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EDEXCEL PHY4, PHY5 questions (Old Spec)

Hey guys just thougt it would be a good idea for people who need help with a particular question from the old spec to post it up and see if we cant answer it. (Also because I have a question I cant get my head around :woo: lol)

Q) (b)(i) A girl standing at the equator is in circular motion about the Earth’s axis. Calculate the angular speed of the girl.

Angular speed = 7.27 x 10^-5 rads-1

(ii) The radius of the Earth is 6400 km. The girl has a mass of 60 kg. Calculate the resultant force on the girl necessary for this circular motion.

Force = 2.03N

(iii) If the girl were to stand on weighing scales calibrated in newtons, what reading would they give?

Scale reading = .................................
(Answer given in ms is 568-567N)

But I dont understand why you minus the resultant force from the weight. Do they not act in the same direction i.e. weight and centripetal force that is.There is another question that askes a similar thing and its kind of thrown me.

Thanks in advance...

Ash
Reply 1
Avatar for logic123
logic123
nah its newtons third law the scales push you up
Reply 2
Avatar for cajohare
cajohare
Yeah the scales measure the normal reaction force by the ground on the person, and because centripetal force is a resultant force:

(taking downwards as positive)

(mv^2/r) = mg - R

and rearrange to find R.

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