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Neglecting friction at the axle and the inertia of the two step pulley shown in figure find the acceleration 'a' of the falling weight P in (m//s^(2)) (assume P=2 kg Q=2 kg & r_(2)=2r_(1)) |
Answer»  `a_(1)=ralpha , a_(2)=2ralpha` ….(1) `T_(2) 2r-T_(1)r=Ialpha` …..(2) `:.(I=10 r^(2))` `T_(1)-2g=2a_(1)` ……(3) `2g-T_(2)=2a_(2)` …..(4) on solving above EQUATION `a_(2)=4 m//s^(2)` |
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