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  • A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length l. The other end is fixed. The system is given an angular speed 

A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length l. The other end is fixed. The system is given an angular speed \omega about the fixed end of the spring such that it rotates in a circle in gravity-free space. Then the stretch in the spring is :     
Option: 1 \frac{ml\omega ^2}{k-\omega m}
 
Option: 2 \frac{ml\omega ^2}{k-m\omega ^2}

Option: 3 \frac{ml\omega ^2}{k+m\omega ^2}  

Option: 4 \frac{ml\omega ^2}{k+m\omega }
 

Answers (1)

best_answer

 

 

 

 

As natural lentgh=l

Let elongation=x

Mass m is moving with angular velocity \omega in a radius r

where r=l+x

Due to elongation x spring force is given by F_s=Kx

And F_C=m\omega ^2r=m\omega ^2(l+x)

as F_C=F_s

So

 Kx =m\omega ^2(l+x)\\ \Rightarrow x=\frac{m\omega ^2l}{K-m\omega ^2}

So the correct option is 2.

 

Posted by

vishal kumar

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