Maximum velocity at the top of a roller coaster?

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SUMMARY

The maximum speed a roller-coaster car can achieve at the top of a loop without leaving the track is determined using the equation derived from the balance of forces. For a roller-coaster car with a mass of 509 kg and a loop radius of 15 meters, the maximum speed is calculated by setting the normal force to zero at the point of leaving the track. The formula used is v = sqrt(gr), where g is the acceleration due to gravity (9.8 m/s²). This results in a maximum speed of approximately 12.12 m/s.

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  • Understanding of Newton's laws of motion
  • Basic knowledge of circular motion dynamics
  • Familiarity with gravitational force calculations
  • Ability to manipulate algebraic equations
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  • Study the principles of centripetal force in circular motion
  • Learn about the effects of mass and radius on speed in roller-coaster design
  • Explore advanced dynamics involving friction and air resistance
  • Investigate safety measures in roller-coaster engineering
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Physics students, mechanical engineers, amusement park designers, and anyone interested in the mechanics of roller coasters.

pcandrepair
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Homework Statement



A roller-coaster car has a mass of 509 kg when fully loaded with passengers.
-The radius of the circle is 15m. (See picture below)

What is the maximum speed the vehicle can have at point B and still remain on the track?
_________m/s


Homework Equations



Sum of forces in the radial direction = -m(v^2)/R



The Attempt at a Solution


There would be a normal force pointing up and mg pointing down
 

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pcandrepair said:
There would be a normal force pointing up and mg pointing down
Right. And what will the normal force become just before the car leaves the track? Use that to calculate the maximum speed. (You already know the radius of the curve at point B.)
 
So I set the normal force equal to zero just before it would leave the track and use..

normal - mg = m(v^2)/r masses would cancel.
0 + g = (v^2)/r
gr = (v^2)
sqrt(gr) = v

when i take the sqare root of 9.8*15 I'm getting 12.1244m/s which sounds reasonable.
 
Sounds good to me.
 

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