I'm stuck on 1aii. The mark scheme says the answer is R=mg - (mv^2/r) but I don't understand why it isn't mg + (mv^2/r) since the centripetal force also acts downwards towards the centre of the circle?
Physics Circular Motion Help
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- 25-09-2016 13:07
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- 25-09-2016 13:25
Hey I think you've made a common mistake that happens in circular motion. The centripetal force is the RESULTANT force and that is given by (mv^2)/r. That is to say it is not it's own separate force that is acting on the body. The only forces that are acting on the parcel is the weight (mg) and the reaction/normal force (R).
They act in opposite directions and if we take the weight to be the positive direction then we can see that the resultant force is given by (mg - R). But remember what I said in the first paragraph? "the RESULTANT force and that is given by (mv^2)/r " so in other words we can say:
mv^2/r = mg - R
Rearraging this gives R = mg - mv^2/r.
No you may say "Well this only works if you choose the weight as acting in the positive direction, other wise you'd end up with the resultant force being R - mg"
And the truth is that I didn't randomly choose the weight to be in the positive direction. I knew that the parcel was undergoing circular motion and so I know that it must be experiencing a net force towards the centre of the circle at all times - hence I knew that the resultant force was acting "downwards" (i.e. towards the centre of the circle). Why is this important? Well since mv^2/r is always a positive quantity - it must equal a positive quantity therefore I had to choose the direction which made the resultant force a positive quantity.
If this is unclear please tell me which bits are and I will attempt to clarify.