Looping the loop
Discussion
This brain teaser was on radio 4's Puzzle for Today yesterday, puzzle 605.
"A parent has purchased a vertical loop the loop track for a toy car. Assuming the track and car are frictionless, how much higher than the top of the loop the loop must the car start in order for it to go round without falling off?"
The answer is on the website today, but doesn't explain how it was calculated.
How do you work it out?
"A parent has purchased a vertical loop the loop track for a toy car. Assuming the track and car are frictionless, how much higher than the top of the loop the loop must the car start in order for it to go round without falling off?"
The answer is on the website today, but doesn't explain how it was calculated.
How do you work it out?
To prevent the car "falling off downwards", the speed of the car must be that which provides a centripetal acceleration equal to the force of gravity. So as the car passes the top of the loop, the upwards acceleration is equal to gravity, which means that although the tyres are not being pressed into the loop, they are not "falling off" it either.
The easiest way to work it out is to do an "energy balance" because without any losses, the total energy of the system must be constant, and we trade potential energy for kinetic energy, ie height for speed.
Potential energy is mgh ie mass (kg) x acceleration due to gravity (m/s/s) x height (m)
Kinetic energy is 0.5mv^2 ie 0.5 x mass (kg) x Velocity (m/s) x Velocity (m/s)
The final formula is for centripetal acceleration Ac = v^2 / r, where the centripetal acceleration (in the radial direction) is the circuferencial velocity (m/s) squared multiplied by the radius (m) of the circle around which the body is moving
When you boil all that down, things like vehicle mass cancel, and you are left with the answer that the height of the starting point must be 25% greater than the height of the loop (i'll leave you to do the dimensional analysis to show this is the case!)
What isn't taken into account by this simple questions is the fact that the vehicles centre of gravity is NOT actually at the same height as the plane where it's wheels touch the ground, and the inertial forces act through the CofG, but the geometric ones act at the tyres surface.........
The easiest way to work it out is to do an "energy balance" because without any losses, the total energy of the system must be constant, and we trade potential energy for kinetic energy, ie height for speed.
Potential energy is mgh ie mass (kg) x acceleration due to gravity (m/s/s) x height (m)
Kinetic energy is 0.5mv^2 ie 0.5 x mass (kg) x Velocity (m/s) x Velocity (m/s)
The final formula is for centripetal acceleration Ac = v^2 / r, where the centripetal acceleration (in the radial direction) is the circuferencial velocity (m/s) squared multiplied by the radius (m) of the circle around which the body is moving
When you boil all that down, things like vehicle mass cancel, and you are left with the answer that the height of the starting point must be 25% greater than the height of the loop (i'll leave you to do the dimensional analysis to show this is the case!)
What isn't taken into account by this simple questions is the fact that the vehicles centre of gravity is NOT actually at the same height as the plane where it's wheels touch the ground, and the inertial forces act through the CofG, but the geometric ones act at the tyres surface.........
BTW a simple way to help you understand how the physics interacts is to do the calcs with "unity" values, ie 1kg of mas, 1m/s/s or acceleration, and a 1m loop diameter
Here, the velocity at the top of the loop is 0.707 m/s the sqrt (1/2)
The PE is 1 J (1 x 1 x 1)
The KE is 0.25 J (0.5 x 1 x 0.707^2)
It's clear that the KE is 25% if the PE, and that ratio is set by the physics of the formulas, effectively half of the square of 0.707, which is half of a half, ie 0.25
And if the energies are always split by that fixed constant, then the ratio of heights between the top of the loop and the height of the release is also fixed in that ratio.
ie for any fixed mass of car, and fixed force or gravity, the release height is 1.25 times that of the top of the loop.
Here, the velocity at the top of the loop is 0.707 m/s the sqrt (1/2)
The PE is 1 J (1 x 1 x 1)
The KE is 0.25 J (0.5 x 1 x 0.707^2)
It's clear that the KE is 25% if the PE, and that ratio is set by the physics of the formulas, effectively half of the square of 0.707, which is half of a half, ie 0.25
And if the energies are always split by that fixed constant, then the ratio of heights between the top of the loop and the height of the release is also fixed in that ratio.
ie for any fixed mass of car, and fixed force or gravity, the release height is 1.25 times that of the top of the loop.
Kawasicki said:
The solutions that say the kinetic energy at the top of the loop must only be more than zero are wrong. A normal force greater than the weight of the car must be maintained until just before the car enters the last quadrant of the Loop.
correct, the KE must be more than 25% of the PE and not "more than zero"Gassing Station | Science! | Top of Page | What's New | My Stuff