Circular motion question (1 Viewer)

[Etude]

This question is from example 10 in the fitzpatrick textbook

A car of mass 1 tonne passes over a bridge formed by the arc of a circle of radius 10 meters.
i) Find the force exerted by the car on the road at the top of the bridge if the car is traveling at 8m/s.
ii) What speed would cause the car to be on the point of leaving the bridge at its highest point.


I understand thier solution for parts i), but for part ii) it says: 'The car leaves the bridge when N=0'.

Thats what i dont understand, why N=0? inst there always a force from gravity pulling the car down, and therefore a normal force pushing it up? Or isnt the car already traveling at 8m/s and so the car will already pass over the bridge? Why does the car leave when N=0 when its already moving at 8m/s?

Any help is appreciated thanks.

 

[vds700]

This question is from example 10 in the fitzpatrick textbook

A car of mass 1 tonne passes over a bridge formed by the arc of a circle of radius 10 meters.
i) Find the force exerted by the car on the road at the top of the bridge if the car is traveling at 8m/s.
ii) What speed would cause the car to be on the point of leaving the bridge at its highest point.


I understand thier solution for parts i), but for part ii) it says: 'The car leaves the bridge when N=0'.

Thats what i dont understand, why N=0? inst there always a force from gravity pulling the car down, and therefore a normal force pushing it up? Or isnt the car already traveling at 8m/s and so the car will already pass over the bridge? Why does the car leave when N=0 when its already moving at 8m/s?

Any help is appreciated thanks.

In (i), they calculate the value of N, which is equal and opposite to the downwards forve of the car on the road (Newton's 3rd law). So when N = 0, the force of the car on the road must be 0, and hence it leaves the bridge. ( I think)

 

[Etude]

In (i), they calculate the value of N, which is equal and opposite to the downwards forve of the car on the road (Newton's 3rd law). So when N = 0, the force of the car on the road must be 0, and hence it leaves the bridge. ( I think)

But isnt there always a force applied to the road when a car is travelling caused by gravity?

 

[hon1hon2hon3]

Ok . . i think the question means that , a car of one tone drove accross the bridge . . .

so for i) it will have to do with centural pedal forces . . (mv^2) / r or mr(omega)^2 . . . (omega = angular velocity)

and what they meant by N=0 , is that they mean the net force on the car is zero . sooo when N=0 , its the point where it loses contact with the bridge surface .

And your right, theres always gravity pulling the car down , and they question is asking , at what speed does the car need to drive to over come the centural pedal force u calculated at part (i).

Hope it make sense . . . N=0 means net force on the car is zero , it doesnt mean, theres no force acting on the car . . its just that they cancel out.

Just sharing my thoughts , might be wrong , Peace ~

 

[Etude]

Well the forces radially is mg=mv^2/r+N
If N=0 then mg=mv^2/r

Does this mean that a particle moving in a circle will only move if weight is equal to centripedal force?? totally lost

Any thoughts? (btw answer to part 1 is 3400N)

 

[hon1hon2hon3]

what u said should be right .
mg=mv^2/r+N

If N=0 then mg=mv^2/r

this is ture, but then u dont quite understand the concpet , now try to rearrange the formula. mg - mv^2/r =0 . Now think of it, like what i said before , if the total force is equal to zero , then the net force on the car is zero, meaning that will be the moment when the car loses its contact with the bridge .

From solving mg - mv^2/r =0 , should be able to find the answer of part (ii) .

Peace again , trying to help dont flame me if i got it wrong

 

[vds700]

N is the reaction force of the bridge on the car (upwards). So if N = 0, then the car will have lost contact, as it is not touching the bridge, then the bridge cannot exert any force on the car, hence N = 0.

 

[Etude]

I think i understand it now. When N=0 the bridge previously supporting it is no longer there, so the car will fall. But since its accelerating forward, the car will fall while moving forward. Hence a circular like motion.

At least i think it works like that.

Thanks for the help really appreciate it

Even still tho i cant imagine that happening in real life...i mean we hit the accelerator and the car moves, does that mean when we travel on a slanted path theres no normal pushing us up?