# Physics U4 Question! HELP!

Have just been doing Physics Jan04 Unit4 paper and I'm confused on something (it's question2 if anyonhe has the paper).

It's talking about a car travelling over a hump-backed bridge in the road. When the car is driven over the bridge it follows part of a vertical circle of radius (25m) at the point O below the bridge. At the top of the bridge when the car is at rest (ie. equil.) it has normal reaction force acting upwards and the weight downwards.

The part I'm stuck on is: "If the car is driv en across the bridge repeatedly at gradually inc speeds it is found that above a critical speed the car loses contact with the road at A and "takes off". Explain why this happens."

I would have thought this would occur only when R is greater than the weight. The markscheme says "The centripetal force required is greater than the weight, which is impossible". Could someone explain this 2 me (tonight!!!).

Thank you sooooo much!
when it gets to a certain speed, the equation for centripetal force (mv^2/r) will become larger than the force that is providing it (the weight). can't think of any other way to put it
Hoofbeat
Have just been doing Physics Jan04 Unit4 paper and I'm confused on something (it's question2 if anyonhe has the paper).

It's talking about a car travelling over a hump-backed bridge in the road. When the car is driven over the bridge it follows part of a vertical circle of radius (25m) at the point O below the bridge. At the top of the bridge when the car is at rest (ie. equil.) it has normal reaction force acting upwards and the weight downwards.

The part I'm stuck on is: "If the car is driv en across the bridge repeatedly at gradually inc speeds it is found that above a critical speed the car loses contact with the road at A and "takes off". Explain why this happens."

I would have thought this would occur only when R is greater than the weight. The markscheme says "The centripetal force required is greater than the weight, which is impossible". Could someone explain this 2 me (tonight!!!).

Thank you sooooo much!

I would say a higher velocity when the car reaches the peak means more upward momentum when the road turns - as the car's weight is constant, and there is no road, the resultant force is the car's weight, but as this is a constant, larger upward momentum takes longer to be changed to downward momentum. Therefore above a higher velocity, when the road slopes downward the car's wheels leave the road - below that velocity, the car still has upward momentum, but not enough for the car to leave the road before it is reduced by gravity..

I think. What's centripetal?
mik1a

I think. What's centripetal?

We started it today, it is unit 4
Centripetal force is the force towards the centre of the circle, be it gravitational, tension in a string, etc.
Rustyk1
Centripetal force is the force towards the centre of the circle, be it gravitational, tension in a string, etc.

So what is causing the centripetal force in this case? And what is the third law pair?
correct me if i'm wrong but the velocity required for circular motion can be found by v=root(gr) where g is the acceleration due to gravity and r is the radius of the circle (in this case 25m). i may be wrong. meaning that the required velocity is 15.66m/s. any greater than this and it will take off (lets be honest, thats what we all want!). like i say i could be wrong
Rustyk1
correct me if i'm wrong but the velocity required for circular motion can be found by v=root(gr) where g is the acceleration due to gravity and r is the radius of the circle (in this case 25m). i may be wrong. meaning that the required velocity is 15.66m/s. any greater than this and it will take off (lets be honest, thats what we all want!). like i say i could be wrong

So the centripetal force is gravity?

How does that work, since gravity is always acting downwards on the car. And the answer states: "The centripetal force required is greater than the weight"... which makes me think they are two seperate things?
mik1a
So the centripetal force is gravity?

How does that work, since gravity is always acting downwards on the car. And the answer states: "The centripetal force required is greater than the weight"... which makes me think they are two seperate things?

friction can cause centripetal force too...so that probably helps
yeah the car will leave the road when the centripetal force required (mv^2/r) is greater than that present (the weight acting towards the centre of the circle = mg) and so the car will not stay in circular motion. discounting friction, the speed of takeoff can be found by v=root(rg).
The centripetal force is whatever force is keeping the object moving in a circle - in this case the weight of the car is the centripetal force.

The centripetal force required to keep the car moving in a circle depends on the velocity of the car (F=V^2/r). When it says "the centripetal force becomes greater than the weight" what it means is more force than is provided by the weight of the car would be required to keep the car following its circular path, therefore there is not enough weight to act as a centripetal force to the circular motion and so the circular motion is broken.
Rustyk1
correct me if i'm wrong but the velocity required for circular motion can be found by v=root(gr) where g is the acceleration due to gravity and r is the radius of the circle (in this case 25m). i may be wrong. meaning that the required velocity is 15.66m/s. any greater than this and it will take off (lets be honest, thats what we all want!). like i say i could be wrong

thanks every1 it makes sense now. And yes 15.6m/s is the correct answer for the subsequent part of the question!

Good Luck to anyone doing Edexcel A2 Physics this morning!