# Question 4 a i (1 Viewer)

#### ghoolz

##### New Member
Did anyone else find this incredibly easy???

I done it in three lines and i think i done it right here goes:

F= (mv^2)/r centripital force
we know that v=rw circluar velocity
subbing in gets

F=(mr^2w^2)/r
=mrw^2
there fore shown

I hope thats it.

It was worth 3 marks and compared to others they were giving the marks away there.

#### underthesun

##### N1NJ4
eh? i dont think you can assume that.

you have to do it terry lee way i think differentiating and stuff.

#### ghoolz

##### New Member
what the hell is the terry lee way???

it only said to show it

#### Amish

##### Member
What i did was differentiate x and y then find v^2 by -
v^2 = (vx)^2 + (vy)^2
which found v to by like r^2 . w^2
and then sub that in for v^2 to the centrip force which gave the answer mrw^2

But knowing my dumbass status...i probably did it totally wrong and as it was out of 3 marks it kinda seemed we had to do more then like 3 lines...but it could happen #### jims

##### Member
i differentiated x and y twice then drew a triangle and used pythagoras to find a

?? : ( i failed

N

#### ND

##### Guest
Originally posted by Amish
What i did was differentiate x and y then find v^2 by -
v^2 = (vx)^2 + (vy)^2
which found v to by like r^2 . w^2
and then sub that in for v^2 to the centrip force which gave the answer mrw^2

But knowing my dumbass status...i probably did it totally wrong and as it was out of 3 marks it kinda seemed we had to do more then like 3 lines...but it could happen Nah you can't just do that. What you've gotta do is find d2y/dt2, d2x/dt2, then resolve the normal force.

#### sugamama

##### Coldplay Fan
Originally posted by ND
Nah you can't just do that. What you've gotta do is find d2y/dt2, d2x/dt2, then resolve the normal force.
How many marks would you get doing it the other way?

#### Fosweb

##### I could be your Doctor...
They give you x=rcos@ and y=rsin@, so you have to use them to differentiate and get two forces which then add to the centripetal force, which is mrw<sup>2</sup>

Fitzpatrick book also has the proof.

#### smeyo

##### Member
yeah you have to differentiate twice and then resolve by pathagoras, it actually was an easy question as long as you knew the process, if you didnt you would kind of been lost...

#### VanCarBus

##### ~--> Quincy <--~
x = rcos@
y = rsin@
but @ = wt
x = rcos(wt) --> differentiate twice --> ddx
y = rsin(wt) --> differentiate twice --> ddy
F = ma
but a = squareroot (ddx squared + ddy squared)

thus

F =mrw^2

bloody html, can't write maths number properly

#### walla

##### Satisfied Customer
vancarbus has the right way....it required implicit differentiation. i didn't resolve the direction though, which was probably 1 of the 3 marks.

#### Mathematician

##### Member
...

Oooo thats an interesting way(easy too)

Mine was probably half right(and crap long way)

x = rcos@
y = rsin@
i d/dt ( ) both sides twice for x and y

x* = -rsin@ d@/dt = -r@*sin@

x** = -r[@**sin@ + @*cos@.d@/dt ]
= {-r@**sin@} - {r@**cos@ }

did the same for y and took the second { } bracket from the x** and y** and tried to add them like vectors to get r@** but was getting confused and it pointed north west ????? its not meant to

i didnt get why but then i just said this is acceleration and times it by the mass to get force..

Pretty crap explainer.

I got this method from arnold the day before and only skim read it , thats why i was confused in the exam. I DIDNT THINK IT'LL COME.

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