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M3 June 2012 Q5

Hi, I'm doing Q5 and got part a correct but part b is what's confusing me. According to the mark scheme, as I understand it, the Reaction force on the particle (R) is acting in the same direction as the Centripetal Force which to me makes no sense considering that the particle is on the outer surface of the hemisphere :/ I'd get grateful if someone could clear this up for me :smile:

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Reply 1
Original post by Ash8991
Hi, I'm doing Q5 and got part a correct but part b is what's confusing me. According to the mark scheme, as I understand it, the Reaction force on the particle (R) is acting in the same direction as the Centripetal Force which to me makes no sense considering that the particle is on the outer surface of the hemisphere :/ I'd get grateful if someone could clear this up for me :smile:

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hello

according to the mark scheme

R:frown:towards centre) = mgcos(x) - R = ma

The centripetal acceleration acts towards the centre of the circle, and the reaction force is in the opposite direction?

I think you are making the mistake of forming a statics-like equation, since if you rearrange the above formulae it would appear that R and ma (cenforce) are in the same direction.

When you look at problems like this, consider the reaction force and the force of gravity, are experiencing centripetal acceleration, don't consider it as a force. Think back to an M1 dynamics problem, you wouldn't make that mistake there.

I realise i havent explained this too well but if you have any questions just quote me :smile:
Reply 2
Original post by xyz9856
hello

according to the mark scheme

R:frown:towards centre) = mgcos(x) - R = ma

The centripetal acceleration acts towards the centre of the circle, and the reaction force is in the opposite direction?

I think you are making the mistake of forming a statics-like equation, since if you rearrange the above formulae it would appear that R and ma (cenforce) are in the same direction.

When you look at problems like this, consider the reaction force and the force of gravity, are experiencing centripetal acceleration, don't consider it as a force. Think back to an M1 dynamics problem, you wouldn't make that mistake there.

I realise i havent explained this too well but if you have any questions just quote me :smile:


As soon as I saw "Don't consider it as a force", it then hit me. Fc is just the overall force and is equal to both mv^2/r and the difference between the reaction and the weight forces while not being an actual force itself. Thank you so so much for helping me, I appreciate it :smile:
Reply 3
Original post by Ash8991
As soon as I saw "Don't consider it as a force", it then hit me. Fc is just the overall force and is equal to both mv^2/r and the difference between the reaction and the weight forces while not being an actual force itself. Thank you so so much for helping me, I appreciate it :smile:


No worries, I made the same mistake as you when I was learning it in class a few weeks ago :smile:

Good luck with the exams.

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