resisted motion q's (1 Viewer)

[Grey Council]

here are a few questions that I'm stuck on:
A partical of mass m is projected vertically upwards from a point high above the ground. The particle experiences a resistance of magnitude mkv². During its downward motion, the terminal velocity of the particle is V. Its initial velocity of projection is half this terminal velocity.

during upward motion:
V²a = -g(V^2 + v^2)
x = (V^2/2g)ln((5V^2)/(4(V^2 + v^2))
during downward motion:
V²a = g(V^2 + v^2)

Find the position of the particle relative to its projection point when it attains 60% of its terminal velocity.
I got an answer, but it looks heaps dodgy to me. I got:
(25 - 9V²)/ 20
hrm

second question (this one is a killer )
A particle of unit mass is projected vertically upwards against a constant gravitational force g and a resistance v/c, where v is the velocity of the particle and c is a constant. s is the distance travelled in time t; at t = 0, s=0, and v = c(h-g) where h is a constant. Write down the equation of motion of the particle. (ooo, btw, what exactly does it mean when it says equation of particle? Just what acceleration is?)
Find the time taken by the particle to reach its highest point, and find the height of that point.
The particle falls to its original position under gravity and under the same law of resistance. Will the time of descent be reater or les than the time of ascent? Give reasons for your answer.

and third question:
A particle of mass m kilograms is dropped from rest in a medium blah blah blah. resistance = 1/10 . mv^2
particle hits ground at ln(1 + rt2) seconds after dropping.
g = 10
Express v as a function of t. Hence find the speed with which the particle hits the ground. Give your answer in simplest exact form.
Find, in simplest exact form, the distance fallen by the particle before it hits the ground.

lol, out of approx 20 trial paper questions, these were the only resisted motion questions I couldn't do, so I guess i'm getting the hang of this. Blah to harder 3u motion though.

 

[Grey Council]

hello?




I need help...

-.-

 

[Xayma]

Originally posted by Grey Council
Write down the equation of motion of the particle. (ooo, btw, what exactly does it mean when it says equation of particle? Just what acceleration is?).

No I think equation of motion would be a velocity equation.

Sorry I cant help further but my Engineering Mechanics textbook only deals with kinetic friction against a plane

 

[Grey Council]

hrm, I asked friend, and he said it IS the acceleration equation.

hrm hrm

thanks for the thought though. You really hsould have done 4u.

 

[George W. Bush]

yes, it is

i've done all those questions before, but since I just had my 4u test I don't really feel like doing them again right now, sorry. if no-one else has done it ill get back to you in a couple of days

 

[Grey Council]

you know, I get the feeling you go to Ruse

they just had their exams

....



and lazy bugger! I need solutions soon! lol, don't you find this fun? Maths is FUN! relax, by doing maths

*turns back to business studies assignment*

 

[George W. Bush]

Originally posted by Grey Council
you know, I get the feeling you go to Ruse

they just had their exams

I shall not comment.


*shifty eyes*

and lazy bugger! I need solutions soon! lol, don't you find this fun? Maths is FUN! relax, by doing maths

*turns back to business studies assignment*

Fine fine, solutions will be posted today
*turns back to economics assignment*

 

[Grey Council]

lol, fair enough



I know a few people at Ruse. Hrm, maybe I know you. Although you seem to be obsessed by Bush.

*defeated look*
and I do economics too. WTF, thats the worst subject ever. so goddamn boring. humph, your giving preference to your economics assy over my maths questions.
:-\

 

[George W. Bush]

Originally posted by Grey Council
here are a few questions that I'm stuck on:
A partical of mass m is projected vertically upwards from a point high above the ground. The particle experiences a resistance of magnitude mkv². During its downward motion, the terminal velocity of the particle is V. Its initial velocity of projection is half this terminal velocity.

during upward motion:
V²a = -g(V^2 + v^2)
x = (V^2/2g)ln((5V^2)/(4(V^2 + v^2))
during downward motion:
V²a = g(V^2 + v^2)

Find the position of the particle relative to its projection point when it attains 60% of its terminal velocity.
I got an answer, but it looks heaps dodgy to me. I got:
(25 - 9V²)/ 20
hrm[/quote

Upwards
x'' = -g-kv^2
dx = - (vdv/g+kv^2)
Downwards
x'' = g - kv^2
at terminal velocity (V), x''=0
i.e. V = sqrt(g/k)
From upwards
Letting h be maximum height,
h= 1/2k [ln |(g+ 1/4 * kV^2)/g|
but V = sqrt g/k
therefore h=1/2k * ln 5/4

Now, for downwards
dx = -1/2k * vdv/g-kv^2
Letting d be the distance from maximum height where v = 3V/5
d= 1/2k ln |g/(g- 9g/25) (using V = sqrtg/k again)
d= 1/2k * ln 25/16

Clearly displacement from starting position will be d - h
1/2k (ln 25/16 - ln 5/4) = 1/2k (2ln5/4 - ln 5/4) = 1/2k ln 5/4

I have a feeling I may have made a mistake somewhere, but i can't find it. hrm.

