A formula to calculate speed on loop de loop

AI Thread Summary
The discussion focuses on calculating the average speed of a Hot Wheels car navigating a loop de loop. The user initially calculated the minimum speed at the top of the loop as 0.431 m/s but later questioned the average speed of 2.91 m/s, which seemed too fast. Contributors suggested using the conservation of energy principle to derive the correct formula, noting that the minimum speed at the top should be around 1.4 m/s for a 19 cm radius loop. The user acknowledged an error in their calculations and expressed a desire for clarification on applying the conservation of energy formula. The conversation highlights the importance of accurate calculations in physics-related problems.
Jaja
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I am trying to calculate the average speed of a car while traveling around a loop de loop in m/s and km/h can anybody help? Thanks
 
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There will be several willing to help ... only if they see some initial attempt
 
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Jaja said:
I am trying to calculate the average speed of a hotwheels car while traveling around a loop de loop in m/s and km/h can anybody help? Thanks
I have been watching loop de loops on youtube and was wanting to see if I could calculate it on the hotwheels track. I worked out the minimum speed the car has to be at the top of the track to be 0.431m/s or 1.55 km/hr but calculating the average speed below , it seems too fast. Wondering if I have the formula right.

Loop radius: 19cm. Loop Time= 0.041 secs

Circumference = 2pi * 19cm radius = 0.038pi (meters)
0.038pi / 0.041 secs = avg speed in m/s
0.038pi / 0.041 = 2.91 m/s
To convert to km/h
2.91m/s * 3600 (seconds in hours) / 1000 (1000meters = km)
2.91 * 3.6 = 10.5 km/h

So the hotwheels car was traveling 2.91m/s or 10.5km/h
 
Jaja said:
Wondering if I have the formula right.

One can use the principle of the conservation of energy to derive the required formula.
 
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The limiting case of a car 'just' keeping on the track when at the top of the invert will be when the centripetal acceleration (v2/r ) is equal to g. (i.e. it is just in contact with the track) At the bottom, the extra KE will be the same as the gravitational potential difference between top and bottom (mgh). The total KE gives you the velocity, which will give you the effective 'g' at the bottom. Believe in the formula and apply it correctly and the numbers will be right.
 
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How do you know the time to go around the loop?
And your minimum speed at the top sem to be off. For a radius of 19 cm should be around 1.4 m/s I believe.
 
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Ok thank you very much
 
nasu said:
How do you know the time to go around the loop?
And your minimum speed at the top sem to be off. For a radius of 19 cm should be around 1.4 m/s I believe.
If it was, can you show me how you did the formula
Thanks
 
The formula can be obtained by using the principle of the conservation of energy... as I said earlier!
 
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grzz said:
The formula can be obtained by using the principle of the conservation of energy... as I said earlier!
So can you show me how to do the formula as I looked on the Internet of conservation of energy I didn't understand how to use it in my question
 
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Th
nasu said:
How do you know the time to go around the loop?
And your minimum speed at the top sem to be off. For a radius of 19 cm should be around 1.4 m/s I believe.
thank you, yes I made an error now this is starting to make sense
 
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