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`

Answer

(b) `(2gsqrt(3))/(5)ms^(-2)`

Explanation

When ring slides form A and A' in time interval `Delta t`, the block descends from C to C' as shown in fig. A' D is perpendicular by dropped from A' on AB. For small displacement , `A'B=DB`
Now as length of string is constant
`AB+BC=A'B+BC'`
or `(AD+DB)+BC=A'B+BC'`
` AD=BC'-BC ( because A'B=DB)`
or `A A' cos theta=C C'`
`implies (A A')/(Delta t) cos theta=(C C')/(Delta t)`
i.e. velocity of ring `xx cos theta=` velocity of block
i.e., velocity of block C, `v_(c)=v cos theta` ...(i)
(b) If a is the initial acceleration of ring, then acceleration of block `= a cos theta`.
Let T be the tension in the string at this instant. Then the free-body diagram of block is shown in fig.
(c ) Equation of motion of block M is
`Mg-T-Ma cos theta` ..(ii)
The free-body diagram of ring is shown in fig. Resolving T along horizontal and vertical, we have
`T cos theta=ma implies T=(ma)/(cos theta)`...(iii)
Equation (ii) and (iii)give
`a=(Mg cos theta)/(m+M cos^(2)theta)`
`=(2mg cos 30^(@))/(m+2m cos^(2)30^(@)) = (2g((sqrt(3))/(2)))/(1+2xx((3)/(4)))=(2gsqrt(3))/(5)ms^(-2)`.