help !! (1 Viewer)

[Napoleon]

A particle of mass 1kg moves in a straight line with velocity v against a resistance force equal to v[rt [1-v^2]]. the displacement, x, of the particle from a fixed origin O is initially ero and its velocity at that time is R.

show that x= arcsin [R[rt [1-v^2] - v[rt [1-R^2]]

 

[KFunk]

A particle of mass 1kg moves in a straight line with velocity v against a resistance force equal to v[rt [1-v^2]]. the displacement, x, of the particle from a fixed origin O is initially ero and its velocity at that time is R.

show that x= arcsin [R[rt [1-v^2] - v[rt [1-R^2]]

The resistive force gives you the equation of motion which is:

acceleration = v(dv/dx) = -v√(1 - v²)

dv/dx = - √(1 - v²)
dx/dv = -1/√(1 - v²) then integrate w.r.t. v to get

x = cos^-1v + c ... when x=0, v=R so c = -cos^-1R

x = cos^-1v - cos^-1R then do the inverse trig thing

sinx = sin(A - B) where A = cos^-1v and B = cos^-1R

sinx = sinAcosB - cosAsinB = R√(1 - v²) - v√(1 - R²) by doing all the triangle stuff

∴ x = sin^-1(R(√(1 - v²) - v√(1 - R²))

 

[Antwan23q]

a = -v Sqrt (1-v^2)

vdv/dx = -v Sqrt (1-v^2) (v's cancle out)

dv/dx = - sqrt (1-v^2)

dv/ Sqrt(1-v^2) = -dx (integrate both sides)

sin^-1 (v/1) = -x +c

x=0, v= R

sin^-1(R) = c

.: x = sin^-1(R) - sin^-1(v) (put both sides in sine function)

sin x = sin [sin^-1(R) - sin^-1(v)]
sin x = sin(sin^-1(R))cos(sin^-1(v)) - cos(sin^-1(R))sin(sin^-1(v))
sin x = Rcos(sin^-1(v)) - vcos(sin^-1(R))

now if u draw a right angle triangle, with v being one side, 1 being the hypotenuis, and thefore the last side is Sqrt (1-v^2)
thefore, cos(sin^-1(v))= Sqrt (1-v^2)
so...
sin x = RSqrt (1-v^2) - vSqrtSqrt (1-R^2) then inverse the sine fucntion

x = sin^-1[RSqrt (1-v^2) - vSqrtSqrt (1-R^2)]

 

[Antwan23q]

arrgh beaten!

 

[Napoleon]

whoa nice!
ummm say F=ma=m[v rt [1-v^2]]
why do you have to put a negative i would of thought the forces balanced them all out?

Btw, you guys going to pwned tomorrow aren't ya?

 

[Antwan23q]

resistance force equal to v[rt [1-v^2]].

since its RESISTED force, it means that the force is in the oppisite direction.
it doesnt say that it already has a force on it. so this is the only force and is slowing it down.

kfunk will own, i will pass... (hopefully)

 

[100percent]

kfunk will own, i will pass... (hopefully)

your definition of passing 100+?

 

[Antwan23q]

no, like 60-70ish

 

[100percent]

no, like 60-70ish

60-70 is what i get, (hopefully i don't make silly mistake in 1st 2 questions)
you should be like 90+ atleast

 

[KFunk]

kfunk will own, i will pass... (hopefully)

Haha, yeah, except for the fact that I got nothin' when it comes to circle geometry and conics. Those two topics, along with my stupid mistakes, kill me.

 

[Napoleon]

any of you dudes want to help me with some miscellaneous questions privately? i.e. via msn or google chat

 

[KFunk]

I would if I had the time at the moment (HSC and whatnot). If you're cool with posting your questions on the board then there are more than enough people to tackle them. I geuss right now the board is good because we can be selective in what we do.

 

[Antwan23q]

yeh ok, hit me up, whats ur email. im usually always on.
i hate circle geometry, biggest bitch, conics im still workin on. hard 3 unit is whats going to fuck me up

 

[robbo_145]

agreed, just post em up in the board. new posts every minute anyway

 

[Napoleon]

check your private msg aswan

 

[Napoleon]

Hey fellas here's another one

 

[Trev]

bi)
Vertically:
Ncosө-mg=F; F=0 (no frictional force)
Ncosө=mg (1)
Horizontally:
Nsinө=mv²/r (2)

(2)/(1) gives tanө = (mv²/r )*(1/rg)
rgtanө=v²
v=√(rgtanө)

ii) {not too sure about this part because I really suck at this lol}
When x=√(rgtanө)
Nsinө=mgtanө
N=mg/cosө [magnitude]
Now i'm stuck

 

[ishq]

The first part is bookwork - no friction
Draw the diagram

Where A is the banked angle
NsinA = mv^2/r
NcosA=mg

N=mg/cosA

mg/cosA=mv^2/rsinA
v^2 = 2gtanA
v = sqrt rgtranA

 

[ishq]

For the second part, do you assume friction to have a direction and then try and find magnitude? Or it is something to do with the fact that for small angles tanA = SinA?

 

[Napoleon]

your the beast =P not me, thanks for doing on
is it easy for you?
mind helping me with a few more?