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

`0m//s`

Explanation

Let x is the vertical distance covered by the ring. Then x=L tan `37^(@) =0.7xx3/4`
`Deltal=I sec 37^(@) -L=L(sec37^(@)-1) Rightarrow L/4=Deltal`
`Deltal`=distance moved by block M Now, from constraint relation
`v_(M) =v_(r)cos 37^(@) =4/5v_(r)` ....(1)
`v_(r)` = velocity of ring `v_(M)` = velocity of the block at this instant From mechanical energy conservation
`DeltaPE+DeltaKE=0`
`-mgx+Mg Deltal sin 37^(@) +1/2mv_(r)^(2)+1/2Mv_(M)^(2)=0` .....(2)
on solving eq (1) and (2) we get
`v_(r)=0m//s`