A sphere of mass M and radius R is attached by a light rod of length l to a point P. The sphere rolls without slipping on a circular track as shown. It is released from the horizontal position. The angular momentum of the system about P when the rod becomes vertical is then?

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Solution

$C o n s i d e r i n g t h e a n g u l a r v e l o c i t y o f s p h e r e i s ω, w h e n t h e r o d b e c o m e s v e r t i c a l . v i s t h e l i n e a r v e l o c i t y o f t h e s p h e r e . F o r p u r e r o l l i n g, v = ω R, W h e r e R i s t h e r a d i u s o f c i r c u l a r p a t h = l . B y c o n s e r v a t i o n o f e n e r g y, M g l = \frac{1}{2} M v^{2} + \frac{1}{2} \times \frac{2}{5} M R^{2} \times \frac{v^{2}}{R^{2}} g l = (\frac{1}{2} + \frac{2}{5}) v^{2} = \frac{7}{10} v^{2} v = \sqrt{\frac{10}{7} g l} A n g u l a r m o m e n t u m = I ω + m r v H e r e, R a d i u s (r) = L e n g t h o f t h e r o d (l) S o a n g u l a r m o m e n t u m = \frac{2}{5} M R^{2} \frac{v}{R} + m l \sqrt{\frac{10 g l}{7}} = \frac{2}{5} M R \sqrt{\frac{10}{7} g l} + m l \sqrt{\frac{10 g l}{7}} = M \sqrt{\frac{10}{7} g l} (\frac{2}{5} R + l)$

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