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# centripetal force question watch

1. http://filestore.aqa.org.uk/subjects...W-QP-JUN11.PDF

on question 1 aii , iam confused... the weight (mg) act downwards and the reaction force acts upwards and the cetripetal force acts downwards . so i dont understand how these forces relate to eacherother could someone help ?
2. (Original post by HG1)
http://filestore.aqa.org.uk/subjects...W-QP-JUN11.PDF

on question 1 aii , iam confused... the weight (mg) act downwards and the reaction force acts upwards and the cetripetal force acts downwards . so i dont understand how these forces relate to eacherother could someone help ?
Hi! I moved this to the Physics forum for you - you're more likely to get an answer here
3. In all circular motion questions, the important thing to remember is that "centripetal force" is not some magic new kind of force. It is just the name that we give to the resultant force on an object when it moves in a circle.

I'm sure that you know an equation for this resultant (centripetal) force F on an object, in terms of its mass m, velocity v and radius of circular motion r.

So in this case, you need to work out what forces actually go to making up F. In other words, can you work out an equation for F in terms of the forces acting on the parcel, its weight and the reaction force. If you can, use this equation and the one that always applies for centripetal force to eliminate F. You'll then need to rearange it so that it reads R = ..., but that should be easy.
4. (Original post by Pangol)
In all circular motion questions, the important thing to remember is that "centripetal force" is not some magic new kind of force. It is just the name that we give to the resultant force on an object when it moves in a circle.

I'm sure that you know an equation for this resultant (centripetal) force F on an object, in terms of its mass m, velocity v and radius of circular motion r.

So in this case, you need to work out what forces actually go to making up F. In other words, can you work out an equation for F in terms of the forces acting on the parcel, its weight and the reaction force. If you can, use this equation and the one that always applies for centripetal force to eliminate F. You'll then need to rearange it so that it reads R = ..., but that should be easy.
thanks for this explanation, so mv^2/r act downwards towards the centre of the circle and w=mg also acts downward so how come the reactin force isnt the sum of them both ?
5. As I say, find the resultant of the forces that are acting, and if the object is moving in a circle, we call the resultant the centripetal force (which, as you say, has a magnitude of mv^2/r).

The weight, mg, acts downwards. The reaction force, R, acts upwards. So, taking the positive direction to be towards the centre of the circle, their resultant is mg - R. As the object is moving in a circle, this is the centripetal force, which you can now equate to mv^2/r.

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