mechanics plz help. (1 Viewer)

[ART]

hi. could any please work out the answer to this mechanics question?

A body of mass m is released from a height h above the ground and it experiences a resistive force R given by R=0.1mv2, where v is the velocity achieved by the body at time t.
If the object falls from rest under gravity (g=10m/s2), find the terminal velocity, U.


... sorry, i know this question should be easy, but for some reason, i get U=infinity ?!?

 

[Ogden_Nash]

There are two forces acting on the particle:
1) weight downwards = mg
2) resistance upwards = 0.1mv²

Therefore,
F = ma = mg - 0.1mv²
giving the equation of motion:
a = g - 0.1v²
Now, terminal velocity, U, is approached as a --> 0
So, v² --> g/0.1

Therefore U = 10ms^-1

Hope that helps

 

[who_loves_maths]

hi ART & Ogden_Nash,

i know these generic type of mechanical questions are part of the 4u course, however, my class has not yet reached that part of 4u theory yet;
consequently, i'd like to present an "alternative" method of solving this question purely from the perspective of (only year 11) Physics...
{this probably belongs in the Physics subforum but my love for phys is too great, i can't resist posting it here, so plz forgive me ppl}:

consider the Law of Conservation of Energy:
initial energy = gravitational potential energy = mgh = 10mh ; where 'g' = 10 m/s/s ;
now, the energy (or work) is defined as: E = Fd ; hence, the energy loss (to the atmosphere) due to the retarding force = Fh = (0.1mv^2)h ;
and, the final kinetic energy of the particle = (1/2)mv^2
hence, Final K.E. = Initial P.E. - Energy Loss ---> (1/2)mv^2 = 10mh - 0.1hmv^2 ;
the 'm' cancels ---> h(10 - 0.1v^2)= 0.5v^2 ---> h(100 -v^2) = 5v^2 ---> h = (5v^2)/(100 -v^2) ;
now, the 'terminal' velocity is as 't'--> infinity, or, it's the same as 'h' --> infinity ;
therefore, as h--> infinity, (5v^2)/(100 -v^2) --> infinity; ie. 100 -v^2 --> 0
-----> v^2 = 100 -----> v = U = 10 m/s.

so there you go, year 11 Physics does the trick here without the need for 4u maths

hope this helps, especially if you're more into phys than maths

 

[ART]

thanks

thanks guys. i think i got it wrong because i tried to work it out with the reasoning "terminal velocity occurs when t approaches infinity". for some reason this doesn't work out.

 

[maths > english]

i tried to work it out with the reasoning "terminal velocity occurs when t approaches infinity". for some reason this doesn't work out.

it is possible to work out the terminal velocity using a <i>t → ∞</i> method but it uses <i>l'hopitals</i> rule [EDIT: l'hopitals rule not needed]

this is attached to the message, however the intended way is clearly the <i>a → 0</i> method posted by <i>Ogden_Nash</i>

EDIT: Do NOT use l'hopitals rule, at the end of the first page just divide the numerator and denominator of v by e^{[sqrt(10g)/5]*t}

 

[who_loves_maths]

Originally Posted by ART
thanks guys. i think i got it wrong because i tried to work it out with the reasoning "terminal velocity occurs when t approaches infinity". for some reason this doesn't work out...

there's nothing wrong with that thinking. like i said in my last post 'h' approaches infinity is the same as 't' approaches infinity which is exactly how you go about formally in finding out the terminal velocity. so you are right, i don't know why you think you're wrong...

here's the mathematical version, as far as i can see (using 't' approaches infinity):

Net Force = ma = mg - 0.1mv^2 ---> a = dv/dt = 10 - 0.1v^2
therefore, dt/dv = 1/(dv/dt) = 1/(10 - 0.1v^2) = 10/(100 -v^2)
hence, integrating, t = 10Int[dv/(100 -v^2)] = 10ln[(10+v)/(10-v)] + c
when t = 0, v = 0 ---> c = 0;
and so, we get the expression for 't': t = 10ln[(10+v)/(10-v)]

now, take 't' --> infinity, hence, infinity = 10ln[(10+v)/(10-v)] = -10ln[(10-v)/(10+v)] ---> ln[(10-v)/(10+v)] = -infinity ---> (10-v)/(10+v) = 0 ---> 10-v = 0 ---> v = U = 10 m/s.

and that's how you'd probably do it, so i don't see why taking 't' approaches infinity would be wrong ?!

anyway, hope that helps

 

[who_loves_maths]

Originally Posted by maths > english
it is possible to work out the terminal velocity using a t → ∞ method but it uses l'hopitals rule

this is attached to the message, however the intended way is clearly the a → 0 method posted by Ogden_Nash

l'hopitals rule is not part of the 4u course, and you don't need it in its explicit form to do this question by taking 't' approaching infinity.
it'd be much easier to do it the way i just did in my last post i think, and you can use that method to generalise it into v = U = Sqrt(10g) in this case too. the integral is also only 4u level anyways.

but like you said, the correct mathematical way (that is, the most expedient) would be to consider a --> 0. [although the Physics way isn't bad either ... j/k; zero marks for the use of physics in a maths exam, lol.]

 

[maths > english]

making t the subject before replacing it with infinity is definately easier, didnt think of that when i did it

 

[FinalFantasy]

ahh i used to think it was "l'hoSpitals" rule instead of l'hopitals rule

 

[maths > english]

lol, me too

its really bad when u misspell things with really stupid errors and others read it

e.g.

latus rectum -> lettice rectum

organism -> orgasm lol

 

[who_loves_maths]

why are you always in invisible mode FinalF.? - that's right, i know you're reading this :uhhuh:

 

[FinalFantasy]

hahaha, i chose it to always appear invisible when i first signed up here

ahh damn u! u know im on!

 

[who_loves_maths]

hehe... if only there was an invisible-invisible-mode, then i wouldn't be able to tell

 

[FinalFantasy]

dat was just a lucky guess! u can't find me, im invisible

 

[who_loves_maths]

yeah, ur right, it was just a lucky guess... too bad i'm 99% lucky most of the time, so you're still 99% visible most of the time

 

[FinalFantasy]

hahaha, this mechanics thread turned into some spam now @_@

 

[who_loves_maths]

lol.

Edit: i'm just adding to the spam here...

 

[FinalFantasy]

we're gonna learn mechanics soon, we have to go to a freaking holiday class to get started on mechanics or we "just won't get the course finished" or something, according to our teacher lol

 

[SeDaTeD]

Mechanics (well for motions of falling particles at least) in 4unit is just solving differential equations without going through the theory for differential equations. ie. m(d2x/dt2) +k(dx/dt) = mg (m,g,k constants).

With the conservation of energy arguement, although valid, you'd be introducing theory from physics not covered in the 4u course. The relations between force, work and energy etc are laws of physics, beyond what 4u expects you to know. So I don't think you should be using energy arguements in 4u.

However using things from uni maths such as l'hopital's rule, as long as u explain it (or prove? someone clarify), should be fine.

 

[FinalFantasy]

so do just say " l'hopitals rule, the limit x-->0 f(x)\g(x) =limit x-->0 f'(x)\g'(x)"
den start using it?