A smooth ring A of mass m can slide on a fixed horizontal rod. A string tied to the ring passes over a fixed pulley B and carries a block C of mass `M(=2m)` as shown in figure. At an instant the string between the ring and the pulley makes an angle `theta` with the rod. a. Show that, if the ring slides with a speed v, the block descends with speed `v cos theta`, b. With what acceleration will the ring starts moving if the system is released from rest with `theta= 30^0`

Explanation

a. Suppose in a small time interva `/_\t` the ring is displaced from A to A' and the block from C to C'. Drop a perpendicular A'P from A' to AB. For small displacement `A'B~~PB`. Since the length of the string is costant, we have

`AB+BC=A'B+BC'`
or `AP+PB+BC=A'B+BC'`
`or, AP=BC'-BC=CC' (as A'B~~PB)`
or, `AA'costheta=CC`
or, `(AA'costheta)/(/_\t)=(CC')/(/_\t)`
or, (velocity of the ring ) `cos theta =` (velocity of the block).
b. If the intial acceleration of te ring is a that of the block will be `a cos theta.` Let T be the tensioin in the string at this instant.Consider the block as the system. The forces acting on the block are
i. Mg downward due to the earth and
T upwared due to the string
The equation of motion of the block is
`Mg-T=Macostheta` ............i
Now consider teh ring as the system . The forces onthe ring are
i. Mg downward due to gravity
b. N upward due to the rod,
T along the string due to teh string.
Taking components along the rod, the equation of motin of the ring is
`Tcosthera =ma`............ii
From i and ii
`Mg=ma/(costheta)=Ma cos theta `
`a=(M g cos theta)/(m+M cos^2 theta)`
putting `theta= 30^0, M=2m ad g=9.8 m/s^2, therefore
`a=6.78 m/s^2`