Apparent forces in circular motion

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SUMMARY

The discussion centers on calculating the speed of a roller coaster car at the bottom of a dip with a 30 m radius of curvature, where passengers feel 50% heavier than their actual weight. The normal force (N) is established as 1.5 times the weight (W), leading to the equation N - W = 0.5W, which represents the centripetal force. By applying Newton's second law and the centripetal force equation, the calculated speed at the bottom of the dip is confirmed to be 12 m/s.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Knowledge of centripetal force and its calculation
  • Familiarity with free-body diagrams
  • Basic concepts of weight and normal force
NEXT STEPS
  • Study the derivation of centripetal acceleration formulas
  • Learn how to draw and analyze free-body diagrams in physics
  • Explore the effects of varying radius on centripetal force
  • Investigate the relationship between speed and radius in circular motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators looking for practical examples of force analysis in real-world scenarios.

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Homework Statement


The passengers in a roller coaster car feel 50% havier than their true weight as the car goes through a dip with a 30 m radius of curvature. What is the cars speed at the bottom of the dip?


Homework Equations


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The Attempt at a Solution


I understand that there are two forces affecting the rollercoaster car - the normal force and the weight (mass x gravity)... Also since they feel 50% heavier the normal force must be greater than the weight. so n (normal force) > w (mass x gravity) thus the n = 1.50w.

So sum of force = n - w = 1.50 w - w after this I am completely lost... can anyone help please!
 
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The net force you have correctly calculated as 0.5w is the centripetal force. Calculate v using the centripetal force equation .
 
Draw a free-body diagram of the roller coaster, label all forces, and write out Newton's second law for the vertical direction. What's the acceleration?
 
thank you
I got 12m/s
 

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