Circular Motion and Gravitation HELP

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SUMMARY

The discussion focuses on solving a physics problem involving circular motion and gravitation, specifically calculating the scale reading of a man in a roller coaster at the top of a loop. Key values include a velocity of 33.1 m/s, a radius of 61 meters, and a mass of 55.3 kg. The scale reading is determined using the equation: Scale reading = mg + mv²/r, resulting in approximately 6,563 Newtons. The discussion also highlights that the approach would differ if the man were at the bottom of the loop due to changes in velocity.

PREREQUISITES
  • Understanding of circular motion principles
  • Familiarity with Newton's laws of motion
  • Basic knowledge of force calculations
  • Ability to manipulate equations involving acceleration, mass, and velocity
NEXT STEPS
  • Study the concept of centripetal force in circular motion
  • Learn how to apply Newton's second law in dynamic scenarios
  • Explore the differences in forces at various points in circular motion
  • Practice similar problems involving roller coasters and circular tracks
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Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators seeking to explain these concepts in practical scenarios.

kittn44
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Hey there! Completely new here but thought someone could help. I have this problem where I don't even know where to start from.

The problem:

(Picture of loop with a man in a roller coaster cart hanging upside down from furthest top point of circle)

A man in the roller coaster is sitting on a bathroom scale. If he is traveling at 33.1 m/s at the point shown, the radius of the vertical coaster track is 61 meters, and the man has a mass of 55.3 kg, to the nearest Newton, what does the scale read?

Ok, first off, the answer isn't something I'm concerned with anyway because these values will not be the same as the ones I end up having to work, so I really just need someone to explain HOW I do this and WHY. My class is only 5 weeks long and I'm taking too much in and I just don't grasp circular motion quite yet. Any ideas and help appreciated!

Also, if the man were at the most bottom point of the loop, would you work the problem differently?

THANKS :smile:
 
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The acceleration for circular motion can be defined as:

[tex]a = \frac{v^2}{r}[/tex]

where "a" is acceleration, "v" is velocity, and "r" is the radius.

Now once the acceleration is known, how might one find the force acting on the scale?
 


Hi there! Don't worry, circular motion and gravitation can be a bit tricky at first, but with some practice and understanding of the concepts, you'll get the hang of it!

To solve this problem, you'll need to use the equations for circular motion and Newton's laws of motion. First, let's break down the problem into parts.

1. The man is traveling at 33.1 m/s at the point shown, which means he has a velocity in the vertical direction. This velocity is caused by the centripetal force, which is directed towards the center of the circle.

2. The radius of the vertical coaster track is 61 meters. This is the distance from the center of the circle to the point where the man is hanging upside down.

3. The man has a mass of 55.3 kg. This is important because it will help us calculate the force acting on him.

Now, to find the scale reading, we need to calculate the force that is acting on the man. This force is the sum of the weight of the man (mg) and the centripetal force (mv^2/r). We can set up an equation:

Scale reading = mg + mv^2/r

Substituting the given values, we get:

Scale reading = (55.3 kg)(9.8 m/s^2) + (55.3 kg)(33.1 m/s)^2/(61 m)

This gives us a scale reading of approximately 6,563 Newtons.

If the man were at the bottom point of the loop, the problem would be slightly different because his velocity would be different. However, the same concept applies and you would still use the same equation to find the scale reading.

I hope this helps you understand the problem better! Remember to always break it down into smaller parts and use the relevant equations to solve it. Good luck with your class! :smile:
 

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