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Forces in Ice-Skating
What are the forces involved when someone is ice-skating (at constant speed) around the rink in circles and which one is the centripetal force requirement?
Force of gravity
Force of friction
The skater is moving counter-clockwise around the rink, so the force of friction is to the left (when drawing a FBD). The force of friction would be the centripetal force requirement. First of all, although the skater is traveling at constant speed, they are still accelerating. They are accelerating because acceleration occurs when an object is changing direction, which is what happens when ice-skating around a rink. The acceleration is perpendicular to the velocity, which is tangential to the circle. The centripetal force is the force applied inward that keeps the skater moving in a circle. The force that keeps the skater moving in a circle is called the centripetal force requirement, which in this case would be friction. The centripetal force requirement is friction because the force of friction opposes the direction of the movement of the skater. In order to find F(F), which is F(C) in this case, you can use the equation: F(C)=mv^2/r.
Friction in ice-skating:
When the skater is going around a flat curve, the force that keeps them moving in a circle is friction. The equation for friction is: u(F(N)). Since the forces in the y-axis of the FBD are in equilibrium, F(N) is equal to F(G). F(G)=mg. If the skater is 60 kg, their weight would be 588 N (since 'g'=9.8). If mu= 0.5 that day, friction would be equal to (0.5)x588. Therefore, the force of friction would be 294 N.
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it all looks good, I like that you put in the centripetal force equation; but you might want to put an explanation next to your fbd so it is more clear. otherwise good job.