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20 Answers
49for cutting spring
mg.Lcos53=Iα
or, a=gcos53=4g/5
that matches with what philip said...now figuring out the other one..:)
49kk got my mistake....here the com is also accelerating...[3]
49ok got it bt....using rotational dynamics T doesnt com into picture does it???
49i have a dbt y r the ones named easy ones the hardest of all??
49nishant bhaiya i feel that u will come soon...so can u please give a small hint to these types of probs???
49do u really kno solution??philip??
and philip if u really kno the soln...no offence...was jst joking...[1][3][4][5][6]
49for cutting the string...
this thing hurts me...:'(
the answer must be 5g/8 indeed but why...ab pata nahi kyun bt lag raha hai philip ko bhi nahi pata y it is 5g/8....just like me he back claculated from CHE's answer ... :P :D
49had the q be find tension in the two cases ans would be different...but this is simple...:)but wait...lets see wat i end up to..:(
49but wait a min this could be done by rotational method also isnt it??
but nt having pen and paper right now...:(
49lemme analyze...
kxsin 53+Tsin 53=mg
also kxcos 53=Tcos 53..
so, kx=T
again mg=2Tsin53=2kxsin53
49but will there be shm taken into consideration as the accn a1 is at the instant when we cut the string...
but it is bound to be differrent from a2..
isnt it??[17]
49a2 is easy but what will be a1??
how to find it?
there will be shm also isnt it??[2][7][11]
1Soumik it can not be 1 actually :P
a2 is gsin53° = 4g/5
while a1 is √(g/2)2+(3g/8)2 = 5g/8
11Point out my mistake....
U see that since the block is in equlibrium, the horizontal omponenets of the spring and the string forces are same. Now as angles are same kx=T.
So at this position, we can safely replace the spring by another string with tension T - i.e. with the same tension as the string present.
Thus immediately after cutting, no matter which one u cut, both the situations will be equivalent, thus ratio should be 1.
