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

Abdul Qadeer , 14 Years ago
Grade 12
anser 1 Answers
Satya Prakash Yadav

Last Activity: 3 Years ago

Answer:  10/3 rad/s
 
 
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
 
 

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