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Mechanics
5 Apr 2008 11:23:23 IST
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5 Apr 2008 14:05:09 IST
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the natural length is 0.4 and let the xtended length be 0.4+x
mg=kxcos@
cos@=mg/kx
which is also equa to 0.4/0.4+x
solving which we get x as 0.1
mgh=1/2mv^2+1/2mv^2+kx^2
since velocity of both the blocks will be the same
solving it we wll get v as 1.5
the natural length is 0.4 and let the xtended length be 0.4+x
mg=kxcos@
cos@=mg/kx
which is also equa to 0.4/0.4+x
solving which we get x as 0.1
mgh=1/2mv^2+1/2mv^2+kx^2
since velocity of both the blocks will be the same
solving it we wll get v as 1.5
5 Apr 2008 17:51:14 IST
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Here is 59 solved by Biki.
let angle made = 
(as shown in figure)
(as shown in figure)let v = velocity just before falling
So at an instant just before falling---
mv2/r = mg.cos _____(1).........where m = mass, r = radius of the sphere
_____(1).........where m = mass, r = radius of the sphereor v2/r = g.cos _____(1)
_____(1)again change in K.E. = (1/2)mv2 - 0 ......as initial velocity = 0
and loss in P.E. mg(r - r.cos)
)So (1/2)mv2 = mg(r - r.cos)
)or v2/2 = gr (1 - cos)
)or v2/r = 2g(1 - cos)
)or g.cos = 2g - 2g.cos
= 2g - 2g.cosor 3g.cos = 2g
= 2gor cos = 2/3
= 2/3or = cos-1(2/3)
= cos-1(2/3)5 Apr 2008 18:12:00 IST
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Here u find another of Biki. Q.56.
Consider the first figure ...
In this fig. we will find out what should be the velocity of the bob at the lowest position so that the string becomes slack ( i,e, T = 0 ) at the horizontal diameter..
Now the weight acts perpendicular to the tension at this position and so has no component along the string....
So taking v =velocity at that point and T = tension and l = length of string....
we have T = mv2/l..
Now let u = velocity of the bob at lowest position so that the bob acquires velocity = v at the horizontal diameter....
So ... 1/2.mu2 = 1/2mv2 + mgl
or v2 = u2 - 2gl
Using this in equ^n of T.... we have
T = m(u2 - 2gl)/l
If string becomes slack.... T = 0
i,e, m(u2 - 2gl)/l = 0
or u2 - 2gl = 0
or u = 
2gl
2glSo we see that if velocity at lowest position should be equal to 2gl such that the string becomes slack at the horizontal diameter...
2gl such that the string becomes slack at the horizontal diameter...And see HCV-1 page 128 , example 8 where you will find that the velocity at lowesrt position such that the bob complets a full circle is 5gl .
5gl .So we conclude that.....
if velocity is less than 2gl, then string becomes slack below the horizontal diameter.
2gl, then string becomes slack below the horizontal diameter.if velocity = 2gl, then string becomes slack at horizontal diameter..
2gl, then string becomes slack at horizontal diameter..if velocity is greater than 2gl but less than 5gl, then string becomes slack in the region between the horizontal diameter and the topmost position....
2gl but less than 5gl, then string becomes slack in the region between the horizontal diameter and the topmost position....Taking this into consideration.... the question says that in the question the string becomes slack between the horizontal diameter and the topmost position...as velocity = 3gl...
3gl...Now consider fig. 2
Let velocity at lowest position = u =3gl
3glvelocity at the point of slackening = v
So ...
taking the angles as shown in fig.
we write.....
T + mg.cos
= mv2/l
= mv2/lfor string to become slack ... T = 0
i,e, mg.cos = mv2/l
= mv2/lor gl.cos = v2 _________ (1)
= v2 _________ (1)And using energy conservation .....
1/2mu2 = 1/2mv2 + mg( l + l.cos )
)i,e, v2 = u2 - 2gl ( 1 + cos )
)or v2 = 3gl - 2gl - 2gl.cos
or v2 = gl - 2gl.cos ________(2)
________(2)Using (2) in (1)...
gl.cos = gl - 2gl.cos
= gl - 2gl.cosor 3gl.cos = gl
= glor cos = (1/3)
= (1/3)or cos (180 - ) = (1/3) .........[ as = 180 - ]
) = (1/3) .........[ as = 180 - ]or - cos = (1/3)
= (1/3)or cos = (-1/3)
= (-1/3)i,e, = cos-1(-1/3)
= cos-1(-1/3)Free Sign Up!
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see cos37=h/L=4/5 L=elongated length of spring
h/L=4/5
let x be the extension in spring
hence, L=h+x
5h=4h+4x
x=h/4
also loss in pe=gain in ke
1/2(kx^2)=1/2(mv^2)
v^2=kh^2/16m
v=h/4 root(k/m)