21
tapanmast Vora
·2009-03-20 23:22:07
iska PERFECT FINAL ANSWER KYA HAI?
106
Asish Mahapatra
·2009-03-19 23:23:00
nishant bhaiyya:
can u solve that question so that atleast i can get some hint how to do them... i cant get head or tail of this one...
21
tapanmast Vora
·2009-03-20 00:01:02
here : F = d(mv)/dt= m(dv/dt) + v(dm/dt)
m will be minst. rite, ie (M - μt + m) and dV/dt = a; dM/dt = μ
thn i guess we'll hav to integrate frm ini. vel V to final inst vel 0
1
Kumar Satyajeet
·2009-03-20 02:21:38
hi.....asish....
Q) Acart moves in horizontal dirn due to a cost. focre F in same dirn. In the process sand spills thru a hole in the bottom with a const. vel m kg/sec. Find accn and vel of the cart at any time instant t.(initial mass of cart+sand=M and vel = 0)
1
skygirl
·2009-03-20 20:38:09
waise i will give a simple problem.
A trolley has a small hole on one side, from which sand falls in the forward direction at a rate of μ kg/ hr at a velocity of v relative to the trolley!
total mass of the sand on trolley is M
mass of trolley is m
initial velocity is V
find the time after which it stops.
doubt it ever stop ?
for that we will get a term ln0 .. which is undefined..
1
skygirl
·2009-03-20 20:43:07
net force on the trolley = mdv/dt = -v μ/(3600)
=> V∫V'dv/v = -μ0∫tdt/(3600m)
=> ln(V/V') = μt/(3600m)
to find out t when V' =0 is impossible.
we can find out t for some particular value of V.
106
Asish Mahapatra
·2009-03-20 20:44:57
@kumar.. i was doing irodov yesterday and saw that prob... but i cudnt solve it.. anyone mind helping out??
1
skygirl
·2009-03-20 21:48:33
bhaiya pls anser post #23.
1
skygirl
·2009-03-20 23:08:22
#22.)
let at time t, mass of the trolley = m = [mo-μt]
and its velocity be v.
afetr time Δt, mass =m-Δm [Δm=μΔt]
v ---> v+Δv
by momentum conservation:
(m-Δm)(v+Δv) + Δm(v-u) = mv [u=velocity of the sand wrt trolley]
=> (m-Δm)Δv = Δm.u
=> Δv = Δm.u/(m-Δm) = μΔt.u/(m-μΔt)
=> Δv/Δt = μ.u/m-μΔt
[since Δm --->0 as delt-->0]
so, dv/dt = μu/m = μu/(mo-μt)
=> 0∫vdv = 0∫tμu/(mo-μt) dt.
=> v = u ln[mo/(mo-μt)]
now u =Ft/m. = F/(m/t) = F/μ = velocity of sand wrt trolley.
put this value in place of u..........
21
tapanmast Vora
·2009-03-21 00:06:13
HMMMMM........ wonderful proof!!! [50]
1
skygirl
·2009-03-21 00:12:02
arey yeh same hai .. just like rocket ...
21
tapanmast Vora
·2009-03-21 00:13:31
haan!!!
par mene socha ki Pink wink kuch nahi hai to shayad rong hoga isliye dekha hi nahi tha, abhi DEKHA
[78]
1
skygirl
·2009-03-21 00:20:17
[3]
woh... bhaiya sat-sunday very busy...
isiliye i wont disturb him by doing jhagra for pink [3]
62
Lokesh Verma
·2009-03-21 00:25:58
it will never stop sky :)
24
eureka123
·2010-01-05 10:00:47
I believed this one was in syllabus...but today again 3 people told me this topic is not in syllabus.....
what do tiitians say ??
62
Lokesh Verma
·2009-03-16 00:18:07
ye typing error tha... solved got replaced by loved (:)
nothing like that in my mind :P
waise i will give a simple problem.
A trolley has a small hole on one side, from which sand falls in the forward direction at a rate of μ kg/ hr at a velocity of v relative to the trolley!
total mass of the sand on trolley is M
mass of trolley is m
initial velocity is V
find the time after which it stops.
24
eureka123
·2009-03-10 21:59:32
2)Impulse method
Δp=J
=>pf-pi=F.t
(here both initial and final momentum will contain variable mass term.....)
Write correct initial and final momentum taking rocket +fuel as system or any other thing like truck with sand......[1][1]
106
Asish Mahapatra
·2009-03-10 22:00:09
eureka can u explain using sum example...
suppose.. there is a heap of rope on a table...of mass M and length L .... what will be the valocity of te rope when the rope has slid down by a dist x??
24
eureka123
·2009-03-10 22:05:18
that can be done using energy method......it wont require all this.......
btw i have to go now,,,,wil reply later
11
Mani Pal Singh
·2009-03-10 22:05:37
@asish
THIS IS NOT IN SYLLABUS
SO I RECOMMEND NOT TO WASTE TIME ON IT
24
eureka123
·2009-03-11 01:07:53
It is very much in the syllabus.......[1][1]
106
Asish Mahapatra
·2009-03-14 23:49:23
yes eureka.. sry for such a late reply..kyonki net ka wire kho gayaa tha
can u solve the question i had given.. and how it wud have been different. if the chain was not heaped but was laid on the table on a single line.... wud it have been different?
62
Lokesh Verma
·2009-03-15 21:57:17
Variable mass is solved by momentum conservation in a small time...
If there is an external force you just have to consider that too..
In theory it is very simple. .but a lot of people and sometimes the books themselves will make you believe that it is very difficult!
The most basic thing to do is F=d(mv)/dt
106
Asish Mahapatra
·2009-03-15 21:59:26
Variable mass is loved by momentum conservation in a small time... [7]
abd bhaiyya can u plz div me some problems so that i can try...
24
eureka123
·2009-03-10 21:56:41
One is which involves log (common method) and other one does not involve log but will involve correct selection of system
1)Instantaneous mass=Force other than impulses + Thrust due to impulses
=>m.dv/dt=Fext+vrelativedm/dt
24
eureka123
·2009-03-16 02:38:34
sir....i also wrote the same..[1]
24
eureka123
·2009-03-16 02:39:11
@ asish should i do this question for u.. or do u want to try???????[7]