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physics

mechanics

Easy

Question

$text Consider two objects with end text m subscript 1 greater than m subscript 2 text connected by a light string that passes over a pulley having a moment of inertia end text o f space I space a b o u t space i t s space a x i s space o f space r o t a t i o n space a s space s h o w n space i n space f i g u r e. space T h e space s t r i n g space d o e s space n o t space s l i p space o n space t h e space p u l l e y space o r space s t r e t c h. space T h e space p u l l e y space t u r n s space w i t h o u t space f r i c t i o n space. space T h e space t w o space o b j e c t s space a r e space r e l e a s e d space f r o m space r e s t space s e p a r a t e d space b y space a space v e r t i c a l space d i s tan c e space 2 h. space T h e space t r a n s l a t i o n a l space s p e e d s space o f space t h e space o b j e c t s space a s space t h e y space p a s s space e a c h space o t h e r space i s space$

$\sqrt{\frac{2 (m_{1} + m_{2}) g h}{m_{1} + m_{2} + \frac{I}{R^{2}}}}$
$\sqrt{\frac{2 (m_{1} - m_{2}) g h}{m_{1} + m_{2} + \frac{I}{R^{2}}}}$
$\sqrt{\frac{(m_{1} - m_{2}) g h}{m_{1} + m_{2} + \frac{I}{R^{2}}}}$
$\sqrt{\frac{(m_{1} + m_{2}) g h}{m_{1} + m_{2} + \frac{I}{R^{2}}}}$

The correct answer is: $\sqrt{\frac{2 (m_{1} - m_{2}) g h}{m_{1} + m_{2} + \frac{I}{R^{2}}}}$

$text Take the two objects, pulley and earth as the system. If we neglect friction in the system, then end text text mechanical enemy is conserved and we can state that the increase in kinetic energy of the system end text text equals the decrease in potential energy since end text straight K subscript i equals 0 text (the system is initially at rest), we have end text straight capital delta K equals K subscript f minus K subscript i equals 1 half m subscript 1 v squared plus 1 half m subscript 2 v squared plus 1 half I omega squared text Where end text m subscript 1 text and end text m subscript 2 text have common speed. But end text v equals R text coso that end text straight capital delta K equals 1 half open parentheses m subscript 1 plus m subscript 2 plus I over R squared close parentheses V squared straight capital delta K plus straight capital delta U equals 0 comma text gives end text 1 half open parentheses m subscript 1 plus m subscript 2 plus I over R squared close parentheses v squared plus m subscript 2 g h minus m subscript 1 g h equals 0 v equals square root of fraction numerator 2 open parentheses m subscript 1 minus m subscript 2 close parentheses g h over denominator m subscript 1 plus m subscript 2 plus I over R squared end fraction end root$

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