Question

A ring of mass m slides on a smooth vertical rod. A light string is attached to the ring and is passing over a smooth peg distant a from the rod, and at the other end of the string is a mass M(M>m). The ring is held on a level with the peg and released. Show that it first comes to rest after falling a distance $\frac{2mMa}{M^2-m^2}$

Answer 1

The system starts from rest. Suppose that after it is released the system comes to rest again (instantaneously) when the ring has fallen a distance y below its initial position and the block has been raised by distance h. The PE lost by the ring must equal the PE gained by the block, since neither mass has KE at this instant. Therefore

$mgy=Mgh.$

Now we have to find how y is related to h.

When the ring has fallen a distance y, the length of the string between the peg and the ring is L where

$L^2=a^2+y^2.$

Here $h=L-a$ so

$L^2=(h+a{)}^2$

$a^2+y^2=h^2+2ah+a^2$

$y^2=h^2+2ah.$

Let $k=\frac{m}{M}$

Then $h=ky$

So substituting in $mgy=Mgh$ we get

$(ky{)}^2=k^2.y^2$

$=h^2$

$=y^2-2ah$

$=y^2-2aky$

Thus,

$(1-k^2)y=2ak$

$y=\frac{2ak}{(1-k^2)}$

$=\frac{2amM}{(M^2-m^2)}$