Determining Minimum Height to Complete a Loop

Click For Summary
To determine the minimum height required for a rider to complete a loop with a radius of 10 meters, the Conservation of Energy principle is applied. At the top of the hill, the potential energy is expressed as mgh, while at the top of the loop, the velocity must be derived using Newton’s Second Law. The key insight is that the minimum height h should equal 2.5 times the loop's radius, resulting in a height of 25 meters. The discussion emphasizes the complexity of the problem and suggests that the method outlined in the hints may be more straightforward than initially perceived. Understanding circular motion is crucial for solving such physics problems effectively.
crushedcorn
Messages
7
Reaction score
0
This question was posted on Khan Academy. I attempted it after watching all of their great Work and Energy videos, but the way they arrived at the answer was way above my skill level even after reviewing the videos a number of times. Here is the question:

What is the minimum height h of a slope that must be constructed in order for a rider in the cart with a combined mass m = 80.0 kg to complete the loop with a radius r = 10.0 meters safely, assuming a frictionless trip?

4e906cd94e8d150a1b45f273793579c77b38f5e6.png

How would you answer this question? The way that they arrived at their answer is below. Is there any alternative/more concise way to arrive at the same answer? I would have never guessed to proceed in the way noted in Hint #3.

We can use the Conservation of Energy formula to evaluate these two points: at the top of the hill and at the top of the loop.

caffd013cb948bf84803ad975be9a21722638d0a.png

Hint #2
At the top of the hill, the kinetic energy is zero, and the potential energy is mgh.

863284b2d565dbf72b029fb3e30cc3989ddc2627.png

Hint #3
At the top of the loop, we must derive another formula to connect ideas together by using Newton’s Second Law to describe the forces acting on the cart and then to solve for v:

d26b49c5ae2f46c757736dc22d56742fe0011f3c.png

Hint #4
This expression is for the velocity at the top of the loop since there is no normal force only at that point. Additionally, h = 2r such that the potential energy is mg(2r).

40419bee3dd2001dd8df7d462b1a36a7f664028c.png

Hint #5
The correct answer is 2.5. Since the radius of the loop is 10 meters, then 2.5r is 25 meters.
 
Physics news on Phys.org
crushedcorn said:
This question was posted on Khan Academy. I attempted it after watching all of their great Work and Energy videos, but the way they arrived at the answer was way above my skill level even after reviewing the videos a number of times.

I believe this is the easiest and smallest way to answer this question - I would certainly have solved the question this way .

Does this answer your doubt ?

crushedcorn said:
I would have never guessed to proceed in the way noted in Hint #3.

Do you know your circular motion ?
 
I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question. Ex. 1 - A moving object impacts a static object at a constant velocity. Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact. Assuming the mass of the objects is the same and the velocity at the moment of impact...

Similar threads

  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 19 ·
Replies
19
Views
6K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 12 ·
Replies
12
Views
12K
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
4K
Replies
24
Views
7K
Replies
9
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K