Circular Motion. Roller-coaster mass of car and max speed

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SUMMARY

The discussion focuses on calculating the mass of a roller-coaster car and determining its maximum speed at a subsequent point on the track. At point A, the car's speed is 20 m/s, and the normal force exerted on it is 2.06x10^4 N. Using the equation for centripetal force, Fc = (mv^2)/r, participants derive the mass of the car by accounting for gravitational force. For point B, the maximum speed is calculated under the condition that only gravitational force acts on the car, emphasizing the importance of free-body diagrams in solving such problems.

PREREQUISITES
  • Understanding of circular motion principles
  • Familiarity with centripetal force equations
  • Ability to draw and interpret free-body diagrams
  • Knowledge of gravitational force calculations
NEXT STEPS
  • Learn how to derive mass using centripetal force equations
  • Study the effects of gravitational force on objects in circular motion
  • Explore advanced topics in dynamics related to roller-coaster physics
  • Practice solving problems involving free-body diagrams in circular motion scenarios
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of roller-coaster design and motion analysis.

AStupidHippo
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Homework Statement


A roller-coaster car speeds down a hill pas point A and then rolls up a hill past point B.
A. Car speed = 20 m/s at point A. If the track exerts a normal force on the car of 2.06x10^4N at this point, what is the mass of the car? (account for gravitational force)
B. What is the maximum speed the car can have at point B for the gravitational force to hold it on the track?
Diagram(I made it myself): http://gyazo.com/0415a6ba75f87f0229e08aab0f68fbe1

Homework Equations


Circular motion
Fc= (mv^2)/r
Fn-mg= (mv^2)/r

The Attempt at a Solution


Completely lost.
 
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AStupidHippo said:

Homework Statement


A roller-coaster car speeds down a hill pas point A and then rolls up a hill past point B.
A. Car speed = 20 m/s at point A. If the track exerts a normal force on the car of 2.06x10^4N at this point, what is the mass of the car? (account for gravitational force)
B. What is the maximum speed the car can have at point B for the gravitational force to hold it on the track?
Diagram(I made it myself): http://gyazo.com/0415a6ba75f87f0229e08aab0f68fbe1

Homework Equations


Circular motion
Fc= (mv^2)/r
Fn-mg= (mv^2)/r

The Attempt at a Solution


Completely lost.
I don't know why you are lost because your relevant equations are quite relevant for part A, assuming point A is at the bottom of dip. Solve for m.
 
I should have clarified.We did most of A in class, B. is where I'm struggling at most.
 
AStupidHippo said:
I should have clarified.We did most of A in class, B. is where I'm struggling at most.
Ok, got it. For part B, the question asks for the max speed at point B if the only force in the vertical direction acting on the coaster at that point on the top of the curve is the gravity (weight) force that is keeping it on the track in centripetal motion, without flying off it. You should be able to manipulate your relevant equations to solve for that speed.
 
Can you draw a free-body diagram for the cart at point B to help you out? What forces are acting on the cart?
 
I don't think I can solve this problem using fn-mg=(mv^2)/r. The Ms would just cancel out?
 
"Can you draw a free-body diagram for the cart at point B to help you out? What forces are acting on the cart?"
It would be G only. Correct?
Jason Healy said:
Can you draw a free-body diagram for the cart at point B to help you out? What forces are acting on the cart?
It would be G only, correct?
 
AStupidHippo said:
I don't think I can solve this problem using fn-mg=(mv^2)/r. The Ms would just cancel out?
if only the gravity force is acting , what is fn? Let the m's cancel if they do. Now watch your signage. Draw a free body diagram. Net centripetal force is always in direction of the centripetal acceleration,
 

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