Mathematical Physics Vol 1

Chapter 5. Series Solutions of Differential Equations. Special functions

298

From the law of change of kinetic energy we obtain

d A = d E k ⇒ A = E k − E k 0 ⇒ mgz − mgR = m 2 v 2

0 ⇒

v 2 − v 2

v 2 0

2 − 0 − 2 gR ≡ 2 h , where h is a constant that can be determined from the initial conditions ( t 0 = 0 : v 0 = R ˙ ϕ 0̸ = 0): h = v 2 0 2 − gz 0 = v 2 0 2 − Rg cos ϕ 0 . Thus, we are interested in the equation v 2 = 2 gz + 2 h (5.273) where z = R cos ϕ , z 0 is the initial height of the pendulum, and v = R ˙ ϕ . From previous equations, we obtain R 2 ˙ ϕ 2 = 2 Rg cos ϕ + h Rg ⇒ ˙ ϕ 2 = 2 g R ( cos ϕ + h Rg | {z } f ( ϕ ) ) . (5.274) As the left hand side is the square of a real value (always non-negative), in order for the point to move, the right hand side must satisfy the condition Analysis: 1. The function f ( ϕ ) has a first order zero, i.e. there exists a zero ϕ ∗ of this function, where cos ϕ ∗ = − h Rg . At that point the pendulum stops, and in order for it to continue moving, an acceleration different from zero is needed, i.e. f ′ ( ϕ )̸= 0 f ′ ( ϕ )= − sin ϕ̸ = 0 . Thus, cos ϕ ∗ = − h Rg , sin ϕ ∗ = q 1 − ( h / Rg ) 2̸ = 0 , (5.276) and the necessary condition is that the expression under the square root is positive, i.e. h Rg < 1 . (5.277) From (5.276) it follows that there exist two positions in which the pendulum stops: ϕ ∗ 1 = − arccos h Rg i ϕ ∗ 2 = arccos h Rg . (5.278) These positions are symmetric with respect to ϕ = 0 and ϕ = π . From the conditions f ( 0 )= 1 + h / Rg and | h / Rg | < 1 it is visible that f ( 0 ) > 0, and the gz = 2 − gR ⇒ v 2 − 2 gz = v 2 f ( ϕ )= cos ϕ + h Rg ≥ 0 . (5.275)