = 0 , and only the gravity moment needs to be calculated. The gravity has the direction of the z -axis, and thus the projection of the moment to the z -axis M z is equal to 0, i.e. d L z d t = M x = 0 ⇒ L z = const . = C . From

i j k x y z m ˙ x m ˙ y m ˙ z

L 0 = r × m v =

= i ( y · m ˙ z − z · m ˙ y )+ j ( z · m ˙ x − x · m ˙ z )+ k ( x · m ˙ y − y · m ˙ x )

with respect to Cartesian coordinates, we obtain for the motion moment, for the z -axis

L z = m ( x ˙ y − y ˙ x )= const = C .

Given that

x = ρ cos θ ⇒ ˙ x = ˙ ρ cos θ − ρ ˙ θ sin θ , y = ρ sin θ ⇒ ˙ y = ˙ ρ sin θ + ρ ˙ θ cos θ ,

with respect to polar coordinates, we obtain L z = m ρ cos θ ˙ ρ sin θ + ρ ˙ θ cos θ − ρ sin θ ˙ ρ cos θ − ρ ˙ θ sin θ = const = C ⇒ L z = ρ 2 ˙ θ = C . The task is to find the position of the point M ( ρ , θ , z ) , i.e. to determine the final equations of motion ρ = ρ ( t ) , θ = θ ( t ) , z = z ( t ) from the system of differential equations - the law of change in kinetic energy ˙ ρ 2 + ρ 2 ˙ θ 2 + ˙ z 2 = 2 gz + h (1)

- the law on motion moment

C ρ 2

˙ θ =

(2)