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A block is tied to one end of a light st...

A block is tied to one end of a light string of length 1 whose other end is fixed to a rigid support. The block is given a speed of `sqrt(7gl//2)` from the lowermost position. Find the height and speed at which the block leaves the circle. Also find the maximum height to which it rises finally.

Text Solution

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As velocity of the block `v_bsqrt(2gl)ltv_bltsqrt(5gl)`.
The block will leave the circle at some point P, where the radius OP makes an angle `theta` with the upward vertical.

From A to P,
`DeltaK+DeltaU=0` or `DeltaK=-DeltaU`
Loss in KE=Gain in GPE
`1/2mv_b^2-1/2mv^2=mg(l+lcostheta)` (i)
From the force diagram: `T+mgcos=(mv^2)/(l)`
As the block leaves the circle at P, `T=0`.
`mgcostheta=(mv^2)/(l)` (ii)
Putting the value of `v^2` from (ii) in (i) ,
`1/2mv_b^2-1/2m(glcostheta)=mgl(1+costheta)`
`implies1/2m(7/2gl)=(mgl)/(2)(2+3costheta)`
`impliescostheta=1/2impliestheta=60^@`
Hence, the block leaves the circle and string slack at a height `h=l+lcos60^@=1.5l` from the bottom. The velocity at that moment v is `v=sqrt(l_gcostheta)=sqrt(l_g/2)`
After the string becomes slack, the block moves as a projectile in parabolic path. Now, further height attained
`=(v^2sin^2theta)/(2g)=(lg)/(4g)(3/4)=(3/16)l`
Thus, total height attained from the bottom is
`3/2l+3/16l=27/16l`
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