IIT-JEE problems

pauldmps

pauldmps

Banned
^^ It is called "reduced mass concept" in which one side of the spring can be fixed while the other side is attached to a single block of resultant mass of the two blocks (not equal to sum of masses) & the velocity of the blocks is replaced by relative velocity between the blocks. The answer comes out to be the same.
 
A

abhijangda

Padawan
Oh so many post in such a short time.
Ok solve this now

Q. In triangle ABC, the equation of the perpendicular bisector of AC is 3x-2y+8=0. If A=(1,-1) and B=(3,1), find the equation of BC and the coordinates of the circumcentre??
 
nims11

nims11

BIOS Terminator
pauldmps said:
^^ It is called "reduced mass concept" in which one side of the spring can be fixed while the other side is attached to a single block of resultant mass of the two blocks (not equal to sum of masses) & the velocity of the blocks is replaced by relative velocity between the blocks. The answer comes out to be the same.
Click to expand...
its not the reduced mass concept i am talking about. it is a different one.
in the "cutting the spring" method, the spring is divided in two parts across the center of mass. the blocks are given the velocities relative to the the center of mass.
if K is the spring constant of the original spring, then the K of the new spring is = K*(ratio of the original and new length of the spring).
the centre of mass behaves as a fixed wall and the two blocks can be observed independently. It might be unclear what i said. But i will be happy to explain if you guyz want to know it.
 
pauldmps

pauldmps

Banned
@abhijangda

Coordinate is my weak point. Is the equation for BC
4x -5y - 7 =0
? Sorry I couldn't find the circumcentre.

@nims11

Got it.
 
nims11

nims11

BIOS Terminator
circumcentre - -4/5,14/5
BC - 29y-16x+18
my answers are strange so they might be wrong...
 
A

abhijangda

Padawan
pauldmps said:
@abhijangda

Coordinate is my weak point. Is the equation for BC
4x -5y - 7 =0
? Sorry I couldn't find the circumcentre.

@nims11

Got it.
Click to expand...

calculus, co-ordinate and algebra are important units for maths. They should be practiced very much and specially calculus.
Also answer to previous question is
circumcentre
(-4/5,14/5)
and
equation of BC is
x + 4y = 7. This was not a difficult one i would say. Solution can be done in this way. Assume C(x1,y1) then find midpt of AC by midpt formula then this midpt should lie on perpendicular bisector of AC. Also slope of AC multiplied by its perpendicular bisector is -1. Solve these equations to get pt C. I guess finding circumcentre easy.
I will post next question shortly.

Here's another one,
Q. The extremities of a diagonal of a square are (1,1) and (-2,-1). Find the other vertices and the equation of the other diagonal.
Solve it.:)
 
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nims11

nims11

BIOS Terminator
OK... here's a physics question which i came across yesterday.

A string is wrapped around a cylinder of mass M and radius R. The string is pulled vertically upward to prevent the center of mass from falling as the cylinder unwinds the string.If the length of the string unwound when the cylinder has reached the angular speed 'w' is (Rw)^2/(Z*g). what is the value of Z?(answer in 0-9).
 
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nims11

nims11

BIOS Terminator
@paul right!!
@Jaskanwar
Let the force on the cylinder applied through the string be T.
so according to the question, T=mg
torque=Iα=TR
α=TR/I
given that final angular speed=w
angular distance=(w^2)/2α=(w^2)I/2TR
putting I=(MR^2)/2 and T=mg
angular distance=(w^2)R/4g
thus, distance=angular distance * r=(Rw)^2/(4g)
I think you took I=MR^2 as i didnt specify the cylinder as a "solid cylinder". But if not specified in any question, it is to be taken as a solid one.
 
OP
Jaskanwar Singh

Jaskanwar Singh

Aspiring Novelist
oh another mistake. :-(

tell me whats wrong with this now -

S=1/2gt^2 and v=gt. v=wr. s=(Rw)^2/(Z*g). t is same at that instant..
 
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nims11

nims11

BIOS Terminator
from what i can make out, you are applying kinematics on a string which is massless, which is against the rules....

Actually i am confused now after seeing your solution, lets see what paul has to say about it.
 
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pauldmps

pauldmps

Banned
Ok so this is how I did it:

Since the C.M of the cylinder is at rest, net force acting on the cylinder is zero.

For vertical equilibrium:

T (upwards due to string) = Mg (downwards)

However since T does not pass through the C.M of the cylinder, there is a net Torque due to T. The net torque due to mg is zero.

Torque = Iɑ = TR
=> ɑ = TR/I = MgR/I
=> ɑ = 2MgR/MR^2
=> ɑ = 2g/R

Using w = w(initial) + ɑt (t is time)
=> w = 0 + 2gt/R = 2gt/R
=> t = Rw/2g

Using θ = θ(initial) + 1/2ɑt^2
=> θ = 1/2 * 2g/R * (Rw/2g)^2
=> θ = R*(w)^2/(4*g)

reqd expr. = s = θR
=> (R*w)^2/4g

Hence Z = 4
 
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