# Please help me with this problem....A ring of mass m free to slide on a fixed smooth horizontal wire is attached to one end of a string of length 60cm. The other end of the string is attached to a particle of mass 2m. The string and the particle are initially held at the same horizontal level with the string taut. If the system is now released, find the angular velocity of the string when it makes an angle of 60 degrees with the vertical.

14 Points
3 years ago

step1: first find the relation between velocity of ring and velocity of block with help of centre of mass.

As the acceleration in horizontal direction is 0  therefor the velocity of centre of mass along horizontal direction will be zero.
step 2. Resolve the velocity component of block of mass 2m in horizontal dierection and vertical direction at an angle of 60°.
step3: using conservation of energy,
you will get

$\frac{1}{2}mu^{2}+\frac{1}{2}2mv^{2}-2mg(0.6)\cos 60=0$

and  on puting u=v
we will get v= 2 m/s

thus
$\omega =\frac{v}{r}$

$\omega =\frac{2}{0.6}$

thus  angular velocity is 10/3 rad/s