- #1
galadriel3562
- 12
- 12
Homework Statement
The problem is as follows:
A skater started at the top of a halfpipe that makes a 40 degree angle with the horizontal. He is now about halfway down the pipe and represented by the rectangle. His center of gravity is represented by Z, the midpoint of the circular movement he's completing in the halfpipe is represented by M. The radius of said circle is 2,8 meters, his center of gravity is 0.7 meters off the surface of the ramp. The question is to calculate the normal force exerted by the ramp on the skater. The skater weighs 61 kg and his radial speed at the moment represented in the figure is 3.2 rad/s
Homework Equations
Fz = m*g*cos α
Fmpz = (m*v2)/r = m*ω2*r
The Attempt at a Solution
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The centripetal force would be the one allowing the skater to complete a circular motion, the gravity is pulling the skater straight down but has components that aid the skater in his downward motion (not important here) and a component that is perpendicular to the surface of the ramp. In my opinion the neutral force would then need to be added to the centripetal force so both counteract the perpendicular component of the gravity in essence:
Fn + Fmpz = Fz-component leads to: Fn = Fz-component - Fmpz
and the radius is the total radius of the circle minus the elevation of the center of gravity of the skater = 2.8-0.7 = 2.1 m
Fz = 61 * 9.81 (gravitational acceleration) * cos 40 = 484 N
Fmpz = 61 * 3.22*2.1 = 1311 N
However this results in a negative neutral force: Fn = 484-1311 = -827
The answer provided however states that the neutral force is equal to the perpendicular gravity component added to the centripetal force (Fn = Fmpz + Fzcomponent) but that would mean the centripetal force is actually pointed away from the center of the circle (more like a centrifugal force?), I've always been taught not to work with the centrifugal force as it is not a 'real force'? Is that wrong?