# A formula to calculate speed on loop de loop

I am trying to calculate the average speed of a car while travelling around a loop de loop in m/s and km/h can any body help? Thanks

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There will be several willing to help ... only if they see some initial attempt

• Jaja
I am trying to calculate the average speed of a hotwheels car while travelling around a loop de loop in m/s and km/h can any body 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

Wondering if I have the formula right.
One can use the principle of the conservation of energy to derive the required formula.

• Jaja
sophiecentaur
Gold Member
2020 Award
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.

• Jaja
nasu
Gold Member
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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• Jaja
Ok thank you very much

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!

• Jaja
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

Th
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