Driving an RC car in a vertical loop, What formula to use?

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SUMMARY

The discussion focuses on calculating the minimum speed required for an RC car to complete a vertical loop in a concrete drain pipe. The key formula derived is Mg < MV²/r, where M is the mass of the vehicle, g is the acceleration due to gravity, r is the radius of the loop, and V is the velocity. Notably, mass cancels out, indicating it does not affect the minimum speed required. The conversation also distinguishes between gravitational force and centripetal force, clarifying that the formula is not applicable for planetary orbits, which require the use of the gravitational constant G.

PREREQUISITES
  • Understanding of basic physics concepts such as gravitational force and centripetal force.
  • Familiarity with the variables in the formula: mass (M), gravitational acceleration (g), radius (r), and velocity (V).
  • Knowledge of the difference between gravitational constant (G) and gravitational acceleration (g).
  • Basic algebra skills for rearranging equations.
NEXT STEPS
  • Research the derivation of centripetal force and its applications in circular motion.
  • Study the implications of mass in different physical scenarios, particularly in dynamics.
  • Explore gravitational forces in planetary motion and the significance of the gravitational constant (G).
  • Learn about the effects of varying radius on the minimum speed required for circular motion.
USEFUL FOR

Physics students, hobbyists interested in RC vehicles, and anyone looking to understand the dynamics of motion in circular paths.

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I once drove my RC car in a vertical loop in a concrete drain pipe.

If you know the diameter of the pipe, and the weight of the vehicle, what is the formula used to determine the minimum speed required for it to complete the loop?
 
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At the top you want the force of gravity Mg to be less than the centripetal force needed to move in a circle radius r at velocity V. So...

Mg < MV2/r

Mass M cancels

Rearrange it to find V.
 
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Wow I never would have thought that mass is irrelevant, but it makes sense. I thought it would have been way more complicated than that, taking into account the amount of time "at the top" and mass.

Would this formula be for a simple planetary orbit of constant speed and radius? Mg = MV2/r
 
No because gm is only valid at the surface of a planet. You would need to use GmM/r2 instead. There is also the complication that both planets orbit about their common centre of mass (eg the moon does not orbit about the centre of the earth).

In case you haven't studied this: m and M are the masses of the two bodies. G isn't the same as g. "Big G" is the gravitational constant whereas "little g" is the gravitational acceleration on earth.
 
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I have encountered a vertically oriented hydraulic cylinder that is designed to actuate and slice heavy cabling into sections with a blade. The cylinder is quite small (around 1.5 inches in diameter) and has an equally small stroke. The cylinder is single acting (i.e. it is pressurized from the bottom, and vented to atmosphere with a spring return, roughly 200lbs of force on the spring). The system operates at roughly 2500 psi. Interestingly, the cylinder has a pin that passes through its...

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