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Two disc of moments of inertia I_(1)andI_(2) about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds omega_(1)andomega_(2) are brought into contact face to face with their axes of rotation coincident. (a) What is the angular speed of the two-disc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take omega_(1)neomega_(2). |
Answer» Solution : Let `I_(1)andI_(2)` are the moment of inertia of the discs and their angular SPEED are `omega_(1)andomega_(2)`. When they brought into CONTACT the system of two discs having moment of inertia `I_(1)+I_(2)`. The angular speed of combined discs is `omega` Angular moment before collision `=I_(1)omega_(1)+I_(2)omega_(2)` Angular MOMENTUM after collision = `(I_(1)+I_(2))omega` From the law of conservation of angular momentum `I_(1)omega_(1)+I_(2)omega_(2)=(I_(1)+I_(2))omega` `therefore omega=(I_(1)omega+I_(2)omega_(2))/(I_(1)+I_(2))` (a) Rotational kinetic energy before collision `K_(i)=(1)/(2)I_(1)omega_(1)^(2)+(1)/(2)I_(2)omega_(2)^(2)....(1)` Rotational kinetic energy after collision `K_(f)=(1)/(2)(I_(1)+I_(2))omega^(2)` Putting the value of `omega`, `K_(f)=(1)/(2)(I_(1)+I_(2))((I_(1)omega_(1)+I_(2)omega_(2))/(I_(1)+I_(2)))^(2)` `=(1)/(2)((I_(1)omega_(1)+I_(2)omega_(2))^(2))/(I_(1)+I_(2))...(2)` Decrease in rotational kinetic energy = `K_(i)-K_(f)` `=(1)/(2)I_(1)omega_(1)^(2)+(1)/(2)I_(2)omega_(2)^(2)-(1)/(2)((I_(1)omega_(1)+I_(2)omega_(2))^(2))/(I_(1)+I_(2))` `=(1)/(2)[I_(1)omega_(1)^(2)+I_(2)omega_(2)^(2)-((I_(1)^(2)omega_(1)^(2)+I_(2)^(2)omega_(2)^(2)+2I_(1)I_(2)omega_(1)omega_(2)))/(I_(1)+I_(2))]` `=(1)/(2)([(I_(1)^(2)omega_(1)^(2)+I_(2)^(2)omega_(2)^(2)+I_(1)I_(2)omega_(1)^(2)+I_(1)I_(2)omega_(2)^(2)),(-I_(1)omega_(1)^(2)-I_(2)^(2)omega_(2)^(2)-2I_(1)I_(2)omega_(1)omega_(2))])/(I_(1)+I_(2))` `=(1)/(2)((I_(1)I_(2)(omega_(1)^(2)+omega_(2)^(2)-2omega_(1)omega_(2)))/(I_(1)+I_(2)))` `=(1)/(2)[(I_(1)I_(2)(omega_(1)-omega_(2))^(2))/(I_(1)+I_(2))]` `=(I_(1)I_(2)(omega_(1)-omega_(2))^(2))/(2(I_(1)+I_(2)))` It is positive `therefore K_(i)-K_(f)gt0` `therefore K_(i)gtK_(f)` Thus, when two discs are combined, the loss of rotational kinetic energy is due to the friction between the discs but the torque produced due to friction is initerial, hence the angular momentum remains UNCHANGED. |
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