IIT JEE , AIEEE Forum - Physics , Mechanics - A particle of mass m is in equilibrium.......

nitish0991

A particle of mass m is in equilibrium with the help of 2 ideal identical strings. Now one string is cut then the ratio of tension in the other string(the uncut string) just before cutting and just after cutting is ::
1)1:2 2) 2:3 3) 3:1 4) 1:3

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nitish0991

Here's the figure too..Plz help me out..

nitish0991

can any1 help plzzzzzzzzz

rooney

hold on. typing.

rooney

Initially the Tension is mg in both strings (this you must have found out). Now, suppose the right string is cut.

So, now the system behaves as a pendulum. Now since the motion of mass m will be in tangential direction , and the velocity of mass m just after cutting will be zero in that instant, equation along the string
T - mg Cos 60 = mv²/l
Since v= 0
T=mg cos 60 = mg/2

So ratio is 2:1

I think the options given might be wrong.

karthik2007

Yeah. I think 2:1 is the answer Even i got the same. Maybe you mistyped the first option?

nitish0991

yes the answer is 2:1 but the initial tension is mg/2..

rooney

No. It is mg.
2Tsin 30 = gm (along vertical direction)
T=mg

nitish0991

yes it is 2:1....Karthik please post your solution too..

nitish0991

oh yeah yeah rite..but i dint understand the concept after the string is cut..

rooney

After the string is cut, isnt it similar to a pendulum system when the bob is at a extreme position where vel = 0 ???
So, as the pendulum moves tangentially to the string, we equate the forces along the string (along which there is no motion) which is the direction perpendicular to the direction in which the mass m will move

nitish0991

Okay..great..i got it..thanks..

karthik2007

Well there cant be more than one solution. Mine is quite similar to rooney's but i'll try to elaborate.

See initially for equillibrium, the sum of the horizontal forces must be zero, and the sum of vertical forces must be zero.

Consider vertical components:

Tsin30 + Tsin30 = mg

or T = mg.

when the string is cut, the forces on the mass m are:

1) Tension acting inwards

2) component of gravity acting outwards, viz mgcos60.
As it is cut, it will move in a circular path
Now we know that centripetal force is the net attractive force that keeps a body moving in a circle.

So T-mgcos60 = mv^2/l

As rooney said, just after its cut, the body will be almost at rest, so v = 0

which gives T = mg/2 in this case.

So ratio is 2:1

kishan12

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