Problem 7.46: Riding a Loop-the-loop

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SUMMARY

The problem discusses a car on a frictionless loop-the-loop amusement park ride, requiring the calculation of the minimum height for the car to complete the loop without falling off. The minimum height is determined to be 4R, where R is the radius of the loop. Given a height of 4.00 meters and a radius of 25.0 meters, the speed of the passengers at the end of the horizontal diameter is calculated to be approximately 14.1 m/s using the energy conservation equation mgh = 1/2 mv^2.

PREREQUISITES
  • Understanding of gravitational potential energy and kinetic energy principles
  • Familiarity with the conservation of mechanical energy
  • Knowledge of circular motion dynamics
  • Ability to manipulate algebraic equations for solving physics problems
NEXT STEPS
  • Study the principles of conservation of energy in mechanical systems
  • Learn about circular motion and the forces acting on objects in a loop
  • Explore the effects of friction in real-world scenarios
  • Practice solving similar problems involving energy conservation and circular motion
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of circular motion and energy conservation in amusement park rides.

annabelx4
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Homework Statement



A car in an amusement park ride rolls without friction around the track shown in the figure . It starts from rest at point at a height above the bottom of the loop. Treat the car as a particle.

What is the minimum value of (in terms of ) such that the car moves around the loop without falling off at the top (point )?

If the car starts at height 4.00 and the radius is = 25.0 , compute the speed of the passengers when the car is at point , which is at the end of a horizontal diameter.

Homework Equations



mgh = 1/2 mv^2

The Attempt at a Solution



we know that h = 2 R because it's the diameter but other than that i am lost
 
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. For the minimum value, I believe it is 4R. For the speed, we use the equation mgh = 1/2 mv^2. mgh = 1/2 mv^2(4)(9.8)(25) = 1/2 (20)(v^2)1960 = 10v^2196 = v^214.1 m/s = v
 

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