Question

A skateboarder with his board can be modeled as a particle of mass 79.0 kg, located at his center of mass, 0.500 m above the ground. As shown below, the skateboarder starts from rest in a crouching position at one lip of a half-pipe (point circled A). The half-pipe forms one half of a cylinder of radius 6.20 m with its axis horizontal. On his descent, the skateboarder moves without friction and maintains his crouch so that his center of mass moves through one quarter of a circle. (a) Find his speed at the bottom of the half-pipe (point circled B). 11.024 Incorrect: Your answer is incorrect. m/s (b) Find his angular momentum about the center of curvature at this point. magnitude kg · m2/s direction (c) Immediately after passing point circled B, he stands up and raises his arms, lifting his center of gravity from 0.500 m to 0.870 m above the concrete (point circled C). Explain why his angular momentum is constant in this maneuver, whereas the kinetic energy of his body is not constant. This answer has not been graded yet. (d) Find his speed immediately after he stands up. m/s (e) How much chemical energy in the skateboarder's legs was converted into mechanical energy in the skateboarder—Earth system when he stood up? kJ

Answer 1

a )

V_B² = ( 2 g h )^1/2

= ( 2 X 9.8 X ( 6.2 - 0.5 ) )^1/2

= 10.5697 m/s

b )

L_C = m V_B R' sin90

= 79 X 10.5697 X ( 6.2 - 0.5 ) X 1

L_C = 4759.5359 kgm²/sec

c )

some energy lost during effort to make change because of heat loss

and chemical energy consumes

d )

L_i = L_f

4759.5359 = 79 X V_s X ( 6.2 - 0.5 )

V_s = 10.5696 m/s

e )

using law of conservation of energy

0.5 X 79 X 10.5697² + U_{chm , B} = 0.5 X 79 X 10.5696² + 79 X 9.8 X 0.5

U_{chm , B} = 0.5 X 79 X 10.5696² + 79 X 9.8 X 0.5 - 0.5 X 79 X 10.5697²

U_{chm , B} = 387.0164 J

U_{chm , B} = 0.3870164 kJ