Rotational Motion and the Law of Gravity

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SUMMARY

The discussion focuses on the physics of a roller coaster vehicle with a mass of 500 kg, analyzing forces at two critical points: point A at the bottom of a drop and point B at the top of a hill. At point A, the force of the track on the vehicle is calculated to be 15,100 N using the formula Fc = MAc = M(v^2/r). For point B, the maximum speed for the vehicle to remain on the track is determined to be 12 m/s, derived from the equation Vtop = √(gr), where g is 9.80 m/s² and r is 15 m.

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  • Familiarity with Newton's laws of motion
  • Knowledge of gravitational force calculations
  • Basic algebra for solving equations
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Teenytiny1991
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Homework Statement


A roller coaster vehicle has a mass of 500 kg when fully loaded with passengers. a) if the vehicle has a speed of 20.0 m/s at point A, what is the force of the track on the vehicle at this point (point a sits at the bottom of a drop of the roller coaster where the radius is 10 meters) B) what is the maximum speed the vehicle can have at point b for gravity to hold it on the track ( Point b sits at the top of the rollercoaster hill where the radius is 15 meters)


Homework Equations





The Attempt at a Solution


Fc = MAc = M(v^2/r)
n=m(v^2/r)-mg
n=500(20^2/10)-(500x9.80) = Force of track on car at a = 15,100 N is this right?

Vtop= "square roote of" gr
="Square roote of" (9.80 x 15) = max velocity at top at point b is 12 m/s is this right?
 
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For a), if the coaster is at the bottom of the drop, then the centripetal acceleration should be upwards. In that case, the normal force from the rails due to weight (upwards) needs to be added to the centripetal force applied by the track (upwards), not removed.

b looks fine.
 

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