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physics help, about the centripetal force.

ok, why does the centripetal force actually act outwards.

for example in the january 2004 phy 4 paper.
a car is stationary such that R(reaction force) = WEIGHT.
it then goes past the same point with velocity v.

if there is a CENTRIPETAL FORCE which should act towards the centre of motion.

shouldnt the equation be W + C.F = R
as the weight and the centripetal force both act downwards?
Reply 1
don't have the paper and am not entirely sure what you mean in:

"if there is a CENTRIPETAL FORCE which should act towards the centre of motion.

shouldnt the equation be W + C.F = R
as the weight and the centripetal force both act downwards?"

Does it travel around a circle???
Reply 2
mathematician
ok, why does the centripetal force actually act outwards.

for example in the january 2004 phy 4 paper.
a car is stationary such that R(reaction force) = WEIGHT.
it then goes past the same point with velocity v.

if there is a CENTRIPETAL FORCE which should act towards the centre of motion.

shouldnt the equation be W + C.F = R
as the weight and the centripetal force both act downwards?

There must be an unbalanced force downwards (the centripetal force) so when the car is moving, the reaction force pushing upwards will be less, the weight will be the same so there will be a nett force downwards of magnitude mv^2/r
Reply 3
Yannis
don't have the paper and am not entirely sure what you mean in:

"if there is a CENTRIPETAL FORCE which should act towards the centre of motion.

shouldnt the equation be W + C.F = R
as the weight and the centripetal force both act downwards?"

Does it travel around a circle???


car on a road at the top such that weight = R(reaction force)
the centripetal force is supposed to acts towards the centre of motion when the car is moving in circular path.

however shouldnt the reaction force increase in that case?
as there will now be a centripetal force -> mv^2/r added on to the weight of the car which also acts towards the centre and this added together shud equal the reaction force acting upwards.
Reply 4
Bezza
There must be an unbalanced force downwards (the centripetal force) so when the car is moving, the reaction force pushing upwards will be less, the weight will be the same so there will be a nett force downwards of magnitude mv^2/r

yes i understand tht.
a c.force required for UCM.

in that case form an equation for R, when the body is moving at velocity v.
Reply 5
When the car's at rest, R1 = mg
When it's moving, taking downwards as positive F = ma, mg - R2 = mv^2/r so R2 = m(g - v^2/r)
Reply 6
If the car is going round on a banked (inclined) road, then the reaction increases in order to provide the horizontal force needed to keep it goin round in circles:


. Reaction
. /\
. \
. \
. Car
. |
. |
. \/
. weight

You get:

R.sin(alpha)=mv^2/r

where r is the radius of the circle and alpha is the angle at which it is banked. Does that help?
Reply 7
Bezza
When the car's at rest, R1 = mg
When it's moving, taking downwards as positive F = ma, mg - R2 = mv^2/r so R2 = m(g - v^2/r)


so your saying the reaction force will decrease, the force mg will always be the same acting downwards.
Reply 8
Yannis
If the car is going round on a banked (inclined) road, then the reaction increases in order to provide the horizontal force needed to keep it goin round in circles:


. Reaction
. /\
. \
. \
. Car
. |
. |
. \/
. weight

You get:

R.sin(alpha)=mv^2/r

where r is the radius of the circle and alpha is the angle at which it is banked. Does that help?


nope im talkin about something different sorry.
Reply 9
Bezza
When the car's at rest, R1 = mg
When it's moving, taking downwards as positive F = ma, mg - R2 = mv^2/r so R2 = m(g - v^2/r)


shouldnt it be

R2 = W + MV^2/R
Reply 10
mathematician
so your saying the reaction force will decrease, the force mg will always be the same acting downwards.

Yep, because when it's travelling in the circle, it's accelerating so there is an unbalanced force downwards. The weight is constant because its only affected by the mass of the car and the gravitational field strength.
Reply 11
Bezza
Yep, because when it's travelling in the circle, it's accelerating so there is an unbalanced force downwards. The weight is constant because its only affected by the mass of the car and the gravitational field strength.


OH shit i get it, cheers m8.
wot examing board r u on
Reply 12
mathematician
OH shit i get it, cheers m8.
wot examing board r u on

Edexcel - I actually did this paper in january!
Reply 13
yer same
wot did u get
im resittin it again though
Reply 14
I got 81/90
Reply 15
Bezza
I got 81/90


really
i got 69/90
i need to resit
wot did u get for phy 1 2 3 and how did your phy 5 practical go
Reply 16
mathematician
really
i got 69/90
i need to resit
wot did u get for phy 1 2 3 and how did your phy 5 practical go

I got 300/300 at AS so dropping 9 in phy4 doesn't really bother me! I thought the practical was ok but my tray was on a slope and I found it hard to count the oscillations with the 100g mass on the ruler cos they were so quick
Reply 17
Bezza
I got 300/300 at AS so dropping 9 in phy4 doesn't really bother me! I thought the practical was ok but my tray was on a slope and I found it hard to count the oscillations with the 100g mass on the ruler cos they were so quick

wot uni u wanna go 2, wot course and wot module marks have u got for maths
Reply 18
mathematician
wot uni u wanna go 2, wot course and wot module marks have u got for maths

Magdalen college, Oxford, Physics, P1 100, P2 94, P3 100, M1 98, M2 92 M3 100, S1 86 :frown:, S2 92
Reply 19
Bezza
Magdalen college, Oxford, Physics, P1 100, P2 94, P3 100, M1 98, M2 92 M3 100, S1 86 :frown:, S2 92


pretty good maths marks, lol i got 100 in s1, d1,p1,p4,d2

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