 

[George W. Bush]

Originally posted by Grey Council
second question (this one is a killer )
A particle of unit mass is projected vertically upwards against a constant gravitational force g and a resistance v/c, where v is the velocity of the particle and c is a constant. s is the distance travelled in time t; at t = 0, s=0, and v = c(h-g) where h is a constant. Write down the equation of motion of the particle. (ooo, btw, what exactly does it mean when it says equation of particle? Just what acceleration is?)
Find the time taken by the particle to reach its highest point, and find the height of that point.
The particle falls to its original position under gravity and under the same law of resistance. Will the time of descent be reater or les than the time of ascent? Give reasons for your answer.

x'' = g - v/c
dv/dt = (cg-v)/c
Letting t be time taken, t = c[ln (cg - c(h-g))/cg]
t= c ln |(2g-h)/g|

dx = -c (-v/cg-v dv)
= -c (1 - cg/cg-v)dv
i.e. h = c(c(h-g) +cg lnc(2g-h) - cg lncg)
= c^2(h-g) + c^2 g ln ((2g-h)/h)

Downwards will take longer as resistance will be in upward direction, against gravity.

and third question:
A particle of mass m kilograms is dropped from rest in a medium blah blah blah. resistance = 1/10 . mv^2
particle hits ground at ln(1 + rt2) seconds after dropping.
g = 10
Express v as a function of t. Hence find the speed with which the particle hits the ground. Give your answer in simplest exact form.
Find, in simplest exact form, the distance fallen by the particle before it hits the ground.

lol, out of approx 20 trial paper questions, these were the only resisted motion questions I couldn't do, so I guess i'm getting the hang of this. Blah to harder 3u motion though. [/B][/QUOTE]

 

[George W. Bush]

q3

x'' = g- v^2/10 = (100 - v^2)/10
dt = 10dv/(100-v^2)
t= 1/2 ln |(10+v)/(10-v)| + c, c=0
e^2t = (10+v)/(10-v)
v= 10((e^2t - 1)/(e^2t + 1)
subbing in t=1+sqrt2
v= 5sqrt2
vdv/dx = (100-v^2)/10
dx = -5 (-2v/(100-v^2)dv
x = -5 (ln |100-5sqrt2| - ln100)
= -5 ln | 1- sqrt2/20|
= 5 ln |(20 + 10sqrt2)/9|

 

[Grey Council]

Once again, I'm impressed.


Damn Ruse'ers


hanyway, I got the third question out yesterday, on the train, so I'm right on that one.
First question is dodgy. It doesn't say what form to put your answer in, so
Anyway, how the hell did you get:

Letting h be maximum height,
h= 1/2k [ln |(g+ 1/4 * kV^2)/g|
but V = sqrt g/k
therefore h=1/2k * ln 5/4

?!

I'm quite sure you made a mistake there somewhere. Hrm. My initial answer was wrong, me thinks. My revised answer is:
1/2k.(ln ((25V - 9v^2)/(20V + 20v^2))
I asked teacher, he said that since I hadn't been asked for a specific form in which to put the answer, I shouldn't bother with it anymore.
hrm

actually, wtf, looking at my solution again, I used the x value. Not the maximum height. But at the maximum height, v = 0. So then it IS 1/2k. ln 5/4
lol, so even my equation is right. Just let v = 0, and i get the same answer as you.



*sigh*
stupid mistakes
*kills himself*

blahthird question:
lmao
Downwards will take longer as resistance will be in upward direction, against gravity
well put!
I didn't quite approach the question that way. lol
I was thinking, upwards motion, resistance and gravity both add up, BUT speed of projection > 0. However, downwards, speed of projection = 0, yet resistance is working against gravity.

Hrm, would your reasoning be accepted in an exam?

 

[George W. Bush]

well, it's the one my teacher keeps hammering in, so I'd suppose so

I guess you could expand it a bit further saying the magnitude of the acceleration and thus the rate of change of velocity will be smaller on the downhill blah blah but I don't think it should be necessary
hrm I seem to remember an example in Cambridge saying pretty much the same thing, but I can't find it at the moment.

 

[Grey Council]

ah, fair enough.
Our teacher hasn't even mentioned this.

:-S

and I'll look at cambridge book tommorow

 

[Grey Council]

huh, hang on. What if the initial velocity of projection is very very small?

 

[George W. Bush]

Doesn't matter.

If we assume that air resistance is neglibile, then the times will be the same. Air resistance provides a force downward when it's moving up, meaning it will stop earlier than otherwise. It provides a force upward when moving down, meaning it will be accelerated slower.

Ergo, time of descent is longer.

 

[Grey Council]

"If we assume that air resistance is neglibile, then the times will be the same"
not true.

acceleration of gravity and initial projection not the same.
hrm

if initial velocity is LESS than terminal velocity, and particle reaches/doesn't reach terminal velocity, there'll be a difference, no?

damn, now I'm totally confused.
>.<
sorry Bush

 

[Constip8edSkunk]

without resistance, the only acceleration on a projected particle is gravity, the initial projection isnt an acceleration but a velocity. w/o resistance, particle reaches max height at pertiod T/2 and reaches initial level with u

the air resistance mucks things up because its direction switches when the particle comes down. This means the magnitude of acceleration as the particle is moving up will be greater than the magnitude of the acceleration of the particle moving down, (think of the motion in 2 parts might help...initial velocity just determines how high the particle will drop from) so it will gain speed at a slower rate (unlike the symmetrical example w/o resistance), hence longer time.

terminal velocity doesnt really matter.

 

[George W. Bush]

As above.

 

[Grey Council]

*feels very dumb atm*

thanks mates

now i hafta learn integration