6well u hav to assume it to be dimensionless otherwise the dimensional formula can itself go wrong!!!!
COPYING DOWN THE WHOLE QUESTION:
plz do this question just like you have just entered 11th and don't know anything about acceleration of a particle in circular motion
Q- Suppose we are told that the acceleration 'a' of a partcle moving with uniform speed 'v' in a circle of radius 'r' is proportional to some power of 'r' say rn and some power of 'v' say vm.How can we determine value of n and m ?
here's the doubt:
in the solution they have assumed that
a = k vmrn (ok upto this step)
(where k is a constant)
but after that the statement is:
"As k is a dimensionless constant so
[L/T2] = [Ln(L/T)m]
and by this way they have calculated the values of n and m but
how can they assume k as dimensionless until it is mentioned in the question??
(the question i have typed is exactly the same as given in book)
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10 Answers
49look consider the different cases in which accn might depend..
they will surely be r and v as given by the question...
now is there any other logical parameters possible?
well umm...it can be mass...and that we can assume though wrongly that the thing also depends on mass!!
so let the accn depend on m , r and v
suppose a is proportional to mx , rn and vm
thus a is proportional to (mxrnvm)
or, a=k.(mxrnvm) where k is a dimensionless constant...
now apply dimensional analysis
u wud get x=0 m=2 n=-1
thus a=k.v2/r
11subho bhayiya you didn't understand my doubt.......there's no problem in solving the question whatever no of dependent quantities are given(whether mass or g or any other such thing) ......the doubt is - how can you assume 'k' to be dimensionless until it is said that in the qustion
suppose a is proportional to mx , rn and vm
thus a is proportional to (mxrnvm)
or, a=k.(mxrnvm) where k is a dimensionless constant...
now apply dimensional analysis
this is your solution but how can you assume 'k' to be dimensionless(we all know by studying circular motion that it is dimensionless but we can't apply that knowledge here)
611no convincing answers till now
@akhil
the dimensional formula won't go wrong but the actual formula will go wrong...just for example let us assume 'k' has dimensions of time
k = [T] then the formula will become(by solving using dimensional analysis):
a = k.v3r2 (whereas you know that actual formula is a = k.v2r )
dimensionally the formula is still correct but this is not the actual formula .
i remember one line regarding this given in many books:
A DIMENSIONALLY CORRECT EQUATION NEED NOT BE ACTUALLY CORRECT BUT A DIMENSIONALLY WRONG EQUATION MUST BE WRONG ...
so this is my only doubt that even when we know that all physical constants are not dimensionless(example is universal gravitational constant) ,still why do we assume these constants to be dimensionless while solving such problems
49nahi vaibhav...
k is dimensionless because in considering the "physical entities" in which the quantity depends we include ALL POSSIBLE dimensions in which the quantity depends so that we CAN actually consider a dimensionless k!!
actually,
k is just a numeric constant just to give right answers from equations and have to be experimentally found out!!
i hope i am clear![1]
11yes subhomoy i got a hint that you wanted to say this when you wrote the line
look consider the different cases in which accn might depend..in your last to last post in this thread and trust me, initially i was really convinced
but when I see the inverse square equations:
1. F= Gm1m2r2 (universal gravitation law)
2. F = kqr2
and other inverse square equations
i sense a big difference in what you say and what the reality is
just taking example of 1.(ie universal gravitation law)
in this equation also m1, m2 and 'r' are the only terms that 'F' depends upon but still the constant is not dimensionless
if you try formulate any of inverse square laws using dimensional analysis like in the topmost question you'll surely get all the formulas wrong
now there can be only two possibilities:
1. either all the inverse square laws are a big exception to dimensional analysis
or
2. there is a very good and convincing reason behind all this
49sorry !
here u are a bit wrong!
this was my class 9 doubt when i asked the teacher why
in F=kma we have dimensionless k and in
H=mcθ, c has a dimension!
:D
the answer lies here!
for inverse square relations like coulomb's law the value k is medium dependent! :D
and thus has to be considered during doing the dimensional analysis! ;)
that is true for coulomb's law and biot-savart law
as of newton's law of gravity i am trying to think that now! [1]
11subhomoy not just newton's gravity law but any law which contains universal constants (universal gas constant etc..etc ) would have to be explained for complete removal of this doubt