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  • A ring of mass and radius r rolls down (without slipping) down an inclined plane starting from r

A ring of mass \mathrm{m} and radius r rolls down (without slipping) down an inclined plane starting from rest. If the centre of the ring is at a height \mathrm{h}, the angular velocity when the ring reaches the base is -
 

Option: 1

\mathrm{\sqrt{\frac{g(h-R)}{r^2}}}
 


Option: 2

\mathrm{\sqrt{\frac{g(h+R)}{r^2}}}

 


Option: 3

\mathrm{\sqrt{\frac{h+R}{r^2}}}


Option: 4

None.


Answers (1)

best_answer

\mathrm{m g(h-R) =\frac{1}{2} m v^2+\frac{1}{2} I \omega^2 }

\mathrm{v =\gamma r, \quad I=m r^2 }

\mathrm{m g(h-R) =\frac{1}{2} m r^2 \omega^2+\frac{1}{2} m r^2 \omega^2 }

\mathrm{m g(h-R) =m r^2 \omega^2 }

\mathrm{\omega =\frac{m g(h-R)}{m r^2} }

\mathrm{\omega =\sqrt{\frac{g(h-R)}{r^2}} }

Hence option 2 is correct.



 

Posted by

Pankaj Sanodiya

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