Mathematical Physics Vol 1

5.10 Examples

303

R At the first glance, the interval − R ≤ z ≤ R seems impossible, as a hemisphere is in question, and thus 0 < z < R . However, this would be the case when the point emerges from the hemisphere.

Figure 5.14: The range of possible solutions − R < z 2 ≤ z ≤ z 1 < R .

Thus, the square of the velocity is determined by the polynomial P ( z ) , and it follows that

R 2 ˙ z 2

= P ( z 1 )= P ( z 2 )= 0 ⇒ ˙ z = 0 (velocity =0) .

z = z 1 , z 2

In order for the motion to continue, it is necessary that the acceleration (derivative of motion by time) be different from zero. In our case ˙ P ( z )= − 2 g [( z − z 2 )( z − z 3 )+( z − z 3 )( z − z 1 )+( z − z 1 )( z − z 2 )] · ˙ z ⇒ ˙ P ( z 1 )= − 2 g [( z 1 − z 2 )( z 1 − z 3 )+( z 1 − z 3 )( z 1 − z 1 )+( z 1 − z 1 )( z 1 − z 2 )] · ˙ z = = − 2 g ( z 1 − z 2 ) | {z } > 0 ( z 1 − z 3 ) | {z } > 0 · ˙ z < 0 . The acceleration is different from zero, and it is negative, so the point moves upward (the axis is directed downward, and the acceleration has the opposite sign (direction)!) until the next point where it stops, and that is point z 2 . At that moment the acceleration is ˙ P ( z 2 )= − 2 g [( z 1 − z 2 )( z 1 − z 3 )+( z 1 − z 3 )( z 1 − z 1 )+( z 1 − z 1 )( z 1 − z 2 )]= = − 2 g ( z 2 − z 2 ) | {z } > 0 ( z 2 − z 3 ) | {z } < 0 > 0 . and the motion is downward. Observe now the differential equation of motion, whose solution gives the position of the point with respect to its height ( z coordinate). R 2 ˙ z 2 = − 2 g ( z − z 1 )( z − z 2 )( z − z 3 ) , with initial conditions: z 0 = z 1 , ˙ z 0 = 0 , ˙ z = − 2 ( z 1 − z 2 ) u · ˙ u (5.288) and substitution z = z 1 − ( z 1 − z 2 ) u 2 . The initial value for this variable is z 0 = z 1 = z 1 − ( z 1 − z 2 ) u 2 0 ⇒ 0 =( z 1 − z 2 ) u 2 0 ⇒ u 0 = 0 .