Re: how do you not fly out of your seat during a loop or a corkscrew?

Hello Alison,

Here is some information to help you in your project

Movement in a curved path needs a force. This is called the centripetal force for circular motion. It is always towards the centre of the circle. The required centripetal force depends on the mass(m) of the person, the velocity (v) and the radius of the circle (r).

The formula is F = (mass x velocity x velocity)/ radius = mvv/r

This force does not exist by itself. It must be provided by the sum of the forces already acting on the person.The direction of these forces must be taken into account, when finding the overall force.

Imagine the vehicle on the highest point (top of a vertical circle) There is a reaction force (R) upwards.This is the force on the person given by the seat. The reaction force R acts upwards and the weight (W) acts downwards.The overall force will be W-R . This provides the force (F) needed to keep moving in a circle.

At this point F will be aimed directly downwards ..in the same direction as W.

So                  F = W-R         (1)

Now W = mg   (g = acceleration due to gravity)

and F = (mvv)/r

Substituting these into (1)and re-arranging gives

Reaction Force (R) = mg - (mvv)/r  = m (g-vv/r)

The condition for feeling weightless occurs when g = vv/r

If the velocity is high, so that mvv/r is greater than mg, then an additional force is needed to keep a person in his/her seat.

The actual size of this additional force depends on the values of v and r.If mvv/r = 2mg ,then the downward force needed to keep a person in his/her seat will be mg (equal to the weight of the person,called one G force). Designers must make sure that the additional forces (provided by retraining straps) are not too great so as to cause injury.The radius and velocity must be calculated very carefully to prevent this happening.A large downward force will be result of too high a velocity,or too small a radius..or both !

An estimate of the maximum G force (the value of R) is around 3mg.This will happen when mvv/r = 4mg (4 times the weight of a person). So,if you know v you can calculate the minimum radius needed to prevent injury.

At the bottom of the circular ride,the Reaction Force R must be greater than the weight W,to keep the person moving in a circle.The reation force is given by

R = mvv/r + mg

when mvv/r = mg the reaction force is twice the value of W, so a person feels twice as heavy. When mvv/r = 2mg the person will feel three time as heavy as normal. Fairground designers must be very careful not to design the rides so that R is too high. There is a limit to which any body can take...without injury.

Some values :

Maximum velocity gained during a fall of h metres = square root of (2gh)
g = 9.8 m/s^2  (metres per second per second)
A drop of 65.5m for a Rollercoaster  gives a velocity of 37.8m/s
Typical maximum G force is when  R = 3.5mg
so 3.5mg = m(vv/r + g) gives r = 58.3 metres.
The Rollercoaster MUST have a large radius at the bottom ,to prevent
injury to passengers (otherwise the G force is too high)

Typical first drop for a Corkscrew =  23m

All very mathematical, I know, but Physics can be interesting.. this is just an example where designers must use their knowledge of Physics to make sure rollercoasters and Cosrkscrew rides are both fun AND safe.

Martyn Overy
North Chadderton School
Oldham
England