Mathematical Physics Vol 1

Chapter 3. Examples

72

Let us first determine its magnitude d ρ d t = lim ∆ t → 0 ∆ ρ ∆ t

| ∆ ρ | ∆ t

AB ∆ t

= lim ∆ t → 0

= lim ∆ t → 0

=

2 ρ sin ( ∆ ϕ / 2 ) ∆ t

sin ( ∆ ϕ / 2 ) ∆ ϕ / 2

∆ ϕ 2 ∆ t

= 2 ρ lim ∆ t → 0

= lim ∆ t → 0

=

sin ( ∆ ϕ / 2 ) ∆ ϕ / 2

∆ ϕ 2 ∆ t

= ρ lim ∆ t → 0

lim ∆ t → 0

=

= ρ ˙ ϕ . Here we used the fact that from ∆ t → 0 it follows that ∆ ϕ → 0, that is, lim ∆ t → 0 sin ( ∆ ϕ / 2 ) ∆ ϕ / 2 = 1. We denoted the derivative of a variable by time t by a point above the variable, which is common in mechanics. Given that ρ 2 = ρ · ρ = ρ 2 = const . , by deriving by t we obtain 2˙ ρ · ρ = 0 , and it thus follows that vectors ρ and ˙ ρ are orthogonal. Finally, for the derivative of a vector with constant magnitude we obtain ˙ ρ = | ˙ ρ | p = ρ ˙ ϕ p = ρω p , where ρ ⊥ p , and ω = ˙ ϕ is the angular velocity, a variable that characterizes the change of the direction of this vector. Bearing in mind that a vector product of two vectors yields an orthogonal vector, it follows that ˙ ρ = ω × ρ . Given that the magnitude of the vector product | ˙ ρ | = ω · ρ · sin ( ω , ρ )= ω · ρ , it fol lows that sin ( ω , ρ )= 1 ⇒ ∠ ( ω , ρ )= π / 2, that is, vectors ρ , ω and ˙ ρ are mutually orthogonal.

Problem30 Represent the velocity of a point of a rigid body as it rotates about a fixed point.

Solution

Observe an arbitrary point of the body M , whose position vector is r . The body rotates about the fixed point C , whose position vector r C is constant. If we denote vector −→ CM by ρ , then (see Figure 3.11)

Figure 3.11:

Position of a point of a rigid

body.