a person standing on rotating platform has his hands lowered . if he suddenly outstretch his arm then what change is expected in angular momentum and in rotational kinetic energy ?

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Solution

$W h e n a p e r s o n o u t s t r r e t c h e s h i s a r m s t h e n t h e m o m e n t o f i n r t i a o f h i s b o d y w i l l i n c r e a s e b u t h i s a n g u l a r s p e e d d e c r e a s e s b u t a s a w h o l e t h e n e t a n g u l a r m o m e n t u m r e m a i n s c o n s e r v e d . O n t h e o t h e r h a n d h i s r o t a t i o n a l k i n e t i c e n e r g y b e c o m e s, R . K . E . = \frac{1}{2} I ω^{2} = \frac{1}{2} I ω \times ω = L \times \frac{ω}{2} . W h i c h s h o w s t h a t r o t a t i o n a l k i n e t i c e n e r g y d e c r e a s e s .$ $W h e n a p e r s o n o u t s t r r e t c h e s h i s a r m s t h e n t h e m o m e n t o f i n r t i a o f h i s b o d y w i l l i n c r e a s e b u t h i s a n g u l a r s p e e d d e c r e a s e s b u t a s a w h o l e t h e n e t a n g u l a r m o m e n t u m r e m a i n s c o n s e r v e d . O n t h e o t h e r h a n d h i s r o t a t i o n a l k i n e t i c e n e r g y b e c o m e s, R . K . E . = \frac{1}{2} I ω^{2} = \frac{1}{2} I ω \times ω = L \times \frac{ω}{2} . W h i c h s h o w s t h a t r o t a t i o n a l k i n e t i c e n e r g y d e c r e a s e s .$

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