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# centripetal force watch

1. part (i) ok

part (ii) why R = mgcos O + mv^2/r ?
isn't that i shud minus centripetal force?
direction of force exertc hoop on P = towards O
direction of centripetal force = towards O

part(iii), projectile motion? can't get the ans
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2. (Original post by Zaq)
part (i) ok

part (ii) why R = mgcos O + mv^2/r ?
isn't that i shud minus centripetal force?
direction of force exertc hoop on P = towards O
direction of centripetal force = towards O

part(iii), projectile motion? can't get the ans
When I teach part (ii) I always explain that this is just another application of F=Ma

i) have a look at the attached diagram ** cannot attach file over 100kb ??
ii) take the positive direction as towards
iii) you have R towards the centre
iii) you have a component of mg away from the centre
iv) the relevant component of the weight has magnitude mg cos (theta)

So applying F=ma

R - mg cos(theta) = ( m v ^ 2) / r

Etc .......

Mrm.

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