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A circular disc of moment of inertia I_(1) is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed omega_(1). Another disc of moment of inertia I_(2) is placed coaxially on the rotating disc. Initially the second disc has zero angular speed. Eventually both the discs rotate with a constant angular speed omega_(2). The energy lost by the initially rotating disc to friction is ............ |
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Answer» `(1)/(2)(I_(2)^(2))/((I_(1)+I_(2)))omega_(1)^(2)`  According to law of conservation of momentum `therefore I_(1)omega_(1)=(I_(1)+I_(2))omega_(2)` `therefore omega_(2)=(I_(1))/((I_(1)+I_(2)))omega_(1)....(1)` Angular velocity of system decreases when another disc is placed on the first disc. HENCE, `therefore omega_(2)ltomega_(1)` Hence decrease in rotational kinetic energy = (initial rotational kinetic energy) - (final rotational kinetic energy) `=(1)/(2)I_(1)omega_(1)^(2)-(1)/(2)(I_(1)+I_(2))omega_(2)^(2)` `=(1)/(2)I_(1)omega_(1)^(2)-(1)/(2)(I_(1)+I_(2))((I_(1))/(I_(1)+I_(2)).omega_(1))^(2)` putting value of equ. (1) `=(1)/(2)I_(1)omega_(1)^(2)-(1)/(2)(I_(1)+I_(2))(I_(1)^(2))/((I_(1)+I_(2))^(2)).omega_(1)^(2)` `=(1)/(2)omega_(1)^(2)(I_(1))/((I_(1)+I_(2)))(I_(1)+I_(2)-I_(1))` `=(1)/(2)omega_(1)^(2)(I_(1)I_(2))/((I_(1)+I_(2)))` |
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