A ring of mass `m=1kg` can slide over a smooth vertical rod. A light string attached to the ring passing over a smooth fixed pulley at a distance of `L=0.7m` from the rod as shown in figure.

At the other end of the string mass `M=5kg` is attached, lying over a smooth fixed inclined plane of inclination angle `37^@`. The ring is held in level with the pulley and released. Determine the velocity of ring when the string makes an angle `(alpha=37^@)` with the horizontal. `[sin 37^@=0.6]`

Answer

`0ms^-1`

Explanation

Let x is the vertical distance covered by the ring. Then
`x=Ltan37^@=0.7xx3/4`
`Deltal=Lsec37^@-L`
`=L(sec37^@-1)`
`impliesL/4=Deltal`

`Delta=` distance moved by block M
Now, from constraint relation
`v_M=v_rcos37^@=4/5v_r` (i)
`v_r=` velocity of ring,
`v_M=` velocity of the block at this instant
From work energy theorem, we get
`W_(gravity)=DeltaKE`
`-mgx+MgDeltalsin37^@+1/2mv_r^2+1/2Mv_M^2=0` (ii)
On solving Eqs. (i) and (ii), we get
`v_r=0ms^-1`