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Two indentical discs are moving with the same kinetic energy. One rolls and the other slides. The ratio of their speeds is |
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Answer» `1:2` `therefore` KINETIC energy for the rolling disc, or `"" KE_(r ) = (1)/(2)mv_(1)^(2) + (1)/(2) I omega^(2)` `""` ( `therefore`MOMENT of INERTIA, I = `(mR^(2))/(2)`) or `"" = (1)/(2) mv_(1)^(2) + (1)/(2) (mR^(2))/(2) ((v_(1))/(R ))^(2) ` or `"" KE_(r ) = (3)/(4) mv_(1)^(2) ""` .... (i) Now, KE for the sliding disc, `therefore "" KE_(s) = (1)/(2) mv_(2)^(2) "" ` ... (ii) Given, KE of rolling disc = KE of sliding disc or,`(3)/(4) mv_(1)^(2)= (1)/(2) mv_(2)^(2) or (v_(1)^(2))/(v_(2)^(2)) = (2)/(3) ` or `"" (v_(1))/(v_(2)) = sqrt((2)/(3))` or `""v_(1) : v_(2) = sqrt(2) : sqrt(3)` |
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