Understanding Motion in a Pulley System

[psykatic]

Homework Statement

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 the given figure. At a instant at 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 start moving if the system is released from rest with \theta=~30^\circ?

Homework Equations

Newtons Equations, free body diagram

The Attempt at a Solution

Well, this question happens to be from a textbook. And its a solved one too.. The solution which they have given is quite complicated! I thought I'd get a brief explanation of "why and how" over it, and yes an alternative method would be highly appreciated!

The solution, is like this (as given in the book),

Suppose in a small time interval \delta 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 displacements A'B~ PB (I didnt get this!), since the length is constant (?), we have,
AB+BC= A'B+BC'
AP+PB+BC=A'B+BC'
AP=BC'-BC=CC' (as A'B=PB)

AA'cos\theta= CC'

or \frac{AA'cos\theta}{\delta t}=~\frac{CC'}{\delta t}

Therefor, (velocity of the ring)cos\theta= (velocity of the block)Please help

 

[rl.bhat]

When the ring moves with a velocity v horizontally, its component along the string is v*cos(theta). Since the length of string between the ring and block is costant, the velocity of the block is equal to v*cos(theta).

 

[psykatic]

Well, did you have a look at the diagram? It hasnt been approved yet..

 

[rl.bhat]

Suppose in a small time interval LaTeX Code: \\delta 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 displacements A'B~ PB (I didnt get this!), since the length is constant (?), we have,
When A and A(dash) are very close, AB and A(dash)B are nearly equal. If you take BA(dash) equal to BP, AP becomes A*A(dash)cos(theta) and A*A(dash)/t = velocity. The length of the string is ABC = A(dash)BC(dash)

 

[psykatic]

okay, that's fine with me, I've several more problems over the free body diagrams! I'll post it in mean time! Thank You!